Question

A) Calculate the pH of the 2.0-L buffer which is 0.05M in butanoic acid, C3H7COOH and 0.04 M in C3H7COONa.

B) Calculate the pH of the buffer after adding 0.02 mol of HNO3

C) Calculate the pH after adding 0.02 mol of Ba(OH)2 to the original buffer.

Answer 1

A) pKa of butanoic acid = 4.82

C3H7 COOH ----------------> C3H7COO- + H+

2x0.05=0.1 2x0.04=0.08 - initial moles

pH of this buffer is calculated using Hendersen equation as

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

= 4.82 + log [ 0.08/0.1] = 4.723

B) after addition of 0.02mol of HNO3

C3H7 COOH ----------------> C3H7COO- + H+

2x0.05=0.1 2x0.04=0.08 - initial moles

---- ------- 0.02 change

0.12 0.06 0 equilibrium moles

Thus new pH = 4.82 + log [ 0.06/0.12]

=4.5189

C) After adding 0.02mol Ba(OH)2

As Ba(OH)2 -----> Ba+2 + 2OH- , the [OH-] added = 2x0.02 = 0.04 mol

C3H7 COOH + OH- ----------------> C3H7COO- + H2O

2x0.05=0.1 0 2x0.04=0.08 - initial moles

---- 0.04 ---- -- change

0.06 0 0.12 -- equilibrium moles

new pH of this buffer = 4.82 + log [0.12/0.06]

=5.1210