Question

A buffer that contains 0.27 M of a base, B and 0.44 M of its conjugate acid BH⁺, has a pH of 9.05. What is the pH after 0.03 mol of NaOH are added to 0.79 L of the solution?

Answer 1

The buffer is

BH+ + NaOH ------------------------------------> B + H2O

0.44x0.79=0.3476 0 0.27x0.79=0.2133 ---- initial moles

------------- 0.03 ------------ ---- change

0.3176 0 0.2433 --- equilibrium

Now

pH of a buffer is given by Henderen equation as

pOH = pKb + log [conjugate acid]/[base]

and pOH = 14-pH = 14-9.05 = 4.95

4.95 = pKb + log 0.44/0.27 initially

Thus pKb = 4.74

Now after adding NaOH

pOH = 4.74 + log 0.3176/0.2433

=4.856

and pH = 14-pOH = 14-4.856 =9.14