Question

Part A: What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.608 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)

Ka = 5.66*10^-7

pKa = - log (Ka)

= - log(5.66*10^-7)

= 6.247

use:

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

pH = pKa + log {mol(conjugate base)/mol(acid)}

= 6.247+ log {0.608/0.405}

= 6.424

Answer: 6.42

B)

mol of HCl added = 0.15 mol

A- will react with H+ to form HA

Before Reaction:

mol of A- = 0.608 mol

mol of HA = 0.405 mol

after reaction,

mol of A- = mol present initially - mol added

mol of A- = (0.608 - 0.15) mol

mol of A- = 0.458 mol

mol of HA = mol present initially + mol added

mol of HA = (0.405 + 0.15) mol

mol of HA = 0.555 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 {0.458/0.555}

= 6.164

Answer: 6.16

C)

mol of LiOH added = 0.195 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.608 mol

mol of HA = 0.405 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.608 + 0.195) mol

mol of A- = 0.803 mol

mol of HA = mol present initially - mol added

mol of HA = (0.405 - 0.195) mol

mol of HA = 0.21 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 {0.803/0.21}

= 6.83

Answer: 6.83