IB Chemistry HL 3.1 Proton Transfer Reactions Question Bank
Practise IB Chemistry HL 3.1 with SL and HL questions on proton transfer, pH, Ka/Kb, buffers, indicators and titration analysis.
- Syllabus
- First assessment 2025
- Course
- Chemistry HL
- Level
- HL
Practise IB Chemistry HL 3.1 with SL and HL questions on proton transfer, pH, Ka/Kb, buffers, indicators and titration analysis.
Bleaches in which chlorine is the active ingredient are the most common, although some environmental groups have concerns about their use.
In aqueous chlorine the equilibrium below produces chloric(I) acid (hypochlorous acid), HOCl , the active bleach.
Chloric(I) acid is a weak acid, but hydrochloric acid is a strong acid. Outline how this is indicated in the equation above.
chloric(I) acid (shown as) a molecule/molecular, but hydrochloric acid (shown as being) split into ions / OWTTE;
Marking guidance:
Accept "chloric(I) acid is partially dissociated and hydrochloric acid is fully dissociated".
Reference needed to both acids for mark.
State a balanced equation for the reaction of chloric(I) acid with water.
HOCl(aq)⇌H+(aq)+ClO−(aq)/HOCl(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq);
Equilibrium sign required for the mark.
Marking guidance:
Ignore state symbols.
Partial neutralization of chloric(I) acid creates a buffer solution. Given that the pKa of chloric(I) acid is 7.53 , determine the pH of a solution that has [HOCl]=0.100 moldm−3 and [ClO−]=0.0500 moldm−3.
Ka=10−7.53=2.95×10−8(moldm−3);
Ka=[HOCl][H+][ClO−]=(0.1)[H+](0.05)≈2[H+]=2.95×10−8(moldm−3);[H+]=2×2.95×10−8=5.9×10−8(moldm−3);
pH=−log(5.9×10−8)=7.23;
Marking guidance:
Accept other methods of carrying out the calculation.
Award [4] for correct final answer.
Describe, using HIn to represent the indicator in its acid form, why an indicator changes colour when excess alkali is added.
HIn⇌H++In−;
Marking guidance:
Do not accept equation without equilibrium arrow.
(weak acid in which the) acid/HIn and conjugate base/In- have different colours / OWTTE;
excess alkali shifts the equilibrium to the RHS/towards the conjugate base;
Nitrous acid, HNO2, is a weak acid which can be used to make acidic buffers.
A 1.00 moldm−3 solution of nitrous acid was prepared. This solution contains two conjugate acid-base pairs.
State the formulas of the conjugate acid and conjugate base in each pair.
Conjugate acid:
Conjugate base:
Conjugate acid:
Conjugate base:
(a)
(i)
conjugate acid H3O+«(aq)» AND conjugate base H2O « (l)»
conjugate acid HNO2 «(aq)» AND conjugate base NO2−«(aq)»
This 1.00 moldm−3 solution of nitrous acid was used to prepare a buffer with pH 3.00 .
Calculate the concentration of the conjugate base of nitrous acid required to make this buffer. The pKa of nitrous acid is 3.25.
Concentration of conjugate base:
(a)
(ii)
Ka<10−3.25>=5.62×10−4
« [H+]=10−3.00 » =0.001 «mol dm −3 »
« [A−]=(5.62×10−4×1.00)/0.001»=0.562 «mol dm−3 »
Alternative solution:
pH=pKa+log10[ salt ]/[ acid ] « log10[ salt ]/[ acid ] 》 =−0.25
OR
«[salt]/[acid]» = 0.562、
《 [A−]=0.562/1.00»=0.562 «mol dm−3 » ↓
Award[3]for correct final answer.
Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula PH3.
Ammonia acts as a weak Brønsted-Lowry base when dissolved in water.
Outline what is meant by the terms "weak" and "Brønsted-Lowry base".
Weak:
Brønsted-Lowry base:
Weak: only partially dissociated/ionized «in dilute aqueous solution»
Brønsted-Lowry base: an acceptor of protons/ H^+/hydrogen ions
Marking guidance:
Accept reaction with water is reversible/an equilibrium.
Accept "water is partially dissociated «by the weak base»".
Phosphine is usually prepared by heating white phosphorus, one of the allotropes of phosphorus, with concentrated aqueous sodium hydroxide. The equation for the reaction is:
The ion H2PO2−is amphiprotic. Outline what is meant by amphiprotic, giving the formulas of both species it is converted to when it behaves in this manner.
can act as both a «Brønsted-Lowry» acid and a «Brønsted-Lowry» base OR can accept and/or donate a hydrogen ion/proton/ H+HPO22− AND H3PO2