Question

500.0 mL of 0.140 M NaOH is added to 565 mL of 0.250 M weak acid (Ka = 8.39 × 10-5). What is the pH of the resulting buffer?

HA(aq)+OH^-(aq)=H2O(l)+ A^-(aq)

pH=?

Answer 1

no. of moles of NaOH = 0.14 x 500 / 1000 = 0.07mol

New conc of NaOH after addition to acid = (0.07 / 1065) x 1000 = 0.0657 M

No. of moles of acid before additon of base = 0.250 x 565 / 1000 = 0.14125 mol

Conc. of acid after addition of base = (0.14125/ 1065) x 1000 = 0.1326M

Since base i s strong it undergoes complete dissociation to form 0.0657 M of OH^- (hydroxide ions)

But acid is weak it undergoes partial ionisation, and according Ostwald's equation,

[H^+] = ( Ka x C)^1/2 = (8.39 x 10 ^-5 x 0.1326)^1/2 = 3.3354 x 10^-3 M

           H+       +          OH- ------>          H2O

initial 0.003354 M             0.0657 M

after   0                             0.06235              0.003354 M

Therefore, pH = 14 + Log [OH-] = 14 + Log (0.06235) = 14 - 1.2052 = 12.79