Question

A 1.00 L solution contains 0.352 M HA (K_a = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA.

What is the pH of this solution?
What is the pH if 0.310 L of 0.138 M KOH is added to the solution be?
What is the pH if 0.620 L of 0.138 M KOH is added to the solution be?

Answer 1

Given solution contain weak acid HA and its salt NaA. Therefore this solution is a buffer solution. We can use Henderson's equation to calculate pH of buffer solution.

We have Henderson's equation, pH = pKa + log [ Salt ] / [ acid ]

$\therefore$ pH = - log Ka + log [ NaA ] / [ HA ]

$\therefore$ pH = - log 4.66 $\times$ 10 -06 + log 0.758 / 0.352

pH = 5.332 + log 0.758 / 0.352

pH = 5.332 + 0.333

pH = 5.665

Question 2

Calculation of moles of NaA, HA and KOH.

We have relation, Molarity = No. of moles of solute / Volume of solution in L

No. of moles of solute = Molarity $\times$ volume of solution in L

No. of moles of HA = 0.352 mol / L $\times$ 1.00 L = 0.352 mol

No. of moles of NaA = 0.758 mol / L $\times$ 1.00 L = 0.758 mol

No. of moles of KOH = 0.138 mol / L $\times$ 0.310 L = 0.04278 mol

Consider reaction of KOH with buffer solution.

HA + OH -  $\rightarrow$ A - + H2O

Let's use ICE table.

Moles	HA	OH -	A -
I	0.352	0.04278	0.758
C	-0.04278	-0.04278	+0.04278
E	0.30922	0.00000	0.80078

Volume of buffer solution after addition of KOH = 1.00 L + 0.310 L = 1.310 L

We have equation, pH = - log Ka + log [ NaA ] / [ HA ]

$\therefore$ pH = - log 4.66 $\times$ 10 -06 + log ( 0.80078 mol / 1.310 L ) / ( 0.30922 mol / 1.310 L )

pH = 5.332 + 0.4132

pH =5.745

Question 3

Calculation of moles of NaA, HA and KOH.

We have relation, Molarity = No. of moles of solute / Volume of solution in L

No. of moles of solute = Molarity $\times$ volume of solution in L

No. of moles of HA = 0.352 mol / L $\times$ 1.00 L = 0.352 mol

No. of moles of NaA = 0.758 mol / L $\times$ 1.00 L = 0.758 mol

No. of moles of KOH = 0.138 mol / L $\times$ 0.620 L = 0.08556 mol

Consider reaction of KOH with buffer solution.

HA + OH -  $\rightarrow$ A - + H2O

Let's use ICE table.

Moles	HA	OH -	A -
I	0.352	0.08556	0.758
C	-0.08556	-0.08556	+0.08556
E	0.26644	0.00000	0.84356

Volume of buffer solution after addition of KOH = 1.00 L + 0.620 L = 1.620 L

We have equation, pH = - log Ka + log [ NaA ] / [ HA ]

$\therefore$ pH = - log 4.66 $\times$ 10 -06 + log ( 0.84356 mol / 1.620 L ) / ( 0.26644 mol / 1.620 L )

pH = 5.332 + 0.5005

pH =5.832