Question

Calculate the pH after 0.017 mole of HCl is added to 1.00 L of each of the four solutions. (Assume that all solutions are at 25°C.) (a) 0.128 M butanoic acid (HC4H7O2, Ka = 1.5 ✕ 10−5) (b) 0.128 M sodium butanoate (NaC4H7O2) (c) pure H2O (d) 0.128 M HC4H7O2 and 0.128 M NaC4H7O2

Answer 1

(a): concentration of HCl in 1L solution = 0.017 mol / 1L = 0.017 M

HCl is a strong acid and it is completely dissociated to form H+(aq) ion. Hence concentration of H+(aq) due to HCl = 0.017 M.

Dissociation of butanoic acid also produces H+ ion that can be calculated as

------------------- C4H8O2 <---> C₄H₇O₂^-(aq) + H+(aq) ; Ka = 1.5*10^-5

Init.Conc(M):0.128, ------------ 0 ------------------ 0.017

Eqm.conc(M):0.128(1 - x),--- 0.128x, --------(0.017 + 0.128x)

Ka = 1.5*10^-5​ = (0.017 + 0.128x) * (0.128x) / 0.128(1 - x)

(1 - x) is nearly equals to 1 =>

=> x = 8.77*10^-4 M

Hence total H+ concentration, [H+(aq)] = (0.017 + 0.128x) = (0.017 + 0.128*8.77*10^-4)M = 0.01711 M

=> pH = - log [H+(aq)] = - log(0.01711 M) = 1.767 (answer)

(b): 0.017 mol HCl will react with 0.017 mol NaC4H7O2 to form 0.017 mol HC4H7O2.

hence [NaC4H7O2] = 0.128 - 0.017 = 0.111 M

[HC4H7O2] = 0.017 M

Now this solution will act as a buffer solution whise pH can be calculated as

pH = pKa + log[NaC4H7O2] / [HC4H7O2]

=> pH = - log(1.5*10^-5 ) + log(0.111M / 0.017) = 5.64 (answer)

(c) When we add 0.017 mol HCl to 1 L water, concentration of HCl, [HCl] = 0.017 mol / 1L = 0.017 M

Hence [H+(aq)] = [HCl] = 0.017 M

=> pH = - log [H+(aq)] = - log(0.017 M) = 1.770 (answer)

(d):  0.128 M HC4H7O2 and 0.128 M NaC4H7O2 act as buffer solution.

When 0.017 M HCl is added to the buffer solution, it will react with 0.017 mol of NaC4H7O2 to form 0.017 mol HC4H7O2.

Hence concentration of NaC4H7O2 after reaction, [NaC4H7O2] = 0.128 - 0.017 = 0.111 M

concentration of HC4H7O2 after reaction, [HC4H7O2] = 0.128 + 0.017 = 0.145 M

Applying Hendersen equation,

pH = pKa + log[NaC4H7O2] / [HC4H7O2]

=> pH = - log(1.5*10^-5 ) + log(0.111M / 0.145) = 4..71 (answer)