Question

a) What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Answer: pH=6.181

b)What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

please help me with part b, as I already solved for part a.

Answer 1

b)What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

HCl is a strong acid, will donate H+ to the solution, in which HA and A- are present

A- + H+ = HA; so A- decreases, HA increases

A- left = 0.608 - 0.150 = 0.458

HA formed = 0.708+0.150 = 0.858

Now, substitute in pH equation

pH = pKa + log(A-/HA)

pKa = -log(5.66*10^-7) = 6.24718

pH = 6.24718 + log(0.458/0.858)

pH = 5.97455

which is more acidic, as expected