Question

An enzymatic reaction is carried out in 100 ml of 0.1M acetate buffer at pH 4.76. During the reaction, 3 mmoles of H+ are produced. What is the final pH of the reaction mixture if the pK of acetic acid is 4.76?

Answer 1

Sol.

Let moles of acetic acid and sodium acetate be x and y respectively

Total moles of buffer = Conc. of buffer × Volume of buffer / 1000

= 0.1 × 100 / 1000 = 0.01 mol

So , x + y = 0.01

Also , pH = 4.76 and pKa = 4.76

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( moles of sodium acetate / moles of acetic acid )

4.76 = 4.76 + log ( y / x )

log ( y / x ) = 0

y / x = 1

y = x

So , x + y = 2x = 2y = 0.01  

x = y = 0.01 / 2 = 0.005 mol  

During reaction , 3 mmol of H+ produced

, that is , 3 / 1000 = 0.003 mol of H+ produced

So , Final moles of acetic acid = 0.005 + 0.003 = 0.008 mol

and Final moles of sodium acetate = 0.005 - 0.003 = 0.002 mol

Therefore , again using Henderson - Hasselbalch equation ,

Final pH = pKa + log ( Final moles of sodium acetate / Final moles of acetic acid )

= 4.76 + log ( 0.002 / 0.008 )

=   4.16