Question

A 1.00 L solution contains 0.311 M HA (K3 = 3.04 - 10 ) and 0.796 M NaA. What is the pH of this solution? What is the pH if 0.318 L of 0.158 M KOH is added to the solution be? What is the pH if 0.636 L of 0.158 M KOH is added to the solution be?

Answer 1

1)

Ka = 3.04*10^-7

pKa = - log (Ka)

= - log(3.04*10^-7)

= 6.517

use:

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

= 6.517+ log {0.796/0.311}

= 6.925

Answer: 6.925

2)

mol of KOH added = 0.158M *0.318 L = 0.05024 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 = 0.311 M *1.0 L

mol of HA = 0.311 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.796 + 0.05024) mol

mol of A- = 0.8462 mol

mol of HA = mol present initially - mol added

mol of HA = (0.311 - 0.05024) mol

mol of HA = 0.2608 mol

Ka = 3.04*10^-7

pKa = - log (Ka)

= - log(3.04*10^-7)

= 6.517

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

use:

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

= 6.517+ log {0.8462/0.2608}

= 7.028

Answer: 7.028

3)

mol of KOH added = 0.158M *0.636 L = 0.1005 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 = 0.311 M *1.0 L

mol of HA = 0.311 mol

after reaction,

mol of A- = mol present initially + mol added

mol of A- = (0.796 + 0.1005) mol

mol of A- = 0.8965 mol

mol of HA = mol present initially - mol added

mol of HA = (0.311 - 0.1005) mol

mol of HA = 0.2105 mol

Ka = 3.04*10^-7

pKa = - log (Ka)

= - log(3.04*10^-7)

= 6.517

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

use:

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

= 6.517+ log {0.8965/0.2105}

= 7.146

Answer: 7.146