Question
Chlorine gas reacts with water to produce hypochlorous acid and hydrochloric acid.
Cl2 (g) + H2O (l) HClO (aq) + HCl (aq)
(i) Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid. [1]
(ii) State the formula of the conjugate base of hypochlorous acid. [1]
(iii) Calculate the concentration of H+ (aq) in a HClO (aq) solution with a pH = 3.61. [1]
Answer/Explanation
Ans
i partially dissociates/ionizes «in water»
ii ClO–
iii «[H+] = 10−3.61 =» 2.5 × 10−4 «mol dm−3»
Question
Water is an important substance that is abundant on the Earth’s surface.
Buffer solutions resist small changes in pH. A phosphate buffer can be made by dissolving \({\text{Na}}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}\) and \({\text{N}}{{\text{a}}_{\text{2}}}{\text{HP}}{{\text{O}}_{\text{4}}}\) in water, in which \({\text{Na}}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}\) produces the acidic ion and \({\text{N}}{{\text{a}}_{\text{2}}}{\text{HP}}{{\text{O}}_{\text{4}}}\) produces the conjugate base ion.
A \({\text{0.10 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) ammonia solution is placed in a flask and titrated with a \({\text{0.10 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrochloric acid solution.
(i) State the expression for the ionic product constant of water, \({K_{\text{w}}}\).
(ii) Explain why even a very acidic aqueous solution still has some \({\text{O}}{{\text{H}}^ – }\) ions present in it.
(iii) State and explain the effect of increasing temperature on the value of \({K_{\text{w}}}\) given that the ionization of water is an endothermic process.
(iv) State and explain the effect of increasing temperature on the pH of water.[7]
(i) Deduce the acid and conjugate base ions that make up the phosphate buffer and state the ionic equation that represents the phosphate buffer.
(ii) Describe how the phosphate buffer minimizes the effect of the addition of a
strong base, \({\text{O}}{{\text{H}}^ – }{\text{(aq)}}\), to the buffer. Illustrate your answer with an ionic equation.
(iii) Describe how the phosphate buffer minimizes the effect of the addition of a
strong acid, \({{\text{H}}^ + }{\text{(aq)}}\), to the buffer. Illustrate your answer with an ionic equation.[7]
(i) Explain why the pH of the ammonia solution is less than 13.
(ii) Estimate the pH at the equivalence point for the titration of hydrochloric acid with ammonia and explain your reasoning.
(iii) State the equation for the reaction of ammonia with water and write the \({K_{\text{b}}}\) expression for \({\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\).
(iv) When half the ammonia has been neutralized (the half-equivalence point), the pH of the solution is 9.25. Deduce the relationship between \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) and \({\text{[NH}}_4^ + {\text{]}}\) at the
half-equivalence point.
(v) Determine \({\text{p}}{K_{\text{b}}}\) and \({K_{\text{b}}}\) for ammonia based on the pH at the half-equivalence point.
(vi) Describe the significance of the half-equivalence point in terms of its effectiveness as a buffer.[11]
Answer/Explanation
Markscheme
(i) \({\text{(}}{K_{\text{w}}}{\text{)}} = {\text{[}}{{\text{H}}^ + }{\text{][O}}{{\text{H}}^ – }{\text{] / (}}{K_{\text{w}}}{\text{)}} = {\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^ + }{\text{][O}}{{\text{H}}^ – }{\text{]}}\);
Do not award mark if [ ] omitted or other brackets are used.
(ii) \({\text{[}}{{\text{H}}^ + }{\text{]}}\) increases, \({\text{[O}}{{\text{H}}^ – }{\text{]}}\) decreases but still some present (\({K_{\text{w}}}\) constant) / \({\text{[O}}{{\text{H}}^ – }{\text{]}}\) cannot go to zero as equilibrium present / \({\text{[O}}{{\text{H}}^ – }{\text{]}} = \frac{{{K_{\text{w}}}}}{{{\text{[}}{{\text{H}}^ + }{\text{]}}}}\), thus \({\text{[O}}{{\text{H}}^ – }{\text{]}}\) cannot be zero / OWTTE;
Accept equilibrium present.
(iii) (changing T disturbs equilibrium) forward reaction favoured / equilibrium shifts to the right;
to use up (some of the) heat supplied;
\({{K_{\text{w}}}}\) increases (as both \({{\text{[}}{{\text{H}}^ + }{\text{]}}}\) and \({\text{[O}}{{\text{H}}^ – }{\text{]}}\) increase);
(iv) (as \({{\text{[}}{{\text{H}}^ + }{\text{]}}}\) increases) pH decreases / \({\text{pH}} < 7\);
No mark for more acidic.
inverse relationship between pH and \({\text{[}}{{\text{H}}^ + }{\text{] / pH}} = – \log {\text{[}}{{\text{H}}^ + }{\text{] / pH}} = {\log _{10}}\frac{{\text{1}}}{{{\text{[}}{{\text{H}}^ + }{\text{]}}}}\);
Accept [H3O+] in place of [H+].
(i) Acid: \({{\text{H}}_2}{\text{PO}}_4^ – \);
(Conjugate) base: \({\text{HPO}}_4^{2 – }\);
No mark for NaH2PO4 or Na2HPO4.
\({{\text{H}}_2}{\text{PO}}_4^ – {\text{(aq)}} \rightleftharpoons {{\text{H}}^ + }{\text{(aq)}} + {\text{HPO}}_4^{2 – }{\text{(aq)}}\);
Accept reverse equation or reaction with water.
Ignore state symbols, but equilibrium sign is required.
Accept OH– (ions) react with H+ (ions) to form H2O.
(ii) strong base/\({\text{O}}{{\text{H}}^ – }\) replaced by weak base (\({\text{HPO}}_4^{2 – }\), and effect minimized) / strong base reacts with acid of buffer / equilibrium in (i) shifts in forward direction;
Accept OH– added reacts with H+ to form H2O.
\({\text{O}}{{\text{H}}^ – }{\text{(aq)}} + {{\text{H}}_2}{\text{PO}}_4^ – {\text{(aq)}} \to {{\text{H}}_2}{\text{O(l)}} + {\text{HPO}}_4^{2 – }{\text{(aq)}}\);
Ignore state symbols, accept equilibrium sign.
(iii) strong acid/\({{\text{H}}^ + }\) replaced by weak acid (\({{\text{H}}_2}{\text{PO}}_4^ – \), and effect minimized) / strong acid reacts with base of buffer / equilibrium in (i) shifts in reverse direction;
\({{\text{H}}^ + }{\text{(aq)}} + {\text{HPO}}_4^{2 – }{\text{(aq)}} \to {{\text{H}}_2}{\text{PO}}_4^ – {\text{(aq)}}\);
Accept reaction with H3O+.
Ignore state symbols.
(i) \({\text{N}}{{\text{H}}_3}\) weak(er) base/partial dissociation;
\({\text{[O}}{{\text{H}}^ – }{\text{]}} < {\text{0.1(0)}}/{\text{pOH}} > 1{\text{ (thus pH}} < 13/{\text{pH}} + {\text{pOH}} = 14{\text{)}}\);
(ii) around \({\text{pH}} = 5\);
Accept a value between 4 and 6.
strong acid–weak base titration, (thus acidic) / at equivalence point, \({\text{NH}}_4^ + \) present is acidic / \({\text{NH}}_4^ + \rightleftharpoons {\text{N}}{{\text{H}}_3} + {{\text{H}}^ + }\);
(iii) \({\text{N}}{{\text{H}}_3}{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{NH}}_4^ + {\text{(aq)}} + {\text{O}}{{\text{H}}^ – }{\text{(aq)}}\);
Ignore state symbols, but equilibrium sign required.
\({K_{\text{b}}} = \frac{{{\text{[NH}}_4^ + {\text{][O}}{{\text{H}}^ – }{\text{]}}}}{{{\text{[N}}{{\text{H}}_3}{\text{]}}}}\);
(iv) \({\text{[N}}{{\text{H}}_3}{\text{]}} = {\text{[NH}}_4^ + {\text{]}}\);
(v) \({\text{pOH}} = 14.00 – 9.25 = 4.75\);
\({\text{p}}{K_{\text{b}}}{\text{ (}} = {\text{pOH)}} = 4.75\);
\({K_{\text{b}}} = 1.78 \times {10^{ – 5}}\);
Ignore units.
Award [3] for correct final answer.
(vi) optimum/most effective/highest buffer capacity/50%–50% buffer/equally effective as an acidic buffer and a basic buffer / OWTTE;
Examiners report
This was the second least commonly answered question. With the exception of the part on buffer chemistry where very few appreciated what was happening, the question was reasonably well done.
While many candidates gave the correct \({K_{\text{w}}}\) expression, it was not uncommon to either find the value of the constant or \({K_{\text{w}}} = {K_{\text{a}}} \times {K_{\text{b}}}\) given as the answers. A few included \({\text{[}}{{\text{H}}_{\text{2}}}{\text{O]}}\) in the expression. Candidates recognised that increasing the temperature shifts the equilibrium to the right, but most did not explain why, namely to use up some of the heat supplied.
Candidates generally concluded that formation of more \({{\text{H}}^ + }\) and \({\text{O}}{{\text{H}}^ – }\) ions gives a higher value of \({K_{\text{w}}}\). A significant number of candidates were able to state the effect of increasing temperature on the pH of water (it decreases) but failed to explain why. Some simply incorrectly stated that the pH would not change.
Many candidates gave the wrong formulas for the acid and the conjugate base ions of the buffer or offered \({\text{Na}}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}\) and \({\text{N}}{{\text{a}}_{\text{2}}}{\text{HP}}{{\text{O}}_{\text{4}}}\) as the answers. Some candidates gave good answers about the effect of adding a small amount of a strong acid or a strong base, but they could not write correct equations to show these two effects.
Nearly all candidates correctly said that the ammonia solution is a weak base because of partial dissociation and \({\text{[O}}{{\text{H}}^ – }{\text{]}}\) would be less than 0.1 to give a pH less than 13. The majority of candidates correctly identified the pH around 4 – 6 because it is a titration between a strong acid and a weak base. When writing the equation for the reaction of ammonia and water some candidates did not write the equilibrium sign. The \({K_{\text{b}}}\) expression was correct in most cases. However, many did not recognise that at the half-equivalence point both the base and the conjugate acid concentrations are equal. The \({\text{p}}{K_{\text{b}}}\) and \({K_{\text{b}}}\) were correctly calculated from the pH of the solution by many candidates. However, most failed to realize that at the half-equivalence point the capacity of the buffer is optimum.
Question
Iron tablets are often prescribed to patients. The iron in the tablets is commonly present as iron(II) sulfate, FeSO4.
Following the experiment, the students proposed the following hypothesis:
“Since sulfuric acid is a strong acid, two other strong acids such as nitric acid, HNO3(aq) or hydrochloric acid, HCl(aq), could also be used in this experiment”.
Suggest one problem with this hypothesis.[1]
The students also explored the role of sulfuric acid in everyday processes and found that sulfuric acid present in acid rain can damage buildings made of limestone. Predict the balanced chemical equation for the reaction between limestone and sulfuric acid, including state symbols.[2]
Answer/Explanation
Markscheme
\({\text{NO}}_{\text{3}}^ – \) and \({\text{C}}{{\text{l}}^ – }\) anions may also react with \({\text{KMn}}{{\text{O}}_{\text{4}}}\) / \({\text{HN}}{{\text{O}}_{\text{3}}}\) is an oxidizing agent / (HCl will not work as) \({\text{C}}{{\text{l}}^ – }\) reacts with \({\text{MnO}}_{\text{4}}^ – \) (to form \({\text{C}}{{\text{l}}_{\text{2}}}\)) / HCl oxidized / OWTTE;
For HCl, allow correctly balanced chemical equation:
2MnO4–+10Cl–+16H+\( \to \)2Mn2++5Cl2+8H2O
Accept NO3– and Cl– may react with KMnO4/Fe2+.
\({\text{CaC}}{{\text{O}}_3}{\text{(s)}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{(aq)}} \to {\text{CaS}}{{\text{O}}_4}{\text{(s)}} + {{\text{H}}_2}{\text{O(l)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}\)
correct chemical equation;
correct state symbols;
Allow CaSO4(aq) instead of CaSO4(s).
M2 can only be scored if M1 is correct.
Award [1max] if H2CO3(aq) is given instead of H2O(l) + CO2(g).
Examiners report
Most candidates were able to state the function of iron in the body but unable to explain why a reading of the top meniscus is taken for the KMnO4 titration, which suggests limited practical experience. Some candidates calculated the moles of Fe using the mass of tablets rather than using mole ratio and titre results. most candidates were able to define ‘reduction’ and determine the oxidation number, although some were penalized for incorrect notation. Many did not know how to prevent the formation of MnO2 precipitate or why HCl or HNO3 are not used in this titration.