Question

A 300mL buffer is prepared in which [HCO3-] = 0.2M and [CO3^2-]=0.075M.

a) What is the pH of this buffer (Ka HCO3-=4.7x10^-11)?

b) What is the pH of the buffer if 0.0100 moles of HCl are added?

c) What is the pH of the buffer if 0.0200 moles of NaOH are added?

Answer 1

a) pH of buffer solution is calculated by using Henderson's equation, pH = - log Ka + log [ Conjugate base ] / [ acid ]

pH = - log Ka + log  [CO ₃ ^2-] / [HCO ₃^-]

We have, [HCO ₃^-] = 0.2 M , [CO ₃ ^2-]= 0.075 M and Ka of HCO ₃^- = 4.7 10 ^-11.

Substituting these values in Henderson's equation, we get pH of buffer solution.

pH = - log ( 4.7 10 ^-11 ) + log 0.075 / 0.2

pH = 10.33 + log 0.075 / 0.2

pH = 10.33 - 0.426

pH = 9.90

b) First calculate no. of moles of HCO ₃^- and no. of moles of CO ₃ ^2-.

We know that , Molarity = no. of moles of solute / volume of solution in L

no. of moles of solute = Molarity   volume of solution in L

No. of moles of HCO ₃^- = 0.2 mol / L 0.300 L = 0.06 mol

No. of moles of CO ₃ ^2- = 0.075 mol / L 0.300 L = 0.0225 mol

Now, consider reaction of HCl with buffer solution.

CO ₃ ^2- + H ⁺ HCO ₃^-

Let's Use ICE table.

moles	CO 3 2- + H + HCO 3-
I	0.0225	0.0100	0.0600
C	-0.0100	-0.0100	+0.0100
E	0.0125	0.0000	0.0700

pH of buffer solution after addition of 0.0100 moles of HCl = 10.33 + log 0.0125 / 0.0700

= 10.33 - 0.748

= 9.58

ANSWER : pH = 9.58

C)

Consider reaction of NaOH with buffer solution.

HCO ₃^- + OH ^- CO ₃ ^2- + H₂O

Let's Use ICE table.

moles	HCO 3- + OH - CO 3 2-
I	0.0600	0.0200	0.0225
C	-0.0200	-0.0200	+0.0200
E	0.0400	0.000	0.0425

pH of buffer solution after addition of 0.0200 moles of NaOH = 10.33 + log 0.0425 / 0.0400

= 10.33 + 0.0263

= 10..36

ANSWER : pH =10.36