The following pictures represent solutions at various stages in the titration of

Question

The following pictures represent solutions at various stages in the titration of a weak acid H2A with aqueous NaOH. (Na^+ ions and water molecules have been omitted for clarity.) To which of the following stages do solutions 1- 4 correspond? Halfway to the first equivalence point At the first equivalence point Halfway between the first and second equivalence points Beyond the second equivalence point Which solution has the highest pH? Which has the lowest pH? Is the pH greater than, equal to. or less than 7 after the neutralization of each of the following pairs of acids and bases? Hl and NaOH HOCI and Ba(OH)^2 HNO_3 and aniline (C6H5NH2) Benzoic acid (C6H5CO2H) and KOH Does the pH increase, decrease, or remain the same on the addition of each of the following? NH4NO3 to an NH3 solution Na2CO3 to an NaHCO3 solution NaClO4 to an NaOH solution Calculate the pH of a buffer solution that is 0.20 M in HCN and 0.12 M in NaCN. Will the pH change if the solution is diluted by a factor of 2? Explain. Calculate the pH and the concentrations of all species present (H3O+. F-. HF, Cl-, and OH-) in a solution that contains 0.10 M HF (Ka = 3.5 times 10-4) and 0.10 M HCl.

Explanation / Answer

(1)

(a) the figure showing,

halfway to the first equivalence point = (2)

at the equivalence point = (4)

halfway between first and second equivalence point = (1)

beyond second equivalence point = (3)

(b) highest pH figure = (3)

lowest pH figure = (2)

2. pH at the neutralization point for,

HI + NaOH would be pH = 7

HOCl + Ba(OH)2 would be pH = 7

HNO3 + aniline would be pH < 7

Benzoic acid + KOH would be pH > 7

3. change in pH upon addition of,

NH4NO3 to NH3 solution = pH would decrease

Na2CO3 to NaHCO3 solution = pH would increase

NaClO4 to NaOH solution = pH would remain the same

4. pH = pKa + log(base/acid)

         = 9.2 + log(0.12/0.2)

         = 8.98

with dilution by a factor of 2, the pH of solution would remain unchanged

5. concentration of species in solution,

[H3O+] = 0.1 M

[Cl-] = 0.1 M

[OH-] = 1 x 10^-13 M

Ka = [H3O+][F-]/[HF]

3.5 x 10^-4 = x^2/0.1

x = [F-] = 5.91 x 10^-3 M

[HF] = 0.1 - 5.91 x 10^-3 = 0.094 M

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