Question

Part A: What is the pH of a buffer prepared by adding 0.506 mol of the weak acid HA to 0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. Part B: 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)

Ka = 5.66*10^-7

pKa = - log (Ka)

= - log(5.66*10^-7)

= 6.247

use:

pH = pKa + log {[conjugate base]/[acid]}

= 6.247+ log {0.406/0.506}

= 6.152

Answer: 6.152

B)

mol of NaOH added = 0.195 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.406 mol

mol of HA = 0.506 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.406 + 0.195) mol

mol of A- = 0.601 mol

mol of HA = mol present initially - mol added

mol of HA = (0.506 - 0.195) mol

mol of HA = 0.311 mol

Ka = 5.66*10^-7

pKa = - log (Ka)

= - log(5.66*10^-7)

= 6.247

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 6.247+ log {0.601/0.311}

= 6.533

Answer: 6.533