Question

A 1.00 L solution contains 0.326 M HA (Kg = 5.09 · 10-7. and 0.796 M NaA. What is the pH of this solution? What is the pH if 0.325 L of 0.219 M KOH is added to the solution be? What is the pH if 0.650 L of 0.219 M KOH is added to the solution be?

Answer 1

1)

Ka = 5.09*10^-7

pKa = - log (Ka)

= - log(5.09*10^-7)

= 6.293

use:

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

= 6.293+ log {0.796/0.326}

= 6.681

Answer: 6.68

2)

mol of KOH added = 0.219M *0.325 L = 0.071175 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.796 M *1.0 L

mol of A- = 0.796 mol

mol of HA = 0.326 M *1.0 L

mol of HA = 0.326 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.796 + 0.071175) mol

mol of A- = 0.8672 mol

mol of HA = mol present initially - mol added

mol of HA = (0.326 - 0.071175) mol

mol of HA = 0.2548 mol

Ka = 5.09*10^-7

pKa = - log (Ka)

= - log(5.09*10^-7)

= 6.293

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

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

= 6.293+ log {0.8672/0.2548}

= 6.825

Answer: 6.82

3)

mol of KOH added = 0.219M *0.65 L = 0.14235 mol

HA will react with OH- to form A-

Before Reaction:

mol of A- = 0.796 M *1.0 L

mol of A- = 0.796 mol

mol of HA = 0.326 M *1.0 L

mol of HA = 0.326 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.796 + 0.14235) mol

mol of A- = 0.9384 mol

mol of HA = mol present initially - mol added

mol of HA = (0.326 - 0.14235) mol

mol of HA = 0.1837 mol

Ka = 5.09*10^-7

pKa = - log (Ka)

= - log(5.09*10^-7)

= 6.293

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

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

= 6.293+ log {0.9384/0.1837}

= 7.002

Answer: 7.00