Question

In: Chemistry

Part A: What is the pH of a buffer prepared by adding 0.405 mol of the...

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.

Solutions

Expert Solution

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


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