Question

Solve for pH:

A) What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7 .

Express the pH numerically to three decimal places.

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.

Express the pH numerically to three decimal places.

C) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Express the pH numerically to three decimal places.

Answer 1

a)

pH = pKa + log(A-/HA)

mol of HA= 0.809

mol of A- = 0.305

pKa = -log(Ka)= -log(5.66*10^-7) = 6.25

substitute in

pH = pKa + log(A-/HA)

pH = 6.25+ log(0.305/0.809)

pH = 5.826

b)

pH after adding 0.15 mol of H+

mol of HA= 0.809 +0.15 = 0.959

mol of A- = 0.305 - 0.15 = 0.155

pH = pKa + log(A-/HA)

pH = 6.25+ log(0.155/0.959)

pH = 5.4585

c)

after

pH after adding 0.195 mol of Oh-

mol of HA= 0.809 -0.195  = 0.614

mol of A- = 0.305 + 0.195 = 0.5

pH = pKa + log(A-/HA)

pH = 6.25+ log(0.5/0.614)

pH = 6.16