A.What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to

Question

A.What is the pH of a buffer prepared by adding 0.708 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.

I finished part A but you use it in parts B and C. The answer to part A is 6.102

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.

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

Explanation / Answer

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 = 0.708 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 = (0.708 + 0.15) mol

mol of HA = 0.858 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 {0.357/0.858}

= 5.866

Answer: 5.866

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 = 0.708 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 = (0.708 - 0.195) mol

mol of HA = 0.513 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 {0.702/0.513}

= 6.383

Answer: 6.383

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