Question

An enzyme-catalyzed reaction was carried out in a solution buffered with 0.03 M phosphate, PH 7.2. As a result of the reaction, 0.004mole/liter of acid was formed

a) What is the pH at the end of the reaction.?

b) what would the pH if No buffer were present?

Answers for questions a and b are pH = 6.9 and pH =2.4 respectively. BUT HOW DID THEY GET THOSE pH?

Answer 1

Solution :-

Calculating the pH at the end of reaction.

The pH of the buffer is 7.2 which is the same as the pka2 of the phosphate H2PO4^- --- >HPO4^2-

As we know at the half equivalence point pH and pka are same

Therefore the moles of H2PO4- and HPO4^2- will be same in the 0.03 M buffer

So their concentrations are half that is (0.03 M/2)=0.015 M each

Assuming 1 L solution the moles of each species will be same as its molarity

Moles of H2PO4^- = 0.015

Moles of HPO4^2- = 0.015

When the 0.004 moles of acid is formed then it increases the concentration of the H2PO4- and decreases concentration of the HPO4^2- by the same number of moles

Therefore

New moles of H2PO4- = 0.015 + 0.004 = 0.019

Moles of HPO4^2- = 0.015 – 0.004 = 0.011

Now using the Henderson equation we can find the pH

pH= pka + log [base]/[acid]

pH= 7.2 + log [0.011]/[0.019]

pH= 6.96

Now calculating the pH when the buffer is not present

When the buffer is not present then 0.004 moles of acid will give 0.004 mol H^+ by complete dissociation

Assuming 1 L volume

Molarity of the [H^+] = 0.004 mol / 1 L = 0.004 M

pH= -log [H^+]

pH= -log [0.004]

     = 2.4