Answer key · Unit 9

Acids and bases.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Acids, bases and ions in water

What makes a solution acidic or basic: hydronium and hydroxide ions, water that splits by itself, and the difference between a strong acid and a concentrated one.

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The 5 steps and their answers

Step 1 of 5
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Two ions
Practice · Choose
Question
pH stands at 3. What is true about the ions in the beaker?

Correct answer
- Correct It holds both $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠, but far more $\mathrm{H_3O^+}$ ⁠
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The Scene at this step
pH
3
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Choices
Shown in a random order.
- Correct It holds both $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠, but far more $\mathrm{H_3O^+}$ ⁠
- Wrong It holds only $\mathrm{H_3O^+}$ ⁠, and no $\mathrm{OH^-}$ ⁠ at all
  Misconception: Thinks an acidic solution holds no OH⁻ at all, or a basic one no H₃O⁺.
  Feedback from the catalogue
  Every solution in water holds both hydronium ions, $\mathrm{H_3O^+}$ ⁠, and hydroxide ions, $\mathrm{OH^-}$ ⁠, because water itself is always splitting a little. What makes a solution acidic is which there is more of: in an acidic solution $[\mathrm{H_3O^+}] > [\mathrm{OH^-}]$ ⁠, and always $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠ at 25 °C.
- Wrong It holds equal amounts of the two ions, but more acid
  Feedback
  Equal amounts of $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠ are what defines a neutral solution, at pH $7$ ⁠.
Feedback for a right answer
At pH $3$ ⁠ there are $10^{-3}$ ⁠ moles of $\mathrm{H_3O^+}$ ⁠ and $10^{-11}$ ⁠ moles of $\mathrm{OH^-}$ ⁠ per litre: a hundred billion times as many, but not zero. An acid is a substance that gives $\mathrm{H_3O^+}$ ⁠ ions in water, and a base gives $\mathrm{OH^-}$ ⁠.

Hint
The second chart shows both ions, each on its own bar.

Worked solution
1. Water always holds both ions, because water splits: $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$ ⁠.
2. An acid adds $\mathrm{H_3O^+}$ ⁠, and the equilibrium shifts left: $\mathrm{OH^-}$ ⁠ thins out, but does not vanish.
3. $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠, so when one grows, the other shrinks.
Step 2 of 5
Open this step
Pure water
Practice · Calculate
Question
In pure water at 25 °C the pH is $7$ ⁠. What is the concentration of $\mathrm{H_3O^+}$ ⁠ ions?

Correct answer
1 × 10⁻⁷ mol/L ± 3%
Setup, choices and feedback
The Scene at this step
pH
7fixed
Labels
offdefault
Fade the water
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Accepted within
± 3%
Computed from the Scene at this step’s setting
'acid-base': H₃O⁺ concentration (10^−pH mol/L)/nodes/beaker/outputs/hydronium
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 mol/L ± 1 × 10⁻¹² mol/L
  Misconception: Thinks pure water contains no ions at all.
  Feedback from the catalogue
  Even in pure water, water molecules pass protons to one another: $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$ ⁠. That equilibrium lies far to the left, but not all the way: at 25 °C there are $10^{-7}$ ⁠ moles of each ion per litre. Because of that little, pure water conducts electricity, very weakly.
- 7 mol/L ± 0.01 mol/L
  Misconception: Gives a pH or pOH number where a concentration in mol/L is asked for.
  Feedback from the catalogue
  pH is not a concentration: it is the concentration’s exponent, with the sign flipped. $\mathrm{pH} = 3$ ⁠ means $[\mathrm{H_3O^+}] = 10^{-3}\ \mathrm{mol/L}$ ⁠. To go from pH to concentration, raise $10$ ⁠ to minus the pH.
Feedback for a right answer
$10^{-7}\ \mathrm{mol/L}$ ⁠: about one molecule in every five hundred million has passed on a proton. And there is exactly as much $\mathrm{OH^-}$ ⁠, so the water is neutral.

Hint
$[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}$ ⁠. You can type 1e-7.

Worked solution
1. $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$ ⁠, so $[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}$ ⁠.
2. $[\mathrm{H_3O^+}] = 10^{-7}\ \mathrm{mol/L}$ ⁠.
3. In pure water each $\mathrm{H_3O^+}$ ⁠ forms with an $\mathrm{OH^-}$ ⁠, so $[\mathrm{OH^-}] = 10^{-7}$ ⁠ too, and $10^{-7} \times 10^{-7} = 10^{-14}$ ⁠.
If you answered $0$ ⁠: even pure water holds ions. If you answered $7$ ⁠, you gave the pH, not the concentration.
Step 3 of 5
Open this step
Strong and weak
Practice · Choose
Question
Two beakers hold acid at 0.1 mol/L: one hydrochloric acid, $\mathrm{HCl}$ ⁠, the other ethanoic acid, $\mathrm{CH_3COOH}$ ⁠, the acid in vinegar. The first has pH $1.0$ ⁠, the second $2.9$ ⁠. Why?

Correct answer
- Correct $\mathrm{HCl}$ ⁠ splits almost entirely into ions in water, and ethanoic acid only a small part of it
Setup, choices and feedback
The Scene at this step
pH
2.9
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Shown in a random order.
- Correct $\mathrm{HCl}$ ⁠ splits almost entirely into ions in water, and ethanoic acid only a small part of it
- Wrong The $\mathrm{HCl}$ ⁠ beaker holds more acid in each litre
  Misconception: Confuses a strong acid with a concentrated one, and a weak acid with a dilute one.
  Feedback from the catalogue
  Strong and concentrated are different properties. A strong acid, like $\mathrm{HCl}$ ⁠, splits almost completely into ions in water; a weak acid, like ethanoic acid, splits only partly, because it reaches an equilibrium. Concentration is how much acid there is per litre. Very dilute hydrochloric acid is still strong, and concentrated vinegar is still weak.
- Wrong Ethanoic acid gives no $\mathrm{H_3O^+}$ ⁠ at all; its pH comes from the water
  Feedback
  Water alone gives pH $7$ ⁠. A pH of $2.9$ ⁠ says the ethanoic acid did give $\mathrm{H_3O^+}$ ⁠, about thirteen thousand times what water gives, only less than hydrochloric acid.
Feedback for a right answer
$\mathrm{HCl}$ ⁠ is a strong acid: every molecule gives a proton to water, $[\mathrm{H_3O^+}] = 0.1$ ⁠. Ethanoic acid is weak: it reaches an equilibrium, $\mathrm{CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+}$ ⁠, in which only about $1\%$ ⁠ of it splits.

Hint
Both have the same concentration. What differs between them?

Worked solution
1. pH $1.0$ ⁠: $[\mathrm{H_3O^+}] = 10^{-1} = 0.1$ ⁠, all the acid.
2. pH $2.9$ ⁠: $[\mathrm{H_3O^+}] = 10^{-2.9} = 0.0013$ ⁠, about $1.3\%$ ⁠ of the acid.
3. The concentrations are equal, so the difference is in strength: how far the acid splits into ions.
“Strong” and “concentrated” are two different questions: how much splits, and how much there is.
Step 4 of 5
Open this step
Bases
Practice · Choose
Question
pH now stands at 11. What is true about the solution?

Correct answer
- Correct It holds more $\mathrm{OH^-}$ ⁠ than $\mathrm{H_3O^+}$ ⁠, and the product of their concentrations is still $10^{-14}$ ⁠
Setup, choices and feedback
The Scene at this step
pH
11
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Choices
Shown in a random order.
- Correct It holds more $\mathrm{OH^-}$ ⁠ than $\mathrm{H_3O^+}$ ⁠, and the product of their concentrations is still $10^{-14}$ ⁠
- Wrong It holds no $\mathrm{H_3O^+}$ ⁠ at all
  Misconception: Thinks an acidic solution holds no OH⁻ at all, or a basic one no H₃O⁺.
  Feedback from the catalogue
  Every solution in water holds both hydronium ions, $\mathrm{H_3O^+}$ ⁠, and hydroxide ions, $\mathrm{OH^-}$ ⁠, because water itself is always splitting a little. What makes a solution acidic is which there is more of: in an acidic solution $[\mathrm{H_3O^+}] > [\mathrm{OH^-}]$ ⁠, and always $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠ at 25 °C.
- Wrong It holds no ions: a base is the opposite of an acid
  Misconception: Thinks pure water contains no ions at all.
  Feedback from the catalogue
  Even in pure water, water molecules pass protons to one another: $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$ ⁠. That equilibrium lies far to the left, but not all the way: at 25 °C there are $10^{-7}$ ⁠ moles of each ion per litre. Because of that little, pure water conducts electricity, very weakly.
Feedback for a right answer
At pH $11$ ⁠: $[\mathrm{H_3O^+}] = 10^{-11}$ ⁠ and $[\mathrm{OH^-}] = 10^{-3}$ ⁠. A base, such as sodium hydroxide, $\mathrm{NaOH \rightarrow Na^+ + OH^-}$ ⁠, or ammonia, $\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$ ⁠, adds $\mathrm{OH^-}$ ⁠, and water’s equilibrium shrinks $\mathrm{H_3O^+}$ ⁠ to match.

Hint
Look at the two ions’ bars in the second chart.

Worked solution
1. $[\mathrm{H_3O^+}] = 10^{-11}$ ⁠.
2. $[\mathrm{OH^-}] = \dfrac{10^{-14}}{10^{-11}} = 10^{-3}$ ⁠.
3. $10^{-3} > 10^{-11}$ ⁠: the solution is basic, and both ions are in it.
Step 5 of 5
Open this step
The ion product
Practice · Calculate
Question
At pH $3$ ⁠, $[\mathrm{H_3O^+}] = 10^{-3}\ \mathrm{mol/L}$ ⁠. What is $[\mathrm{OH^-}]$ ⁠?

Correct answer
1 × 10⁻¹¹ mol/L ± 3%
Setup, choices and feedback
The Scene at this step
pH
3fixed
Labels
offdefault
Fade the water
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Accepted within
± 3%
Computed from the Scene at this step’s setting
'acid-base': OH⁻ concentration (10^(pH − 14) mol/L)/nodes/beaker/outputs/hydroxide
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 11 mol/L ± 0.01 mol/L
  Misconception: Gives a pH or pOH number where a concentration in mol/L is asked for.
  Feedback from the catalogue
  pH is not a concentration: it is the concentration’s exponent, with the sign flipped. $\mathrm{pH} = 3$ ⁠ means $[\mathrm{H_3O^+}] = 10^{-3}\ \mathrm{mol/L}$ ⁠. To go from pH to concentration, raise $10$ ⁠ to minus the pH.
- 0 mol/L ± 1 × 10⁻¹⁵ mol/L
  Misconception: Thinks an acidic solution holds no OH⁻ at all, or a basic one no H₃O⁺.
  Feedback from the catalogue
  Every solution in water holds both hydronium ions, $\mathrm{H_3O^+}$ ⁠, and hydroxide ions, $\mathrm{OH^-}$ ⁠, because water itself is always splitting a little. What makes a solution acidic is which there is more of: in an acidic solution $[\mathrm{H_3O^+}] > [\mathrm{OH^-}]$ ⁠, and always $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠ at 25 °C.
Feedback for a right answer
$\dfrac{10^{-14}}{10^{-3}} = 10^{-11}\ \mathrm{mol/L}$ ⁠. Very little, but not zero: a litre beaker holds about $6 \times 10^{12}$ ⁠ hydroxide ions.

Hint
$[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠.

Worked solution
1. $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠.
2. $[\mathrm{OH^-}] = \dfrac{10^{-14}}{10^{-3}}$ ⁠.
3. $[\mathrm{OH^-}] = 10^{-14 - (-3)} = 10^{-11}\ \mathrm{mol/L}$ ⁠.
If you answered $11$ ⁠, you gave the pOH, the exponent with its sign flipped, not the concentration.

Lesson 2 of 4 · About 25 minutes

The pH scale

pH as a logarithm: minus the exponent of the hydronium concentration, one step that is ten times, working out pH from a concentration and back, dilution, and bases on the same scale.

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The 6 steps and their answers

Step 1 of 6
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Minus the logarithm
Practice · Reach the target
Question
Change pH so that the beaker holds $[\mathrm{H_3O^+}] = 10^{-4}\ \mathrm{mol/L}$ ⁠, then press Check.

Goal
- 0.0001 mol/L ± 5%
  At the start of the step: 1 × 10⁻⁷ mol/L
Setup, choices and feedback
The Scene at this step
pH
7
Labels
offdefault
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Off
Read from the Scene at
'acid-base': H₃O⁺ concentration (10^−pH mol/L)/nodes/beaker/outputs/hydronium
One setting that reaches it
pH
4
Labels
off
Feedback for a right answer
pH $4$ ⁠, about like tomato juice. The minus in the definition turns the exponent $-4$ ⁠ into the positive number $4$ ⁠.

Feedback for a wrong answer
Not yet. The pH is the concentration’s exponent, with the sign flipped.

Hint
$\mathrm{pH} = -\log[\mathrm{H_3O^+}]$ ⁠.

Worked solution
1. $\mathrm{pH} = -\log(10^{-4})$ ⁠.
2. $\log(10^{-4}) = -4$ ⁠.
3. $\mathrm{pH} = 4$ ⁠: set pH to 4.0.
Step 2 of 6
Open this step
Powers of ten
Practice · Predict, then look
Question
pH stands at 4. How many times more $\mathrm{H_3O^+}$ ⁠ ions will the beaker hold at pH 3?

Correct answer
- Correct Ten times
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pH
4fixed
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Choices
- Correct Ten times
- Wrong $\tfrac43$ ⁠ times, because $4 \div 3 = 1.33$ ⁠
  Misconception: Treats the pH scale as linear: pH 4 is twice as acidic as pH 8, one step of pH is a fixed amount.
  Feedback from the catalogue
  The pH scale is logarithmic: each step of one is ten times in $[\mathrm{H_3O^+}]$ ⁠. From pH $8$ ⁠ to pH $4$ ⁠ is four steps, that is $10^4 = 10\,000$ ⁠: a solution at pH $4$ ⁠ holds ten thousand times more hydronium ions, not twice as many.
- Wrong A little more: one step out of fourteen
  Misconception: Treats the pH scale as linear: pH 4 is twice as acidic as pH 8, one step of pH is a fixed amount.
  Feedback from the catalogue
  The pH scale is logarithmic: each step of one is ten times in $[\mathrm{H_3O^+}]$ ⁠. From pH $8$ ⁠ to pH $4$ ⁠ is four steps, that is $10^4 = 10\,000$ ⁠: a solution at pH $4$ ⁠ holds ten thousand times more hydronium ions, not twice as many.
Once the prediction is locked in, the Scene changes
pH
3fixed
Fade the water
Off
The Scene then shows: 0.001 mol/L · 'acid-base': H₃O⁺ concentration (10^−pH mol/L)/nodes/beaker/outputs/hydronium
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Ten times: from $10^{-4}$ ⁠ to $10^{-3}$ ⁠. In the second chart the $\mathrm{H_3O^+}$ ⁠ bar climbs one square, because each square is a power of ten.

Worked solution
1. At pH $4$ ⁠: $[\mathrm{H_3O^+}] = 10^{-4}$ ⁠.
2. At pH $3$ ⁠: $[\mathrm{H_3O^+}] = 10^{-3}$ ⁠.
3. The ratio: $\dfrac{10^{-3}}{10^{-4}} = 10$ ⁠.
The pH scale is a logarithmic scale, like the Richter scale for earthquakes: each step multiplies, it does not add.
Step 3 of 6
Open this step
How many times more acidic
Practice · Calculate
Question
How many times more $\mathrm{H_3O^+}$ ⁠ ions does a solution at pH $4$ ⁠ hold than one at pH $8$ ⁠?

Correct answer
10000 ± 0.1%
Setup, choices and feedback
The Scene at this step
pH
4
Labels
offdefault
Fade the water
Off
Accepted within
± 0.1%
A fixed value in the question
Wrong answers it recognises
- 2 ± 0.01
  Misconception: Treats the pH scale as linear: pH 4 is twice as acidic as pH 8, one step of pH is a fixed amount.
  Feedback from the catalogue
  The pH scale is logarithmic: each step of one is ten times in $[\mathrm{H_3O^+}]$ ⁠. From pH $8$ ⁠ to pH $4$ ⁠ is four steps, that is $10^4 = 10\,000$ ⁠: a solution at pH $4$ ⁠ holds ten thousand times more hydronium ions, not twice as many.
- 4 ± 0.01
  Misconception: Treats the pH scale as linear: pH 4 is twice as acidic as pH 8, one step of pH is a fixed amount.
  Feedback from the catalogue
  The pH scale is logarithmic: each step of one is ten times in $[\mathrm{H_3O^+}]$ ⁠. From pH $8$ ⁠ to pH $4$ ⁠ is four steps, that is $10^4 = 10\,000$ ⁠: a solution at pH $4$ ⁠ holds ten thousand times more hydronium ions, not twice as many.
Feedback for a right answer
$10^{8-4} = 10^4 = 10\,000$ ⁠. Not twice: a solution at pH $4$ ⁠ is ten thousand times more acidic, though the number is only half.

Hint
Four steps of pH, each one ten times.

Worked solution
1. pH $4$ ⁠: $[\mathrm{H_3O^+}] = 10^{-4}$ ⁠. pH $8$ ⁠: $[\mathrm{H_3O^+}] = 10^{-8}$ ⁠.
2. $\dfrac{10^{-4}}{10^{-8}} = 10^{-4-(-8)} = 10^4$ ⁠.
3. $10\,000$ ⁠ times.
If you answered $2$ ⁠, you divided the pH numbers. The pH is an exponent, so subtract the exponents and raise $10$ ⁠ to the difference.
Step 4 of 6
Open this step
From a concentration to pH
Practice · Calculate
Question
In lemon juice $[\mathrm{H_3O^+}] = 2.5 \times 10^{-3}\ \mathrm{mol/L}$ ⁠. What is its pH? (Two decimal places.)

Correct answer
2.6 ± 0.02
Setup, choices and feedback
The Scene at this step
pH
7
Labels
offdefault
Fade the water
Off
Accepted within
± 0.02
A fixed value in the question
Wrong answers it recognises
- -2.6 ± 0.02
  Misconception: Forgets the minus in pH = −log[H₃O⁺], giving a negative pH for an ordinary acid.
  Feedback from the catalogue
  $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$ ⁠. The concentrations are below one, so their logarithm is negative, and the minus makes it positive: $\log(2.5 \times 10^{-3}) = -2.60$ ⁠, so $\mathrm{pH} = 2.60$ ⁠.
- 3 ± 0.02
  Misconception: Gives a pH or pOH number where a concentration in mol/L is asked for.
  Feedback from the catalogue
  pH is not a concentration: it is the concentration’s exponent, with the sign flipped. $\mathrm{pH} = 3$ ⁠ means $[\mathrm{H_3O^+}] = 10^{-3}\ \mathrm{mol/L}$ ⁠. To go from pH to concentration, raise $10$ ⁠ to minus the pH.
Feedback for a right answer
$\mathrm{pH} = -\log(0.0025) = 2.60$ ⁠. Between $2$ ⁠ and $3$ ⁠, because the concentration is between $10^{-3}$ ⁠ and $10^{-2}$ ⁠; nearer $3$ ⁠, because on a log scale $2.5$ ⁠ is nearer $1$ ⁠ than $10$ ⁠.

Hint
$\mathrm{pH} = -\log(2.5 \times 10^{-3})$ ⁠. On a calculator: log(0.0025).

Worked solution
1. $\log(2.5 \times 10^{-3}) = \log 2.5 + \log 10^{-3} = 0.40 - 3 = -2.60$ ⁠.
2. $\mathrm{pH} = -(-2.60) = 2.60$ ⁠.
If you got $-2.60$ ⁠, you forgot the minus in the definition. If you answered $3$ ⁠, you took only the exponent and forgot the $2.5$ ⁠.
Step 5 of 6
Open this step
Dilution
Practice · Calculate
Question
To 10 mL of 0.1 mol/L hydrochloric acid, at pH $1$ ⁠, you add water up to 1000 mL. What is the pH now?

Correct answer
3 ± 0.02
Setup, choices and feedback
The Scene at this step
pH
1
Labels
offdefault
Fade the water
Off
Accepted within
± 0.02
A fixed value in the question
Wrong answers it recognises
- 1 ± 0.02
  Misconception: Thinks diluting an acid with water does not change its pH.
  Feedback from the catalogue
  Dilution lowers the concentration of every ion. Add water to a strong acid up to a hundred times its volume and $[\mathrm{H_3O^+}]$ ⁠ falls a hundred times, that is $10^2$ ⁠, so the pH rises by two. The acid does not get weaker; it gets more dilute.
- 100 ± 0.02
  Misconception: Treats the pH scale as linear: pH 4 is twice as acidic as pH 8, one step of pH is a fixed amount.
  Feedback from the catalogue
  The pH scale is logarithmic: each step of one is ten times in $[\mathrm{H_3O^+}]$ ⁠. From pH $8$ ⁠ to pH $4$ ⁠ is four steps, that is $10^4 = 10\,000$ ⁠: a solution at pH $4$ ⁠ holds ten thousand times more hydronium ions, not twice as many.
Feedback for a right answer
$[\mathrm{H_3O^+}]$ ⁠ fell a hundred times, from $10^{-1}$ ⁠ to $10^{-3}$ ⁠, and the pH rose from $1$ ⁠ to $3$ ⁠. To dilute all the way to pH $7$ ⁠ would take a millionfold dilution, and by then the acid no longer decides: the water itself does.

Hint
How many times does the concentration fall, and how many steps of pH is that?

Worked solution
1. The volume grows from $10$ ⁠ to $1000$ ⁠ mL, $100$ ⁠ times, so the concentration falls $100$ ⁠ times: $\dfrac{0.1}{100} = 10^{-3}\ \mathrm{mol/L}$ ⁠.
2. $\mathrm{pH} = -\log(10^{-3}) = 3$ ⁠.
$100 = 10^2$ ⁠ times in concentration is two steps of pH.
Step 6 of 6
Open this step
Bases on the same scale
Practice · Calculate
Question
What is the pH of a 0.01 mol/L solution of sodium hydroxide, $\mathrm{NaOH}$ ⁠?

Correct answer
12 ± 0.02
Setup, choices and feedback
The Scene at this step
pH
7
Labels
offdefault
Fade the water
Off
Accepted within
± 0.02
A fixed value in the question
Wrong answers it recognises
- 2 ± 0.02
  Misconception: Takes −log[OH⁻] as the pH of a basic solution, giving the pOH instead.
  Feedback from the catalogue
  Minus the logarithm of $[\mathrm{OH^-}]$ ⁠ is the pOH, not the pH. Because $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$ ⁠, $\mathrm{pH} + \mathrm{pOH} = 14$ ⁠. For 0.01 mol/L sodium hydroxide: $\mathrm{pOH} = 2$ ⁠, so $\mathrm{pH} = 12$ ⁠.
Feedback for a right answer
pH $12$ ⁠: $[\mathrm{OH^-}] = 10^{-2}$ ⁠, so $[\mathrm{H_3O^+}] = 10^{-12}$ ⁠. Move pH to 12 and check in the chart.

Hint
$\mathrm{NaOH}$ ⁠ is a strong base: $[\mathrm{OH^-}] = 0.01$ ⁠. From there to $[\mathrm{H_3O^+}]$ ⁠.

Worked solution
1. A strong base splits completely: $[\mathrm{OH^-}] = 0.01 = 10^{-2}$ ⁠.
2. $[\mathrm{H_3O^+}] = \dfrac{10^{-14}}{10^{-2}} = 10^{-12}$ ⁠.
3. $\mathrm{pH} = 12$ ⁠. (Or: $\mathrm{pOH} = 2$ ⁠ and $\mathrm{pH} = 14 - 2 = 12$ ⁠.)
If you answered $2$ ⁠, you gave the pOH.

Lesson 3 of 4 · About 30 minutes

Neutralisation and titration

An acid and a base that give a salt and water, a pH curve that leaps near the equivalence point, and titrating drop by drop to measure an unknown concentration.

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The 5 steps and their answers

Step 1 of 5
Open this step
Neutralisation
Practice · Choose
Question
The flask holds hydrochloric acid, and the burette sodium hydroxide. What forms when they react?

Correct answer
- Correct A salt and water: $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$ ⁠
Setup, choices and feedback
The Scene at this step
Volume added
0 mL
Curve so far / Whole curve
Curve so far
Choices
Shown in a random order.
- Correct A salt and water: $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$ ⁠
- Wrong Nothing: the acid and the base cancel out and disappear
  Misconception: Thinks neutralisation makes the acid and base disappear, leaving nothing behind.
  Feedback from the catalogue
  In neutralisation no atoms vanish: $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠ ions join into water, and the other ions, $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠, stay in the solution. Boil off the water and table salt is left. An acid and a base give a salt and water.
- Wrong Hydrogen, bubbling out of the flask
  Feedback
  Hydrogen forms when an acid reacts with a metal, like magnesium. With a base it gives water.
Feedback for a right answer
This is neutralisation: $\mathrm{H_3O^+ + OH^- \rightarrow 2H_2O}$ ⁠. The $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠ ions take no part and stay in the solution: salt water. The reaction gives out heat, about 57 kJ per mole of water.

Hint
What do $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠ do when they meet? And what happens to the $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠?

Worked solution
1. The acid gives $\mathrm{H_3O^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠; the base gives $\mathrm{Na^+}$ ⁠ and $\mathrm{OH^-}$ ⁠.
2. $\mathrm{H_3O^+}$ ⁠ and $\mathrm{OH^-}$ ⁠ join into water.
3. $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠ are left: a solution of the salt $\mathrm{NaCl}$ ⁠.
Step 2 of 5
Open this step
The curve
Practice · Predict, then look
Question
The flask holds 25 mL of hydrochloric acid, and the burette sodium hydroxide, both at 0.1 mol/L. How will the pH change as you add up to 30 mL?

Correct answer
- Correct Very slowly at first, then a big jump near 25 mL
Setup, choices and feedback
The Scene at this step
Volume added
0 mLfixed
Curve so far / Whole curve
Curve so farfixed
Choices
- Correct Very slowly at first, then a big jump near 25 mL
- Wrong At a steady rate, a few tenths for each millilitre
  Misconception: Expects the pH to rise steadily as base is added during a titration.
  Feedback from the catalogue
  While acid is in excess, the pH is set by what is left of it, and it rises slowly: neutralising half the acid moves it by less than half a unit. Only near the equivalence point, with the acid nearly gone, does one drop change the concentration thousands of times, and the pH leaps several units at once.
- Wrong Very fast at first, then slowly
  Feedback
  At first there is plenty of acid, and the first drops neutralise only a small share of it: the pH hardly moves. Only when the acid is nearly gone does each drop change a lot.
Once the prediction is locked in, the Scene changes
Volume added
30 mLfixed
Curve so far / Whole curve
Whole curvefixed
The Scene then shows: 11.96 · PH of the flask/nodes/rig/outputs/pH
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Up to 24 mL the pH rises only from $1.0$ ⁠ to $2.7$ ⁠. Then, between 24.9 and 25.1 mL, two drops, it leaps from $3.7$ ⁠ to $10.3$ ⁠, and the flask turns pink. After that it flattens again, around $12$ ⁠.

Worked solution
1. At 24 mL there is $2.5 - 2.4 = 0.1$ ⁠ millimole of acid left in 49 mL: $[\mathrm{H_3O^+}] = 0.002$ ⁠, $\mathrm{pH} = 2.7$ ⁠.
2. At 24.9 mL there is $0.01$ ⁠ millimole left: $\mathrm{pH} = 3.7$ ⁠.
3. At 25.0 mL the acid and base are equal: $\mathrm{pH} = 7$ ⁠.
4. At 25.1 mL there is $0.01$ ⁠ millimole of $\mathrm{OH^-}$ ⁠ in excess: $\mathrm{pOH} = 3.7$ ⁠, $\mathrm{pH} = 10.3$ ⁠.
Each drop near the equivalence point changes the concentration ten times or more, so the pH leaps.
Step 3 of 5
Open this step
The equivalence point
Practice · Calculate
Question
Now the flask holds 20 mL of 0.10 mol/L hydrochloric acid, and the burette 0.08 mol/L sodium hydroxide. What volume of base will exactly neutralise all the acid?

Correct answer
25 mL ± 0.1 mL
Setup, choices and feedback
The Scene at this step
Volume added
0 mL
Curve so far / Whole curve
Curve so farfixed
Other settings
Sample volume/nodes/rig/props/sampleVolume = 20
Other settings
Base concentration/nodes/rig/props/baseConcentration = 0.08
Accepted within
± 0.1 mL
Computed from the Scene at this step’s setting
Volume of titrant that exactly neutralises the sample/nodes/rig/outputs/equivalenceVolume
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 16 mL ± 0.1 mL
  Misconception: Turns the ratio of concentrations upside down when working out a titration volume or concentration.
  Feedback from the catalogue
  What becomes equal at the equivalence point is the number of moles: $c_a V_a = c_b V_b$ ⁠. A more dilute solution needs more volume, not less. Check the direction: if the base is more dilute than the acid, its volume must come out larger than the acid’s.
- 20 mL ± 0.1 mL
  Misconception: Thinks equal volumes of acid and base always neutralise each other.
  Feedback from the catalogue
  Equal volumes neutralise each other only when the concentrations are equal. What has to match is the number of moles, $n = c \times V$ ⁠: 20 mL of 0.10 mol/L acid holds $2.0$ ⁠ millimoles, and 0.08 mol/L base needs 25 mL for that.
Feedback for a right answer
25 mL: $\dfrac{0.10 \times 20}{0.08} = 25$ ⁠. The base is more dilute than the acid, so you need more of it. Choose Whole curve under Look inside to see the equivalence point on the chart.

Hint
At the equivalence point the moles of base equal the moles of acid: $c_a V_a = c_b V_b$ ⁠.

Worked solution
1. Moles of acid: $n = c \times V = 0.10 \times 0.020 = 0.0020$ ⁠ mol, that is $2.0$ ⁠ millimoles.
2. $\mathrm{HCl + NaOH}$ ⁠ react $1 : 1$ ⁠, so you need $2.0$ ⁠ millimoles of base.
3. $V_b = \dfrac{n}{c_b} = \dfrac{2.0\ \mathrm{mmol}}{0.08\ \mathrm{mol/L}} = 25\ \mathrm{mL}$ ⁠.
If you answered $16$ ⁠, you turned the ratio of concentrations upside down. If $20$ ⁠, you assumed equal volumes neutralise each other.
Step 4 of 5
Open this step
Running it in
Practice · Reach the target
Question
Again 25 mL of hydrochloric acid and sodium hydroxide, both at 0.1 mol/L. Change Volume added until you have added exactly enough base to neutralise all the acid, then press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Volume added
0 mL
Curve so far / Whole curve
Curve so farfixed
Read from the Scene at
0 before equivalence, 1 at it (within 0.05 mL), 2 past it/nodes/rig/outputs/regime
One setting that reaches it
Volume added
25 mL
Feedback for a right answer
At 25.0 mL: $2.5$ ⁠ millimoles of acid and $2.5$ ⁠ of base, and the pH is exactly $7$ ⁠. Notice the flask is still colourless: phenolphthalein turns pink only with the next drop, at 25.1 mL. That is the end point, a drop after the equivalence point.

Feedback for a wrong answer
Not yet. How many millimoles of acid are in the flask, and how many millilitres of base hold the same number?

Hint
Do not wait for the colour: work out the volume from the moles.

Worked solution
1. In the flask: $0.1 \times 25 = 2.5$ ⁠ millimoles of $\mathrm{HCl}$ ⁠.
2. In the burette: $0.1$ ⁠ millimole in each millilitre.
3. $\dfrac{2.5}{0.1} = 25.0$ ⁠: set Volume added to 25.0 mL.
Step 5 of 5
Open this step
An acid of unknown concentration
Practice · Calculate
Question
The flask holds 25.0 mL of hydrochloric acid of unknown concentration, and the burette 0.100 mol/L sodium hydroxide. Look inside at Whole curve shows the equivalence point. What is the acid’s concentration?

Correct answer
0.08 mol/L ± 0.002 mol/L
Setup, choices and feedback
The Scene at this step
Volume added
20 mLfixed
Curve so far / Whole curve
Whole curvefixed
Other settings
Acid concentration/nodes/rig/props/acidConcentration = 0.08
Other settings
Readings/nodes/rig/props/readings = false
Accepted within
± 0.002 mol/L
Computed from the Scene at this step’s setting
Acid concentration/nodes/rig/props/acidConcentration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.125 mol/L ± 0.002 mol/L
  Misconception: Turns the ratio of concentrations upside down when working out a titration volume or concentration.
  Feedback from the catalogue
  What becomes equal at the equivalence point is the number of moles: $c_a V_a = c_b V_b$ ⁠. A more dilute solution needs more volume, not less. Check the direction: if the base is more dilute than the acid, its volume must come out larger than the acid’s.
- 0.1 mol/L ± 0.002 mol/L
  Misconception: Thinks equal volumes of acid and base always neutralise each other.
  Feedback from the catalogue
  Equal volumes neutralise each other only when the concentrations are equal. What has to match is the number of moles, $n = c \times V$ ⁠: 20 mL of 0.10 mol/L acid holds $2.0$ ⁠ millimoles, and 0.08 mol/L base needs 25 mL for that.
Feedback for a right answer
0.080 mol/L: 20.0 mL of 0.100 mol/L base is $2.00$ ⁠ millimoles, and they sit in 25.0 mL of acid. That is how titration is used in the lab: measuring an unknown concentration with a solution whose concentration is known.

Hint
Read the equivalence volume under the chart, and work out how many moles of base it holds.

Worked solution
1. The equivalence volume from the chart: $20.0$ ⁠ mL.
2. Moles of base: $0.100 \times 20.0 = 2.00$ ⁠ millimoles; at $1 : 1$ ⁠, that is also the millimoles of acid.
3. $c_a = \dfrac{2.00\ \mathrm{mmol}}{25.0\ \mathrm{mL}} = 0.080\ \mathrm{mol/L}$ ⁠.
If you got $0.125$ ⁠, you turned the ratio of volumes upside down: an acid that needs less base than its own volume is more dilute than the base.

Lesson 4 of 4 · About 25 minutes

Titration curves and indicators

A weak acid on the curve: a pH that climbs gradually, an equivalence point above 7 and the pKa half way. And how to choose an indicator that changes colour right on the steep part.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A weak acid
Practice · Predict, then look
Question
This time the flask holds ethanoic acid, a weak acid, at the same concentration and volume. What will the pH be at the equivalence point?

Correct answer
- Correct Above $7$ ⁠
Setup, choices and feedback
The Scene at this step
Acid in the flask
Ethanoic acid (weak)fixed
Indicator
Phenolphthalein
Volume added
0 mL
Curve so far / Whole curve
Curve so farfixed
Choices
- Correct Above $7$ ⁠
- Wrong Exactly $7$ ⁠: the acid and the base are equal
  Misconception: Thinks the pH at the equivalence point is always 7.
  Feedback from the catalogue
  At the equivalence point the amounts of acid and base are equal, but the solution is not always neutral. When a weak acid is neutralised its ion is left behind, ethanoate, $\mathrm{CH_3COO^-}$ ⁠, and that is a weak base that takes protons from water: the pH at equivalence is above $7$ ⁠. With a weak base, like ammonia, it is below $7$ ⁠. Only a strong acid with a strong base gives exactly $7$ ⁠.
- Wrong Below $7$ ⁠: a weak acid leaves acid behind
  Feedback
  At the equivalence point all the acid has reacted. What is left is its ion, ethanoate, and that is slightly basic.
Once the prediction is locked in, the Scene changes
Volume added
25 mL
Curve so far / Whole curve
Whole curvefixed
The Scene then shows: 8.73 · PH at the equivalence point/nodes/rig/outputs/equivalencePH
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$8.7$ ⁠. At the equivalence point the flask holds sodium ethanoate, and the ethanoate ion takes a few protons from water: $\mathrm{CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-}$ ⁠. The start differs too: $2.9$ ⁠ instead of $1.0$ ⁠, and the curve climbs gradually rather than leaping early.

Worked solution
1. At the equivalence point: $2.5$ ⁠ millimoles of $\mathrm{CH_3COO^-}$ ⁠ in 50 mL, that is $0.05\ \mathrm{mol/L}$ ⁠.
2. Ethanoate is a weak base, with $K_b = \dfrac{K_w}{K_a} = \dfrac{10^{-14}}{1.74 \times 10^{-5}} = 5.8 \times 10^{-10}$ ⁠.
3. $[\mathrm{OH^-}] = \sqrt{K_b \times 0.05} = 5.4 \times 10^{-6}$ ⁠: $\mathrm{pOH} = 5.3$ ⁠, $\mathrm{pH} = 8.7$ ⁠.
Only when the acid and base are both strong is the equivalence point at pH $7$ ⁠.
Step 2 of 5
Open this step
Half way
Practice · Calculate
Question
The equivalence volume is 25 mL. Change Volume added to 12.5 mL, half of it, and read the pH. What is the $\mathrm{p}K_a$ ⁠ of ethanoic acid?

Correct answer
4.76 ± 0.05
Setup, choices and feedback
The Scene at this step
Acid in the flask
Ethanoic acid (weak)fixed
Indicator
Phenolphthalein
Volume added
0 mL
Curve so far / Whole curve
Whole curve
Accepted within
± 0.05
Computed from the Scene at this step’s setting
PH half way to equivalence (the weak acid's pKa, or 14 − the weak base's pKb)/nodes/rig/outputs/halfEquivalencePH
Wrong answers it recognises
- 8.73 ± 0.05
  Misconception: Reads the pKa of a weak acid at the equivalence point rather than half way to it.
  Feedback from the catalogue
  The pKa is read half way to the equivalence point. There exactly half the acid has turned into its ion, $[\mathrm{CH_3COOH}] = [\mathrm{CH_3COO^-}]$ ⁠, so by $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ ⁠, $[\mathrm{H_3O^+}] = K_a$ ⁠ and $\mathrm{pH} = \mathrm{p}K_a$ ⁠. At the equivalence point there is hardly any acid left.
Feedback for a right answer
$4.76$ ⁠. Half way, $[\mathrm{CH_3COOH}] = [\mathrm{CH_3COO^-}]$ ⁠, and then $[\mathrm{H_3O^+}] = K_a$ ⁠. It is also the flattest part of the curve: a mixture of a weak acid and its ion is a buffer, which resists a change of pH.

Hint
Half way to equivalence, half the acid has turned into ethanoate.

Worked solution
1. $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ ⁠.
2. At 12.5 mL half the acid has been neutralised: $[\mathrm{CH_3COO^-}] = [\mathrm{CH_3COOH}]$ ⁠, and they cancel.
3. That leaves $K_a = [\mathrm{H_3O^+}]$ ⁠, so $\mathrm{p}K_a = \mathrm{pH} = 4.76$ ⁠.
If you answered $8.7$ ⁠, you read at the equivalence point, where there is hardly any acid left.
Step 3 of 5
Open this step
The wrong indicator
Practice · Predict, then look
Question
Methyl orange changes from red to yellow between pH $3.1$ ⁠ and $4.4$ ⁠. If you titrate the ethanoic acid with it instead of phenolphthalein, when will the flask turn yellow?

Correct answer
- Correct Long before the equivalence point, after a few millilitres
Setup, choices and feedback
The Scene at this step
Acid in the flask
Ethanoic acid (weak)fixed
Indicator
Phenolphthaleinfixed
Volume added
0 mL
Curve so far / Whole curve
Whole curve
Choices
- Correct Long before the equivalence point, after a few millilitres
- Wrong At the equivalence point, at 25 mL: that is what an indicator is for
  Misconception: Thinks any indicator marks the equivalence point, whatever pH it changes colour at.
  Feedback from the catalogue
  An indicator changes colour over a pH range of its own, not at the equivalence point. It suits a titration only when its range falls on the steep part of the curve, where one drop moves the pH several units. With a weak acid the steep part runs from about $7$ ⁠ to $11$ ⁠, so phenolphthalein ( $8.2$ ⁠– $10$ ⁠) suits, and methyl orange ( $3.1$ ⁠– $4.4$ ⁠) changes colour far too early.
- Wrong Never: methyl orange does not work with bases
  Feedback
  Methyl orange changes colour by pH, not by substance. Once the pH passes $4.4$ ⁠, it is yellow.
Once the prediction is locked in, the Scene changes
Indicator
Methyl orangefixed
Volume added
5 mL
Curve so far / Whole curve
Whole curve
The Scene then shows: 1.949 mL · Volume added when the indicator is half changed (pH = pKIn)/nodes/rig/outputs/endPointVolume
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At about 2 mL, before you have added a tenth of the base needed. Ethanoic acid’s pH passes $3.7$ ⁠ early, and on the steep part, between $7$ ⁠ and $11$ ⁠, methyl orange has long been yellow. An end point like that is worthless.

Worked solution
1. An indicator changes colour when the pH crosses its range, here $3.1$ ⁠– $4.4$ ⁠.
2. On the ethanoic acid curve the pH starts at $2.9$ ⁠ and passes $3.7$ ⁠ after only about 2 mL.
3. The end point is far from the equivalence point, so methyl orange does not suit a weak acid.
Step 4 of 5
Open this step
Choosing an indicator
Practice · Choose
Question
Which indicator suits titrating ethanoic acid with sodium hydroxide?

Correct answer
- Correct Phenolphthalein, which changes between pH $8.2$ ⁠ and $10$ ⁠
Setup, choices and feedback
The Scene at this step
Acid in the flask
Ethanoic acid (weak)
Indicator
Phenolphthalein
Volume added
0 mL
Curve so far / Whole curve
Whole curve
Choices
Shown in a random order.
- Correct Phenolphthalein, which changes between pH $8.2$ ⁠ and $10$ ⁠
- Wrong Methyl orange, which changes between pH $3.1$ ⁠ and $4.4$ ⁠
  Misconception: Thinks any indicator marks the equivalence point, whatever pH it changes colour at.
  Feedback from the catalogue
  An indicator changes colour over a pH range of its own, not at the equivalence point. It suits a titration only when its range falls on the steep part of the curve, where one drop moves the pH several units. With a weak acid the steep part runs from about $7$ ⁠ to $11$ ⁠, so phenolphthalein ( $8.2$ ⁠– $10$ ⁠) suits, and methyl orange ( $3.1$ ⁠– $4.4$ ⁠) changes colour far too early.
- Wrong Universal indicator, which changes all along the scale
  Feedback
  Universal indicator changes gradually, so it has no one moment when the colour flips. It is good for estimating pH, not for marking an end point.
Feedback for a right answer
Phenolphthalein. The equivalence point is at pH $8.7$ ⁠, and the steep part runs from about $7$ ⁠ to $11$ ⁠: phenolphthalein’s range sits on it, and the colour flips in one drop, at 25.0 mL.

Hint
Where is the steep part of the curve, and which range falls on it?

Worked solution
1. With a weak acid and a strong base, the equivalence point is above $7$ ⁠ ( $8.7$ ⁠).
2. Choose an indicator whose range lies in the steep part, around the equivalence point.
3. Phenolphthalein, $8.2$ ⁠– $10$ ⁠, suits. Methyl orange, $3.1$ ⁠– $4.4$ ⁠, changes on the flat part, far too early.
Step 5 of 5
Open this step
A weak base
Practice · Choose
Question
Now the flask holds ammonia, a weak base, and the burette hydrochloric acid. The pH falls, and the equivalence point is at pH $5.3$ ⁠. Which indicator suits?

Correct answer
- Correct Methyl orange, $3.1$ ⁠– $4.4$ ⁠
Setup, choices and feedback
The Scene at this step
Acid in the flask
Hydrochloric acid (strong)hidden
Indicator
Phenolphthalein
Volume added
0 mL
Curve so far / Whole curve
Whole curve
Other settings
Base/nodes/rig/props/base = "weak"
Other settings
Titrant/nodes/rig/props/titrant = "acid"
Choices
Shown in a random order.
- Correct Methyl orange, $3.1$ ⁠– $4.4$ ⁠
- Wrong Phenolphthalein, $8.2$ ⁠– $10$ ⁠
  Misconception: Thinks any indicator marks the equivalence point, whatever pH it changes colour at.
  Feedback from the catalogue
  An indicator changes colour over a pH range of its own, not at the equivalence point. It suits a titration only when its range falls on the steep part of the curve, where one drop moves the pH several units. With a weak acid the steep part runs from about $7$ ⁠ to $11$ ⁠, so phenolphthalein ( $8.2$ ⁠– $10$ ⁠) suits, and methyl orange ( $3.1$ ⁠– $4.4$ ⁠) changes colour far too early.
- Wrong Neither: a weak base cannot be titrated
  Feedback
  It can, as long as you choose an indicator whose range falls on the steep part of the curve, here about $3$ ⁠ to $7$ ⁠.
Feedback for a right answer
Methyl orange. The steep part falls from about $7$ ⁠ to $3$ ⁠, and methyl orange changes colour on it, a drop after 25 mL. Phenolphthalein would lose its pink at about 10 mL, on the flat part.

Hint
Compare with the last question: everything is reversed.

Worked solution
1. Strong acid with weak base: the equivalence point is below $7$ ⁠, because the ammonium ion, $\mathrm{NH_4^+}$ ⁠, is a weak acid.
2. The steep part runs from about $7$ ⁠ down to $3$ ⁠.
3. Methyl orange, $3.1$ ⁠– $4.4$ ⁠, falls on it; phenolphthalein falls on the flat part at the start.

Across the unit · 15 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Acids, bases and ions in water

Two ions

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Pure water

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Strong and weak

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Bases

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The ion product

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The pH scale

Minus the logarithm

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Powers of ten

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

How many times more acidic

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

From a concentration to pH

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Dilution

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Bases on the same scale

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Neutralisation and titration