Question

A-

What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Express the pH numerically to three decimal places.

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.

Express the pH numerically to three decimal places.

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.

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 {0.305/0.405}

= 6.124

Answer: 6.12

B)

mol of HCl added = 0.15 mol

A- will react with H+ to form HA

Before Reaction:

mol of A- = 0.305 mol

mol of HA = 0.405 mol

after reaction,

mol of A- = mol present initially - mol added

mol of A- = (0.305 - 0.15) mol

mol of A- = 0.155 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 {0.155/0.555}

= 5.693

Answer: 5.69

C)

mol of NaOH added = 0.195 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.305 mol

mol of HA = 0.405 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.305 + 0.195) mol

mol of A- = 0.5 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 {0.5/0.21}

= 6.624

Answer: 6.62