Questions on pH curves
1. Calculate the pH of the resulting solution at 25 oC when 49.1 cm3 of 0.100 mol dm-3 sodium hydroxide
solution has been added to 50.0 cm3 of (a) 0.100 mol dm-3 hydrochloric acid solution and (b) 0.100 mol
dm-3 ethanoic acid solution. pKa of ethanoic acid = 4.76 at 25 oC.
2. Calculate the pH of the resulting solution at 25 oC when 50.1 cm3 of 0.100 mol dm-3 sodium hydroxide
solution has been added to 50.0 cm3 of (a) 0.100 mol dm-3 hydrochloric acid solution and (b) 0.100 mol
dm-3 ethanoic acid solution. pKa of ethanoic acid = 4.76 at 25 oC.
3. Deduce the pH of the resulting solution obtained at 25 oC when 25.0 cm3 of 0.100 mol dm-3 potassium
hydroxide solution has been added to 50.0 cm3 of 0.100 mol dm-3 methanoic acid solution, HCOOH(aq).
pKa of methanoic acid = 3.75 at 25 oC.
4. Sketch the graph of pH against volume of alkali added that would be obtained when 50.0 cm3 of 0.100
mol dm-3 hydrochloric acid is titrated with 50.0 cm3 of 0.100 mol dm-3 ammonia solution.
5. Acid-base titration is a technique used to find the concentration of an acid or a base where the
concentration is unknown.
(a) Explain why titrating a weak acid with a weak base will not give a reliable result.
(b) Explain why titrating a weak acid with a weak base is not necessary in order to find an unknown
concentration.
6. Explain the difference between equivalence point and end point when applied to acid-base titrations.
7. Explain why the equivalence point for the titration of ethanoic acid with sodium hydroxide does not
occur at pH 7.
8. An aqueous solution of sodium carbonate is alkaline. Give the equations for the reactions taking place
when solid sodium carbonate is dissolved in water to explain this fact.
9. Cyanide ions are very poisonous if swallowed or breathed in as hydrogen cyanide gas. Explain why it is
dangerous to dissolve solid sodium cyanide crystals, NaCN(s), in hot water (even though you have no
intention of drinking the solution).
10. From Section 22 in the IB data booklet it can be seen that methyl red has a pKa value of 5.1 and is red
in acid solution and yellow in alkaline solution.
(a) If methyl red is represented as HIn(aq) and it dissociates to H+(aq) and In–(aq) in water then identify
the colours of HIn(aq) and In–(aq).
(b) Explain why the pH range of methyl red is listed as 4.4 – 6.2.
(c) Explain whether methyl red would be a suitable indicator to use for the titration of ethanoi acid
solution (pKa = 4.76) with sodium hydroxide solution.
11. A student used phenolphthalein as the indicator when titrating 25.00 cm3 of 1.00 x 10-2 mol dm3
hydrochloric acid with 1.00 x 10-2mol dm-3 aqueous ammonia solution. Use information given in Section
22 of the data booklet to explain what will be observed when one drop of the 1.00 x 10-2 mol dm3
ammonia solution is added after the equivalence point has been reached.
12. The following shows a screen-shot of an acid-base simulation when 25.0 cm3 of 0.1 mol dm3
chloroethanoic acid is titrated with 0.1 mol dm-3 sodium hydroxide solution before it is run:
(a) Sketch the graph obtained when bromophenol blue is chosen as the indicator.
(b) Select two indicators from the list that would be suitable to use for this titration.
Answers
1. (a) After 49.1 cm3 of NaOH(aq) has been added there will be 0.9 cm3 of 0.100 mol dm-3 HCl(aq)
remaining in 99.1 cm3 (say 100 cm3) of solution.
Amount of HCl(aq) = 0.9 x 0.100 x 1.0/1000 = 9 x 10-5 mol
HCl(aq)] = [H+(aq)] = 9 x 10-5 x 1000/100 = 9 x 10-4 mol dm-3
pH = – log10 9 x 10-4 = 3.05
(b) After 49.1 cm3 of NaOH(aq) has been added:
[CH3COOH(aq)] = 0.9 x 0.1 x 100 = 9 x 10-4 mol dm-3
[CH3COO–(aq)] = 49.1 x 0.100 x 1/1000 x 1000/100 = 4.91 x 10-2 mol dm-3
Ka = [H+(aq)] x [CH3COO–(aq)] / [CH3COOH(aq)] = 10-4.76 = [H+(aq)] x 4.91 x 10-2 / 9 x 10-4
[H+(aq)] = 9 x 10-4 x 10-4.76 / 4.91 x 10-2 = 3.19 x 10-7 mol dm-3
pH = – log103.19 x 10-7 = 6.50
2.(a) After 50.1 cm3 of NaOH(aq) has been added there will be 0.1 cm3 of 0.100 mol dm-3 NaOH(aq)
present in 100.1 cm3 (say 100 cm3) of solution.
Amount of NaOH(aq) = 0.1 x 0.100 x 1.0/1000 = 1 x 10-5 mol
NaOH(aq)] = [OH–(aq)] = 1 x 10-5 x 1000/100 = 1 x 10-4 mol dm-3
pOH = – log10 1 x 10-4 = 4.00 so pH = 10.0
3. Since this is the half-equivalence point (when [HCOOH(aq)] = [HCOO–(aq)]) then pH = pKa = 3.75.
4.
6. Equivalence point is when equimolar amounts of H+(aq and OH–(aq) have reacted. End point is when
the acid-base indicator just changes colour.
5. (a) There is no sharp change in pH as the equivalence point is approached, reached and passed.
(b) The unknown solution is always titrated against a solution of known concentration. If a weak acid is
being titrated then a strong base of known concentration can be used and vice-versa.
6. Equivalence point is when equimolar amounts of H+(aq and OH–(aq) have reacted. End point is when
the acid-base indicator just changes colour.
7. At the equivalence point the salt formed will be sodium ethanoate. This salt is completely dissociated
into its ions:
NaCH3COO(aq) → Na+(aq) + CH3COO–(aq). Also present will be water which is very slightly dissociated
into its ions:
H2O(l) ⇌ H+(aq) + OH–(aq). The ethanoate ions and the hydrogen ions will combine to form the weak,
largely undissociated, ethanoic acid, CH3COOH(aq). The hydroxide ions will remain in solution as sodium
hydroxide is a strong base and therefore completely dissociated into its ions in solution. The solution will
therefore have a pH > 7.
8. Na2CO3(s) → 2Na+(aq) + CO32–(aq)
CO32–(aq) + H2O(l) ⇌ HCO3–(aq) + OH–(aq) (although this reaction is reversible the equilibrium lies very
much on the products side as HCO3–(aq) is a weak acid)
9. Sodium cyanide is the salt of a weak acid, HCN, and a strong base, NaOH. When it is dissolved in water
it dissociates completely to form sodium ions and cyanide ions. The cyanide ions react with water to
form hydrogen cyanide.
CN–(aq) + H2O(l) ⇌ HCN(aq) + OH–(aq)
Heating the solution helps to dissolve the salt more quickly but gases are less soluble in hot water than
they are in cold water so the hydrogen cyanide is evolved as a gas and can be breathed in.
10.(a) In acid solution when[H+(aq)] is high [HIn(aq)] >> [In–(aq)] so HIn(aq) is red and In–(aq) is yellow.
(b) Ka = [H+(aq)] x [In–(aq)] / [HIn(aq)]. When [In–(aq)] = [HIn(aq)] Ka = [H+(aq)] and pH = pKa
For there to be a colour change one of the species must dominate so the pH range will be around the
pKa value. Since the pKa value is 5.1 the pH range of the indicator lies just either side of this value, i.e. 4.4
– 6.2.
(c) It will not be a suitable indicator. At the equivalence point the salt formed will be sodium ethanoate
which is a basic salt with a pH greater than 7.0 and this does not lie in the 4.4 – 6.2 range of methyl red.
11. From Section 22 in the data booklet the pH range for phenolphthalein is 8.3 – 10. At the equivalence
point the salt formed is ammonium chloride which is an acidic salt so the solution will have a pH value
less than 7. The solution will be colourless as this is the colour of the indicator in acid solution. When
one more drop of the weak alkali solution is added the pH of the solution will still not reach 8.3 so the
solution will remain colourless.
12. (a)
(b) Any two from bromothymol blue, phenol red, thymol blue or phenolphthalein.