Question

PART A: What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.507 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 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

PART 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

Answer 1

A)

[HA] = mol of HA / volume in L

= 0.809 mol / 2.0 L

= 0.4045 M

[NaA] = mol of NaA / volume in L

= 0.507 mol / 2.0 L

= 0.2535 M

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 {0.2535/0.4045}

= 6.044

Answer: 6.04

B)

mol of HCl added = 0.15 mol

A- will react with H+ to form HA

Before Reaction:

mol of A- = 0.507 mol

mol of HA = 0.809 mol

after reaction,

mol of A- = mol present initially - mol added

mol of A- = (0.507 - 0.15) mol

mol of A- = 0.357 mol

mol of HA = mol present initially + mol added

mol of HA = (0.809 + 0.15) mol

mol of HA = 0.959 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 {0.357/0.959}

= 5.818

Answer: 5.82

C)

mol of NaOH added = 0.195 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.507 mol

mol of HA = 0.809 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.507 + 0.195) mol

mol of A- = 0.702 mol

mol of HA = mol present initially - mol added

mol of HA = (0.809 - 0.195) mol

mol of HA = 0.614 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 {0.702/0.614}

= 6.305

Answer: 6.31