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A Level H1 Chemistry Practice Paper 4

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TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

TuitionGoWhere Secondary School (AI)

Subject: Chemistry
Level: A-Level H1
Paper: Practice Paper (Version 4 of 5)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this question paper.
You may use a scientific calculator.
A Data Booklet is provided for reference.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Structured Questions

Answer all questions in this section.

1. Ethanoic acid, $CH_3COOH$ , is a weak organic acid commonly found in vinegar.

(a) Define the term weak acid.
[1]

(b) Write a balanced equation, including state symbols, for the dissociation of ethanoic acid in water.
[2]

(c) Explain, in terms of bonding and structure, why ethanoic acid has a higher boiling point than ethanol ( $C_2H_5OH$ ), despite having a similar molar mass.
[2]

2. A student performs a titration to determine the concentration of a solution of benzoic acid, $C_6H_5COOH$ .

25.0 $cm^3$ of the benzoic acid solution is titrated against 0.050 $mol \ dm^{-3}$ sodium hydroxide ( $NaOH$ ). The equivalence point is reached when 22.4 $cm^3$ of $NaOH$ has been added.

(a) Calculate the amount, in moles, of $NaOH$ used in the titration.
[1]

(b) The equation for the reaction is:
$C_6H_5COOH(aq) + NaOH(aq) \rightarrow C_6H_5COONa(aq) + H_2O(l)$
Calculate the concentration of the benzoic acid solution in $mol \ dm^{-3}$ .
[2]

(c) Benzoic acid is often used as a food preservative. Suggest why benzoic acid is more effective as a preservative in acidic foods (low pH) than in neutral foods.
[1]

3. Aluminium oxide, $Al_2O_3$ , is described as an amphoteric oxide.

(a) Define the term amphoteric.
[1]

(b) Write balanced ionic equations for the reaction of solid aluminium oxide with:
(i) Dilute hydrochloric acid.
[2]
 (ii) Aqueous sodium hydroxide.
[2]

4. The pH of a 0.10 $mol \ dm^{-3}$ solution of methanoic acid ( $HCOOH$ ) is 2.38 at 298 K.

(a) Calculate the concentration of hydrogen ions, $[H^+]$ , in this solution.
[1]

(b) Write the expression for the acid dissociation constant, $K_a$ , for methanoic acid.
[1]

5. Buffer solutions are important in maintaining pH stability in biological systems.

(a) Explain how a buffer solution composed of ethanoic acid ( $CH_3COOH$ ) and sodium ethanoate ( $CH_3COONa$ ) resists changes in pH when a small amount of strong acid ( $H^+$ ) is added.
[3]

(b) Calculate the pH of a buffer solution containing 0.10 $mol \ dm^{-3}$ ethanoic acid and 0.20 $mol \ dm^{-3}$ sodium ethanoate.
( $K_a$ for ethanoic acid = $1.7 \times 10^{-5} \ mol \ dm^{-3}$ )
[3]

Section B: Data Interpretation and Application

Answer all questions in this section.

6. Carbonic acid ( $H_2CO_3$ ) plays a crucial role in maintaining the pH of blood plasma. It is formed when carbon dioxide dissolves in water.

$CO_2(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$

(a) Identify the conjugate base of carbonic acid in the equilibrium above.
[1]

(b) In a sample of blood plasma, the concentration of $HCO_3^-$ is $2.4 \times 10^{-2} \ mol \ dm^{-3}$ and the concentration of $H_2CO_3$ is $1.2 \times 10^{-3} \ mol \ dm^{-3}$ .
Given that $pK_{a1}$ for carbonic acid is 6.35, calculate the pH of this blood sample.
[3]

(c) Hyperventilation causes a rapid loss of $CO_2$ from the blood. Explain the effect this has on the pH of the blood, referring to the equilibrium equation.
[2]

7. Solubility equilibria are governed by the solubility product constant, $K_{sp}$ .

(a) Write the expression for the solubility product, $K_{sp}$ , for magnesium hydroxide, $Mg(OH)_2$ .
[1]

(b) The $K_{sp}$ of $Mg(OH)_2$ is $2.0 \times 10^{-11} \ mol^3 \ dm^{-9}$ at 298 K.
Calculate the solubility of $Mg(OH)_2$ in $mol \ dm^{-3}$ .
[3]

(c) Explain why the solubility of $Mg(OH)_2$ decreases when it is placed in a solution of sodium hydroxide compared to pure water.
[2]

8. Propanoic acid ( $C_2H_5COOH$ ) reacts with propan-1-ol ( $C_3H_7OH$ ) in the presence of an acid catalyst to form an ester.

(a) Name the ester formed.
[1]

(b) This reaction is reversible. State two conditions that would maximize the yield of the ester.
[2]

(c) The $pK_a$ of propanoic acid is 4.87. The $pK_a$ of ethanoic acid is 4.76.
Explain why propanoic acid is a slightly weaker acid than ethanoic acid, referring to the inductive effect.
[2]

9. Consider the following oxides of Period 3 elements: $Na_2O$ , $MgO$ , $Al_2O_3$ , $SiO_2$ , $P_4O_{10}$ , $SO_2$ .

(a) Which of these oxides has the highest melting point? Explain your answer in terms of structure and bonding.
[3]

(b) Write an equation for the reaction of $P_4O_{10}$ with excess water.
[1]

(c) Describe the pH of the solution formed when $SO_2$ is dissolved in water. Explain your answer with an equation.
[2]

10. An unknown organic compound X has the molecular formula $C_4H_8O_2$ .

(a) Compound X reacts with sodium carbonate to produce effervescence. Identify the functional group present in X.
[1]

(b) Draw the structural formula of one possible isomer of X that contains this functional group.
[1]

(c) Another isomer of X, compound Y, does not react with sodium carbonate but reacts with aqueous sodium hydroxide upon heating to form a carboxylate salt and an alcohol. Identify the functional group in Y.
[1]

Section C: Extended Response and Synthesis

Answer all questions in this section.

11. Lactic acid ( $CH_3CH(OH)COOH$ ) is produced in muscles during intense exercise. It is a weak monoprotic acid with $K_a = 1.4 \times 10^{-4} \ mol \ dm^{-3}$ .

(a) Calculate the pH of a 0.050 $mol \ dm^{-3}$ solution of lactic acid.
[4]

(b) State one assumption made in your calculation in (a) and justify why it is valid for this concentration.
[2]

12. The table below shows the $pK_a$ values of three different acids.

Acid	Formula	$pK_a$
Chloroethanoic acid	$ClCH_2COOH$	2.86
Ethanoic acid	$CH_3COOH$	4.76
Fluoroethanoic acid	$FCH_2COOH$	2.66

(a) Arrange the three acids in order of increasing acid strength.
[1]

(b) Explain the difference in acid strength between ethanoic acid and chloroethanoic acid.
[3]

(c) Predict whether fluoroethanoic acid is a stronger or weaker acid than chloroethanoic acid. Explain your reasoning.
[2]

13. A student wishes to prepare a buffer solution with a pH of 4.00 using ethanoic acid ( $pK_a = 4.76$ ) and sodium ethanoate.

(a) Calculate the ratio $\frac{[CH_3COO^-]}{[CH_3COOH]}$ required to achieve this pH.
[3]

(b) If the student mixes 50 $cm^3$ of 0.10 $mol \ dm^{-3}$ ethanoic acid with 50 $cm^3$ of 0.10 $mol \ dm^{-3}$ sodium ethanoate, will the resulting pH be higher or lower than 4.00? Explain.
[2]

14. Magnesium oxide ( $MgO$ ) and sulfur dioxide ( $SO_2$ ) are both oxides of Period 3 elements.

(a) Describe the observation when universal indicator is added to separate suspensions of $MgO$ and $SO_2$ in water.
[2]

(b) Explain the difference in the acid-base character of these two oxides with reference to the electronegativity of the central atom and the polarity of the bonds.
[4]

15. Aspirin (2-ethanoyloxybenzoic acid) contains both a carboxylic acid group and an ester group.

(a) Draw the structure of aspirin.
[2]

(b) Aspirin can be hydrolyzed by boiling with aqueous sodium hydroxide. Draw the structures of the two organic products formed in this reaction.
[3]

16. The solubility product of calcium sulfate, $CaSO_4$ , is $2.4 \times 10^{-5} \ mol^2 \ dm^{-6}$ at 298 K.

(a) Calculate the maximum concentration of calcium ions, $[Ca^{2+}]$ , that can exist in a saturated solution of $CaSO_4$ .
[2]

(b) Will a precipitate of $CaSO_4$ form if 100 $cm^3$ of 0.010 $mol \ dm^{-3}$ $CaCl_2$ is mixed with 100 $cm^3$ of 0.010 $mol \ dm^{-3}$ $Na_2SO_4$ ? Show your working.
[4]

17. Glycine ( $H_2NCH_2COOH$ ) is an amino acid.

(a) Draw the structure of glycine as it exists in a solution of high pH (alkaline conditions).
[1]

(b) Draw the structure of glycine as it exists in a solution of low pH (acidic conditions).
[1]

(c) Explain why glycine has a high melting point (233 °C) compared to ethanoic acid (16.6 °C), despite having a similar molar mass.
[2]

18. Nitric acid ( $HNO_3$ ) is a strong acid, while nitrous acid ( $HNO_2$ ) is a weak acid.

(a) Write the equation for the dissociation of nitrous acid in water.
[1]

(b) A solution of nitric acid has a pH of 1.0. A solution of nitrous acid has a pH of 1.0.
Compare the concentrations of the two acid solutions. Explain your answer.
[2]

(c) Nitrous acid is unstable and decomposes. One decomposition product is nitrogen monoxide ( $NO$ ).
State the oxidation number of nitrogen in $HNO_2$ and in $NO$ .
[2]

19. The indicator bromothymol blue has a $pK_{In}$ of 7.0. It is yellow in acidic solution and blue in alkaline solution.

(a) Write the equilibrium equation for the indicator $HIn$ in water.
[1]

(b) Explain why the color changes from yellow to blue as the pH increases from 5 to 9.
[3]

(c) Would bromothymol blue be a suitable indicator for the titration of ethanoic acid with sodium hydroxide? Explain.
[2]

20. A diprotic acid, $H_2A$ , has the following dissociation constants:
$K_{a1} = 1.0 \times 10^{-3} \ mol \ dm^{-3}$
$K_{a2} = 1.0 \times 10^{-7} \ mol \ dm^{-3}$

(a) Write the equations for the two dissociation steps.
[2]

(b) Explain why $K_{a1}$ is significantly larger than $K_{a2}$ .
[2]

(c) Calculate the pH of a 0.10 $mol \ dm^{-3}$ solution of $H_2A$ , assuming only the first dissociation contributes significantly to the $[H^+]$ .
[3]

*** End of Paper ***

Answers

TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

Answer Key and Marking Scheme
Paper: Practice Paper (Version 4 of 5)
Topic: Acids, Bases, and Salts

Section A: Structured Questions

1. (a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1] (b) $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$ [1 for equation, 1 for state symbols and reversible arrow] (c) Ethanoic acid forms dimers via strong hydrogen bonding between two molecules (two H-bonds per dimer). Ethanol forms hydrogen bonds but not dimers to the same extent. More energy is required to break the intermolecular forces in ethanoic acid. [1 for dimer/H-bonding detail, 1 for comparison of energy/forces]

2. (a) $n(NaOH) = c \times V = 0.050 \times \frac{22.4}{1000} = 1.12 \times 10^{-3} \ mol$ [1] (b) Ratio is 1:1. So $n(acid) = 1.12 \times 10^{-3} \ mol$ . $c(acid) = \frac{n}{V} = \frac{1.12 \times 10^{-3}}{0.025} = 0.0448 \ mol \ dm^{-3}$ [1 for moles, 1 for conc] (c) Benzoic acid is a weak acid. In acidic conditions, the equilibrium $C_6H_5COOH \rightleftharpoons C_6H_5COO^- + H^+$ shifts to the left, favoring the uncharged molecular form ( $C_6H_5COOH$ ). The uncharged molecule is more lipid-soluble and can penetrate bacterial cell membranes more effectively than the ionized form. [1 for reference to equilibrium/uncharged form]

3. (a) An amphoteric substance can act as both an acid and a base. [1] (b)(i) $Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)$ [1 for species, 1 for balancing] (b)(ii) $Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)$ [1 for species, 1 for balancing] (Note: Accept $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAl(OH)_4$ )

4. (a) $[H^+] = 10^{-pH} = 10^{-2.38} = 4.17 \times 10^{-3} \ mol \ dm^{-3}$ [1] (b) $K_a = \frac{[HCOO^-][H^+]}{[HCOOH]}$ [1] (c) Assume $[H^+] = [HCOO^-]$ and $[HCOOH]_{eq} \approx [HCOOH]_{initial}$ (since dissociation is small). $K_a = \frac{(4.17 \times 10^{-3})^2}{0.10} = \frac{1.74 \times 10^{-5}}{0.10} = 1.74 \times 10^{-4} \ mol \ dm^{-3}$ [1 for substitution, 1 for answer, 1 for units]

5. (a) The buffer contains high concentrations of $CH_3COOH$ and $CH_3COO^-$ . When $H^+$ is added, it reacts with the conjugate base $CH_3COO^-$ to form $CH_3COOH$ : $CH_3COO^- + H^+ \rightarrow CH_3COOH$ . This removes most of the added $H^+$ , keeping the pH relatively constant. [1 for identifying reaction with conjugate base, 1 for equation, 1 for explanation of pH stability] (b) $pH = pK_a + \log_{10} \left( \frac{[salt]}{[acid]} \right)$ $pK_a = -\log(1.7 \times 10^{-5}) = 4.77$ $pH = 4.77 + \log_{10} \left( \frac{0.20}{0.10} \right) = 4.77 + \log_{10}(2) = 4.77 + 0.30 = 5.07$ [1 for pKa, 1 for substitution, 1 for final answer]

Section B: Data Interpretation and Application

6. (a) $HCO_3^-$ (Hydrogencarbonate ion) [1] (b) $pH = pK_a + \log_{10} \left( \frac{[HCO_3^-]}{[H_2CO_3]} \right)$ $pH = 6.35 + \log_{10} \left( \frac{2.4 \times 10^{-2}}{1.2 \times 10^{-3}} \right) = 6.35 + \log_{10}(20) = 6.35 + 1.30 = 7.65$ [1 for formula, 1 for substitution, 1 for answer] (c) Loss of $CO_2$ shifts the equilibrium $CO_2 + H_2O \rightleftharpoons H^+ + HCO_3^-$ to the left to restore $CO_2$ . This consumes $H^+$ ions, causing the pH to increase (become more alkaline). [1 for shift direction, 1 for effect on pH]

7. (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1] (b) Let solubility be $s \ mol \ dm^{-3}$ . Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$ . $K_{sp} = (s)(2s)^2 = 4s^3$ $2.0 \times 10^{-11} = 4s^3$ $s^3 = 5.0 \times 10^{-12}$ $s = \sqrt[3]{5.0 \times 10^{-12}} = 1.71 \times 10^{-4} \ mol \ dm^{-3}$ [1 for expression in s, 1 for calculation, 1 for answer] (c) Common ion effect. Adding $NaOH$ increases $[OH^-]$ . To maintain constant $K_{sp}$ , the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ shifts to the left, precipitating more solid and decreasing solubility. [1 for common ion effect, 1 for shift explanation]

8. (a) Propyl propanoate [1] (b) 1. Use excess alcohol or acid. 2. Remove water/ester as it forms (distillation). [1 for each] (c) The ethyl group ( $C_2H_5-$ ) in propanoic acid is more electron-releasing (positive inductive effect) than the methyl group ( $CH_3-$ ) in ethanoic acid. This increases electron density on the carboxylate anion, destabilizing it slightly and making the O-H bond harder to break, thus making it a weaker acid. [1 for inductive effect comparison, 1 for link to stability/acidity]

9. (a) $SiO_2$ (Silicon dioxide). It has a giant covalent structure with strong covalent bonds throughout the lattice requiring large amounts of energy to break. The others are simple molecular (weak van der Waals) or ionic (strong but generally lower MP than giant covalent network like silica). [1 for SiO2, 1 for giant covalent, 1 for strong bonds] (b) $P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$ [1] (c) Acidic. $SO_2 + H_2O \rightleftharpoons H_2SO_3$ (sulfurous acid). The solution contains $H^+$ ions. [1 for acidic, 1 for equation]

10. (a) Carboxylic acid [1] (b) e.g., $CH_3CH_2CH_2COOH$ (Butanoic acid) or $(CH_3)_2CHCOOH$ (2-methylpropanoic acid) [1] (c) Ester [1]

Section C: Extended Response and Synthesis

11. (a) $K_a = \frac{[H^+]^2}{[HA]}$ $[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.4 \times 10^{-4} \times 0.050} = \sqrt{7.0 \times 10^{-6}} = 2.65 \times 10^{-3} \ mol \ dm^{-3}$ $pH = -\log(2.65 \times 10^{-3}) = 2.58$ [1 for formula rearrangement, 1 for [H+], 1 for pH, 1 for sig figs/units] (b) Assumption: The dissociation of water is negligible / $[H^+]_{eq} \approx [A^-]_{eq}$ . Or: The change in $[HA]$ is negligible ( $[HA]_{eq} \approx [HA]_{initial}$ ). Justification: $K_a$ is small ( $10^{-4}$ ), so the degree of dissociation is low (<5%). [1 for assumption, 1 for justification]

12. (a) Ethanoic < Chloroethanoic < Fluoroethanoic [1] (b) Chlorine is more electronegative than hydrogen. It exerts a negative inductive effect (-I), withdrawing electron density from the carboxylate group. This disperses the negative charge on the conjugate base ( $ClCH_2COO^-$ ), stabilizing it. A more stable conjugate base means the acid is stronger. [1 for electronegativity, 1 for inductive effect, 1 for stabilization] (c) Fluorine is more electronegative than chlorine. It has a stronger -I effect, stabilizing the conjugate base more effectively. Thus, fluoroethanoic acid is stronger. [1 for prediction, 1 for reasoning]

13. (a) $pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$ $4.00 = 4.76 + \log \left( \frac{[A^-]}{[HA]} \right)$ $\log \left( \frac{[A^-]}{[HA]} \right) = -0.76$ $\frac{[A^-]}{[HA]} = 10^{-0.76} = 0.174$ [1 for substitution, 1 for log calc, 1 for ratio] (b) Mixing equal volumes of equal concentration gives a ratio of 1:1. $pH = pK_a + \log(1) = 4.76$ . 4.76 is higher than 4.00. [1 for identifying pH 4.76, 1 for comparison]

14. (a) $MgO$ : Blue/Purple (Alkaline). $SO_2$ : Red/Orange (Acidic). [1 for each] (b) Mg is a metal with low electronegativity. The Mg-O bond is ionic. $O^{2-}$ ions react with water to form $OH^-$ . S is a non-metal with high electronegativity. The S-O bonds are covalent and polar. $SO_2$ reacts with water to form covalent bonds with O, releasing $H^+$ . The high oxidation state and electronegativity of S draw electron density, weakening O-H bonds in the resulting oxoacid. [1 for Mg ionic/basic, 1 for S covalent/acidic, 1 for electronegativity/polarity link, 1 for mechanism]

15. (a) Benzene ring with -COOH at position 1 and -OCOCH3 at position 2. [1 for ring/substituents, 1 for correct positions] (b) Products: Sodium salicylate (2-hydroxybenzoate ion) and Sodium ethanoate (acetate ion). Structures: Benzene ring with -COO- Na+ and -OH; and CH3COO- Na+. [1 for salicylate, 1 for ethanoate, 1 for correct ionic forms] (c) In NaOH, the carboxylic acid group reacts to form a soluble ionic salt ( $COO^- Na^+$ ). Ionic compounds are generally more soluble in polar solvents like water due to ion-dipole interactions. [1 for salt formation, 1 for solubility explanation]

16. (a) $K_{sp} = [Ca^{2+}][SO_4^{2-}]$ . In pure water, $[Ca^{2+}] = [SO_4^{2-}] = s$ . $s^2 = 2.4 \times 10^{-5}$ $s = \sqrt{2.4 \times 10^{-5}} = 4.90 \times 10^{-3} \ mol \ dm^{-3}$ [1 for expression, 1 for answer] (b) Total volume = 200 $cm^3$ . Concentrations halve. $[Ca^{2+}] = 0.005 \ mol \ dm^{-3}$ $[SO_4^{2-}] = 0.005 \ mol \ dm^{-3}$ Ionic Product (IP) = $0.005 \times 0.005 = 2.5 \times 10^{-5}$ $IP (2.5 \times 10^{-5}) > K_{sp} (2.4 \times 10^{-5})$ . Yes, a precipitate will form. [1 for new concs, 1 for IP calc, 1 for comparison, 1 for conclusion]

17. (a) $H_2NCH_2COO^-$ [1] (b) $^+H_3NCH_2COOH$ [1] (c) Glycine exists as a zwitterion ( $^+H_3NCH_2COO^-$ ) in the solid state. This results in strong electrostatic forces (ionic bonding) between ions in the lattice. Ethanoic acid exists as simple molecules held by weaker hydrogen bonds. Ionic forces are stronger than H-bonds. [1 for zwitterion/ionic nature, 1 for comparison of forces]

18. (a) $HNO_2(aq) \rightleftharpoons H^+(aq) + NO_2^-(aq)$ [1] (b) Nitric acid is strong (fully dissociated), so $[HNO_3] = [H^+] = 0.1 \ mol \ dm^{-3}$ . Nitrous acid is weak (partially dissociated), so to achieve the same $[H^+]$ of 0.1, the initial concentration of $HNO_2$ must be much higher than 0.1 $mol \ dm^{-3}$ . [1 for strong/weak distinction, 1 for conc comparison] (c) In $HNO_2$ : H is +1, O is -2. $1 + N + 2(-2) = 0 \Rightarrow N = +3$ . In $NO$ : O is -2. $N - 2 = 0 \Rightarrow N = +2$ . [1 for each]

19. (a) $HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$ [1] (b) At low pH (high $[H^+]$ ), equilibrium shifts left, favoring $HIn$ (yellow). As pH increases, $[H^+]$ decreases, equilibrium shifts right, favoring $In^-$ (blue). At $pH = pK_{In} = 7$ , $[HIn] = [In^-]$ (green/mix). [1 for shift left/right, 1 for species color, 1 for equilibrium principle] (c) No. The titration of a weak acid (ethanoic) with a strong base (NaOH) has an equivalence point at pH > 7 (approx 8-9). Bromothymol blue changes color at pH 7, which is before the equivalence point, leading to an inaccurate endpoint. Phenolphthalein (range 8-10) would be better. [1 for No, 1 for explanation of eq point pH]

20. (a) 1. $H_2A \rightleftharpoons H^+ + HA^-$ 2. $HA^- \rightleftharpoons H^+ + A^{2-}$ [1 for each] (b) It is harder to remove a positive proton ( $H^+$ ) from a negatively charged ion ( $HA^-$ ) than from a neutral molecule ( $H_2A$ ) due to electrostatic attraction. [1 for charge argument, 1 for electrostatic attraction] (c) Use $K_{a1}$ . $[H^+] = \sqrt{K_{a1} \times [H_2A]} = \sqrt{1.0 \times 10^{-3} \times 0.10} = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2} \ mol \ dm^{-3}$ $pH = -\log(1.0 \times 10^{-2}) = 2.00$ [1 for formula, 1 for [H+], 1 for pH]

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