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TuitionGoWhere Practice Paper - Chemistry H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry H1
Level: A-Level (Singapore-Cambridge GCE)
Paper: Practice Paper – Version 2 of 5
Topic Focus: Acids, Bases, and Salts (Section V: Chemistry of Aqueous Solutions)
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 separately (refer to standard constants and $K_a$ values where necessary, though specific values are provided in questions where required).
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 acid commonly found in vinegar.
(a) Define the term weak acid. [1]
....................................................................................................................................................
....................................................................................................................................................

(b) Write an equation, including state symbols, to show the dissociation of ethanoic acid in water. [1]
....................................................................................................................................................

(c) Explain, in terms of bonding and structure, why ethanoic acid has a higher boiling point than propane ( $C_3H_8$ ), despite having a similar relative molecular mass. [2]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

2. A student titrates $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ sodium hydroxide ( $NaOH$ ) with $0.100 \text{ mol dm}^{-3}$ hydrochloric acid ( $HCl$ ).
(a) Calculate the pH of the sodium hydroxide solution before any acid is added. ( $K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$ ) [2]

(b) Sketch the pH curve for this titration on the axes below. Label the equivalence point clearly. [3]

pH
14 |
   |
   |
   |
   |
   |
   |
   |
   |
   |
   |
   |
 0 |________________________________________ Volume of HCl added (cm³)
    0          12.5         25.0         37.5

(c) Suggest a suitable indicator for this titration and state the colour change at the endpoint. [2]
Indicator: __________________________
Colour change: __________________________ to __________________________

3. Ammonia ( $NH_3$ ) acts as a Brønsted-Lowry base in water.
(a) Write an equation for the reaction of ammonia with water. [1]
....................................................................................................................................................

(b) Identify the conjugate acid-base pairs in your equation above. [2]
Pair 1: __________________________ and __________________________
Pair 2: __________________________ and __________________________

(c) The $K_b$ for ammonia is $1.8 \times 10^{-5} \text{ mol dm}^{-3}$ . Calculate the pH of a $0.10 \text{ mol dm}^{-3}$ solution of ammonia. [3]

4. 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 aluminium oxide with:
(i) Dilute hydrochloric acid. [2]
....................................................................................................................................................
(ii) Aqueous sodium hydroxide. [2]
....................................................................................................................................................

5. Consider the buffer solution formed by mixing $0.10 \text{ mol dm}^{-3}$ methanoic acid ( $HCOOH$ ) and $0.10 \text{ mol dm}^{-3}$ sodium methanoate ( $HCOONa$ ). The $K_a$ of methanoic acid is $1.8 \times 10^{-4} \text{ mol dm}^{-3}$ .
(a) Calculate the pH of this buffer solution. [2]

(b) Explain how this buffer solution resists a change in pH when a small amount of strong acid ( $H^+$ ) is added. [2]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

6. The solubility product constant, $K_{sp}$ , for magnesium hydroxide, $Mg(OH)_2$ , is $5.6 \times 10^{-12} \text{ mol}^3 \text{ dm}^{-9}$ at 298 K.
(a) Write the expression for $K_{sp}$ for magnesium hydroxide. [1]
....................................................................................................................................................

(b) Calculate the solubility of magnesium hydroxide in $\text{mol dm}^{-3}$ in pure water. [3]

(c) Explain why magnesium hydroxide is less soluble in a solution of magnesium chloride than in pure water. [2]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

7. Chloric(I) acid ( $HClO$ ) is a weak acid used in swimming pools.
(a) $HClO$ dissociates as follows: $HClO(aq) \rightleftharpoons H^+(aq) + ClO^-(aq)$ .
Write the expression for the acid dissociation constant, $K_a$ . [1]
....................................................................................................................................................

(b) A sample of pool water has a pH of 7.4. Calculate the concentration of $H^+$ ions in the water. [1]

(c) If the concentration of $HClO$ is $0.050 \text{ mol dm}^{-3}$ and the concentration of $ClO^-$ is $0.050 \text{ mol dm}^{-3}$ , calculate the pH of the solution given $K_a = 3.0 \times 10^{-8} \text{ mol dm}^{-3}$ . [2]

8. Distinguish between the terms strength and concentration when applied to acids. [2]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

9. Carbon dioxide dissolves in water to form carbonic acid ( $H_2CO_3$ ), which contributes to the acidity of rainwater.
(a) Write the equation for the formation of carbonic acid from carbon dioxide and water. [1]
....................................................................................................................................................

(b) Carbonic acid is a diprotic acid. Write the equation for the second dissociation step of carbonic acid. [1]
....................................................................................................................................................

10. A student performs a titration of a weak acid ( $HA$ ) with a strong base ( $NaOH$ ). The pH at the half-equivalence point is found to be 4.8.
(a) State the relationship between pH and $pK_a$ at the half-equivalence point. [1]
....................................................................................................................................................

(b) Determine the value of $K_a$ for the acid $HA$ . [2]

Section B: Data-Based and Application Questions

Answer all questions in this section.

11. The table below shows the pH values of four different aqueous solutions, all with a concentration of $0.10 \text{ mol dm}^{-3}$ .

Solution	Compound	pH
A	$HCl$	1.0
B	$CH_3COOH$	2.9
C	$NH_3$	11.1
D	$NaOH$	13.0

(a) Explain why Solution A has a lower pH than Solution B. [2]
....................................................................................................................................................
....................................................................................................................................................

(b) Calculate the percentage dissociation of ethanoic acid in Solution B. [2]

(c) Solution C is a weak base. Explain why its pH is lower than that of Solution D. [2]
....................................................................................................................................................
....................................................................................................................................................

12. Tooth enamel consists mainly of hydroxyapatite, $Ca_5(PO_4)_3OH$ . In the mouth, bacteria produce acid which can dissolve enamel.
$Ca_5(PO_4)_3OH(s) \rightleftharpoons 5Ca^{2+}(aq) + 3PO_4^{3-}(aq) + OH^-(aq)$

(a) Using Le Chatelier’s principle, explain how the production of acid by bacteria leads to tooth decay. [3]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

(b) Fluoride toothpaste contains fluoride ions ( $F^-$ ). These ions replace the hydroxide ions in hydroxyapatite to form fluoroapatite, $Ca_5(PO_4)_3F$ , which is less soluble.
Suggest why fluoroapatite is more resistant to acid attack than hydroxyapatite. [2]
....................................................................................................................................................
....................................................................................................................................................

13. An unknown monoprotic acid, $HX$ , has a concentration of $0.010 \text{ mol dm}^{-3}$ and a pH of 3.0.
(a) Calculate the concentration of $H^+$ ions. [1]

(b) Determine whether $HX$ is a strong or weak acid. Justify your answer with a calculation. [2]

14. Buffer solutions are essential in biological systems. Blood plasma is buffered by the carbonic acid/hydrogencarbonate system.
$H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$

(a) Given that the pH of blood is 7.4 and the $pK_a$ of carbonic acid is 6.1, calculate the ratio $\frac{[HCO_3^-]}{[H_2CO_3]}$ in blood. [3]

(b) Explain the significance of this buffer system in maintaining blood pH during exercise when lactic acid is produced. [2]
....................................................................................................................................................
....................................................................................................................................................

15. The diagram below represents the distribution of species for a diprotic acid $H_2A$ as pH changes.
(Note: Imagine a graph where at low pH, $H_2A$ dominates; at intermediate pH, $HA^-$ dominates; at high pH, $A^{2-}$ dominates.)

(a) At what pH is $[H_2A] = [HA^-]$ ? [1]
....................................................................................................................................................

(b) At what pH is $[HA^-] = [A^{2-}]$ ? [1]
....................................................................................................................................................

(c) If $pK_{a1} = 3.0$ and $pK_{a2} = 7.0$ , sketch the predominant species at pH 5.0. [1]
....................................................................................................................................................

Section C: Extended Response and Synthesis

Answer all questions in this section.

16. A student is asked to prepare $250 \text{ cm}^3$ of a buffer solution with pH 5.0 using ethanoic acid ( $CH_3COOH$ ) and sodium ethanoate ( $CH_3COONa$ ). The $K_a$ of ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ .
(a) Calculate the required ratio of $\frac{[CH_3COO^-]}{[CH_3COOH]}$ to achieve pH 5.0. [3]

(b) If the student uses $0.10 \text{ mol dm}^{-3}$ ethanoic acid, calculate the mass of solid sodium ethanoate ( $M_r = 82.0$ ) required to add to $250 \text{ cm}^3$ of this acid to create the buffer. Assume the volume change upon adding the solid is negligible. [4]

17. Magnesium hydroxide, $Mg(OH)_2$ , is used as an antacid to neutralize excess stomach acid ( $HCl$ ).
(a) Write the balanced equation for the neutralization reaction. [1]
....................................................................................................................................................

(b) Explain why $Mg(OH)_2$ is preferred over $NaOH$ for this purpose, referring to solubility and strength. [2]
....................................................................................................................................................
....................................................................................................................................................

(c) Calculate the volume of $0.10 \text{ mol dm}^{-3}$ $HCl$ that can be neutralized by $0.50 \text{ g}$ of $Mg(OH)_2$ . ( $M_r$ of $Mg(OH)_2 = 58.3$ ) [3]

18. Propanoic acid ( $C_2H_5COOH$ ) is a weak acid.
(a) Compare the electrical conductivity of $0.1 \text{ mol dm}^{-3}$ propanoic acid with $0.1 \text{ mol dm}^{-3}$ hydrochloric acid. Explain your answer. [2]
....................................................................................................................................................
....................................................................................................................................................

(b) Describe and explain the effect of diluting propanoic acid with water on its pH and its percentage dissociation. [4]
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................

19. The salt ammonium chloride ( $NH_4Cl$ ) dissolves in water to form an acidic solution.
(a) Write the equation for the dissociation of ammonium chloride in water. [1]
....................................................................................................................................................

(b) Write the equation for the hydrolysis of the ammonium ion. [1]
....................................................................................................................................................

(c) Explain why the resulting solution is acidic. [2]
....................................................................................................................................................
....................................................................................................................................................

20. A titration curve is obtained for the reaction between $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ammonia ( $NH_3$ ) and $0.10 \text{ mol dm}^{-3}$ hydrochloric acid ( $HCl$ ).
(a) State the pH nature (acidic, alkaline, or neutral) of the solution at the equivalence point. [1]
....................................................................................................................................................

(b) Explain your answer in (a) by referring to the salt formed. [2]
....................................................................................................................................................
....................................................................................................................................................

(c) Why is phenolphthalein (pH range 8.3–10.0) not a suitable indicator for this titration? [2]
....................................................................................................................................................
....................................................................................................................................................

End of Paper

Answers

TuitionGoWhere Practice Paper - Chemistry H1 A-Level

Answer Key and Marking Scheme

Subject: Chemistry H1
Topic: Acids, Bases, and Salts
Version: 2 of 5

Section A: Structured Questions

1.
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$
Note: Must use reversible arrow $\rightleftharpoons$ and state symbols. [1]
(c) Ethanoic acid molecules can form hydrogen bonds between molecules due to the presence of the -OH group in the carboxyl group. Propane only has weak van der Waals forces (instantaneous dipole-induced dipole). Hydrogen bonds are stronger than van der Waals forces, requiring more energy to break. [2]

2.
(a)
$[OH^-] = 0.100 \text{ mol dm}^{-3}$
$pOH = -\log(0.100) = 1.0$
$pH = 14.0 - 1.0 = 13.0$ [2]
(b)

Start pH high (~13).
Gradual decrease, then steep drop around 25.0 cm³.
Equivalence point at pH 7 (strong acid + strong base).
End pH low (~1).
Label equivalence point at volume = 25.0 cm³. [3]
(c)
Indicator: Methyl orange OR Bromothymol blue.
Colour change:
Methyl orange: Yellow to Orange/Red.
Bromothymol blue: Blue to Yellow.
(Phenolphthalein is also acceptable but less ideal due to steep range, though technically works for strong/strong. Methyl orange is standard for strong acid into base if reversing, but here acid is added to base. Methyl orange changes 3.1-4.4. The vertical section is pH 10-4. So MO works. Phenolphthalein 8.3-10 also works. Accept either with correct colour change for Acid into Base).
Correction: Titration is NaOH (in flask) + HCl (burette). Start Alkaline. End Acidic.
Phenolphthalein: Pink to Colourless.
Methyl Orange: Yellow to Orange. [2]

3.
(a) $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ [1]
(b)
Pair 1: $NH_3$ (base) and $NH_4^+$ (conjugate acid).
Pair 2: $H_2O$ (acid) and $OH^-$ (conjugate base). [2]
(c)
$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$
Assume $[NH_4^+] = [OH^-] = x$ and $[NH_3]_{eq} \approx 0.10$ .
$1.8 \times 10^{-5} = \frac{x^2}{0.10}$
$x^2 = 1.8 \times 10^{-6}$
$x = [OH^-] = 1.34 \times 10^{-3} \text{ mol dm}^{-3}$
$pOH = -\log(1.34 \times 10^{-3}) = 2.87$
$pH = 14.0 - 2.87 = 11.13$ [3]

4.
(a) An amphoteric substance can act as both an acid and a base. [1]
(b)
(i) $Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)$ [2]
(ii) $Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)$
Note: Accept $2AlO_2^- + H_2O$ depending on syllabus variant, but tetrahydroxoaluminate is preferred in modern A-Level. [2]

5.
(a)
$pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)$
$pK_a = -\log(1.8 \times 10^{-4}) = 3.74$
Since $[salt] = [acid]$ , $\log(1) = 0$ .
$pH = 3.74$ [2]
(b)
When $H^+$ is added, it reacts with the conjugate base ( $HCOO^-$ ) to form undissociated acid ( $HCOOH$ ):
$H^+(aq) + HCOO^-(aq) \rightarrow HCOOH(aq)$
This removes the added $H^+$ , keeping the pH relatively constant. [2]

6.
(a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]
(b)
Let solubility be $s \text{ mol dm}^{-3}$ .
$[Mg^{2+}] = s$ , $[OH^-] = 2s$ .
$K_{sp} = (s)(2s)^2 = 4s^3$
$5.6 \times 10^{-12} = 4s^3$
$s^3 = 1.4 \times 10^{-12}$
$s = \sqrt[3]{1.4 \times 10^{-12}} = 1.12 \times 10^{-4} \text{ mol dm}^{-3}$ [3]
(c)
Common ion effect. $MgCl_2$ provides $Mg^{2+}$ ions.
According to Le Chatelier’s principle, increasing $[Mg^{2+}]$ shifts the equilibrium position to the left (precipitate side), reducing the solubility of $Mg(OH)_2$ . [2]

7.
(a) $K_a = \frac{[H^+][ClO^-]}{[HClO]}$ [1]
(b)
$[H^+] = 10^{-pH} = 10^{-7.4} = 3.98 \times 10^{-8} \text{ mol dm}^{-3}$ [1]
(c)
$pH = pK_a + \log \left( \frac{[ClO^-]}{[HClO]} \right)$
$pK_a = -\log(3.0 \times 10^{-8}) = 7.52$
Ratio is 1 ( $0.050/0.050$ ). $\log(1) = 0$ .
$pH = 7.52$ [2]

8.
Strength refers to the degree of dissociation/ionization of the acid in water (strong acids fully dissociate, weak acids partially dissociate).
Concentration refers to the amount of acid (moles) per unit volume of solution ( $\text{mol dm}^{-3}$ ). [2]

9.
(a) $CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)$ [1]
(b) $HCO_3^-(aq) \rightleftharpoons H^+(aq) + CO_3^{2-}(aq)$ [1]

10.
(a) At half-equivalence point, $pH = pK_a$ . [1]
(b)
$pK_a = 4.8$
$K_a = 10^{-4.8} = 1.58 \times 10^{-5} \text{ mol dm}^{-3}$ [2]

Section B: Data-Based and Application Questions

11.
(a) HCl is a strong acid and fully dissociates, producing a high $[H^+]$ ( $0.1 \text{ mol dm}^{-3}$ ). Ethanoic acid is a weak acid and only partially dissociates, producing a much lower $[H^+]$ . Lower $[H^+]$ means higher pH. [2]
(b)
$[H^+] = 10^{-2.9} = 1.26 \times 10^{-3} \text{ mol dm}^{-3}$
$\% \text{ dissociation} = \frac{[H^+]_{eq}}{[HA]_{initial}} \times 100$
$= \frac{1.26 \times 10^{-3}}{0.10} \times 100 = 1.26 \%$ [2]
(c) $NH_3$ is a weak base and partially dissociates to produce $OH^-$ . $NaOH$ is a strong base and fully dissociates. Thus, $[OH^-]$ in ammonia is lower than in NaOH, resulting in a lower pOH and therefore a lower pH (less alkaline). [2]

12.
(a) Acid production increases $[H^+]$ . $H^+$ reacts with $OH^-$ from the equilibrium to form water. This decreases $[OH^-]$ . According to Le Chatelier’s principle, the equilibrium shifts to the right to restore $[OH^-]$ , causing more hydroxyapatite to dissolve. [3]
(b) Fluoroapatite is less soluble (has a lower $K_{sp}$ ) than hydroxyapatite. Therefore, the equilibrium concentration of ions is lower, making it harder for acid to shift the equilibrium sufficiently to cause significant dissolution/decay. [2]

13.
(a) $[H^+] = 10^{-3.0} = 1.0 \times 10^{-3} \text{ mol dm}^{-3}$ [1]
(b)
If strong, $[H^+]$ would be $0.010 \text{ mol dm}^{-3}$ (pH 2.0).
Actual $[H^+]$ is $0.001 \text{ mol dm}^{-3}$ .
Since $[H^+] < [HX]_{initial}$ , it is partially dissociated, so it is a weak acid. [2]
(c)
$K_a = \frac{[H^+][X^-]}{[HX]}$
Assume $[H^+] = [X^-] = 10^{-3}$ .
$[HX]_{eq} \approx 0.010$ (since dissociation is small).
$K_a = \frac{(10^{-3})^2}{0.010} = \frac{10^{-6}}{10^{-2}} = 1.0 \times 10^{-4} \text{ mol dm}^{-3}$ [2]

14.
(a)
$pH = pK_a + \log \left( \frac{[HCO_3^-]}{[H_2CO_3]} \right)$
$7.4 = 6.1 + \log (\text{ratio})$
$1.3 = \log (\text{ratio})$
$\text{Ratio} = 10^{1.3} = 19.95 \approx 20$ [3]
(b)
Lactic acid adds $H^+$ . The $HCO_3^-$ (conjugate base) in the buffer reacts with the added $H^+$ to form $H_2CO_3$ . This removes the excess $H^+$ , preventing a significant drop in blood pH. [2]

15.
(a) pH = $pK_{a1}$ [1]
(b) pH = $pK_{a2}$ [1]
(c) At pH 5.0 (between $pK_{a1}=3$ and $pK_{a2}=7$ ), the predominant species is $HA^-$ . [1]

Section C: Extended Response and Synthesis

16.
(a)
$pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$
$pK_a = -\log(1.7 \times 10^{-5}) = 4.77$
$5.0 = 4.77 + \log (\text{ratio})$
$0.23 = \log (\text{ratio})$
$\text{Ratio} = 10^{0.23} = 1.70$ [3]
(b)
$[HA] = 0.10 \text{ mol dm}^{-3}$ .
$\frac{[A^-]}{0.10} = 1.70 \Rightarrow [A^-] = 0.170 \text{ mol dm}^{-3}$ .
Volume = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$ .
Moles of $CH_3COONa = 0.170 \times 0.250 = 0.0425 \text{ mol}$ .
Mass = moles $\times M_r = 0.0425 \times 82.0 = 3.485 \text{ g}$ .
Answer: 3.49 g (or 3.5 g). [4]

17.
(a) $Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)$ [1]
(b) $Mg(OH)_2$ is sparingly soluble and a weak base. It neutralizes acid gradually without causing a sudden, dangerous spike in pH or damaging tissue, unlike NaOH which is a strong, corrosive base and fully soluble. [2]
(c)
Moles $Mg(OH)_2 = \frac{0.50}{58.3} = 0.008576 \text{ mol}$ .
From equation, 1 mol $Mg(OH)_2$ reacts with 2 mol $HCl$ .
Moles $HCl = 2 \times 0.008576 = 0.01715 \text{ mol}$ .
Volume $HCl = \frac{n}{c} = \frac{0.01715}{0.10} = 0.1715 \text{ dm}^3 = 171.5 \text{ cm}^3$ .
Answer: 172 cm³. [3]

18.
(a) Propanoic acid has lower conductivity than HCl. HCl fully dissociates into many ions ( $H^+, Cl^-$ ). Propanoic acid partially dissociates, resulting in fewer ions to carry charge. [2]
(b)
pH: Increases (becomes less acidic) because $[H^+]$ decreases upon dilution.
% Dissociation: Increases. According to Ostwald’s dilution law (or Le Chatelier), diluting adds water. The equilibrium $HA \rightleftharpoons H^+ + A^-$ shifts to the right (more particles) to counteract the decrease in concentration, leading to a higher fraction of dissociated molecules. [4]

19.
(a) $NH_4Cl(s) \rightarrow NH_4^+(aq) + Cl^-(aq)$ [1]
(b) $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$
(Or $NH_4^+ \rightleftharpoons NH_3 + H^+$ ) [1]
(c) The ammonium ion acts as a weak acid, donating protons to water to form $H_3O^+$ (or $H^+$ ). The chloride ion is the conjugate base of a strong acid and does not hydrolyze. The production of $H_3O^+$ makes the solution acidic. [2]

20.
(a) Acidic. [1]
(b) The salt formed is ammonium chloride ( $NH_4Cl$ ). The $NH_4^+$ ion is a weak acid (conjugate of weak base $NH_3$ ) and undergoes hydrolysis to produce $H^+$ ions. The $Cl^-$ ion is neutral. Thus, the solution contains excess $H^+$ . [2]
(c) The equivalence point for a Weak Base + Strong Acid titration is at pH < 7 (approx pH 5-6). Phenolphthalein changes colour at pH 8.3–10.0, which is far from the equivalence point. It would change colour before the equivalence point is reached, leading to a large titration error. [2]