Question

A buffer contains 0.025 mol of acetic acid (pKa = 4.76) and 0.100 mol of sodium acetate per liter. What is the pH, when 150 mL of 0.1M HCl is added to 1.0 L of the buffer.

A. 5.09 CORRECT ANSWER

B. 5.17

C. 4.76

D. 6.32

E. 2.18

Please show how A. is the correct answer

Answer 1

First find out the no of mole of HCl added

No of mole of HCl added =( 0.1mol/1000ml)×150ml =0.015mol

HCl react with the conjucate base of the buffer system i.e with CH3COONa

HCl + CH3COONa ---------> CH3COOH + NaCl

Stoichiometrically,1mol of HCl react with 1mol of CH3COO- to produce 1mol of CH3COOH

Therefore, 0.015 mole of HCl react with 0.015 mole of CH3COONa to produce 0.015mol of CH3COOH

Now , Calculate the initial No of mole of Acetic acid and its conjucate base Sodium acetate

No of mole of CH3COOH = 0.025mol

No of mole of CH3COONa = 0.100mol

Therefore, After addition of HCl

No of mole of CH3COOH = 0.025 + 0.015 = 0.040

No of mole of CH3COONa = 0.100 - 0.015 = 0.085

Total volume = 1000ml + 150ml = 1150ml

Therefore,

[ CH3COOH ] = (0.040mol/1150ml)×1000ml =0.0348M

[ CH3COONa] = (0.085mol/1150ml)×1000ml = 0.0739M

Now , come to the Henderson-Hasselbalch equation

pH = pKa + log([acid]/[Conjucate base])

pKa of Acetic acid = 4.76

[ Acid ] = [ CH3COOH] = 0.0348M

[ Conjucate base ] = [ CH3COONa] = [CH3COO-] = 0.0739M

Applying the values

pH = 4.76 + log(0.0739M/0.0348M)

= 4.76 + 0.33

= 5.09

Therefore,

The answer is A)5.09