Question

An enzymatic reaction takes place in a 10 ml solution that has a total citrate (plus citric acid) concentration of 120 mM and an initial pH of 7.00. During the reaction 0.2 millimoles of H+ are produced. The relevant pKa of citric acid is 6.40. A) Calculate the final pH of the solution. B) What would the final pH of the solution be if the citrate were absent from the solution and no other buffering agents were present (assuming a starting pH of 7.00 again, basically, what would be the pH is you added the same amount of acid to that amount of water)?

Answer 1

Citric acid   ---> citrate   + H+
Initial millimoles of citric acid + citrate = 120 mM*0.01 L = 1.2 mmol
Let the initial moles of citric acid = x mmol
[citric acid] = x/`0 = 0.1 x M

Then initial mole of citrate = 1.2 - x mmol
[citrate ] = (1.2-x) /10 = 0.1 (1.2-x) M

use:
pH = pKa + log [citrate] / [citric acid]
7 = 6.4 + log (0.1(1.2-x) / 0.1x)
0.6 = log ((1-2-x) /x )
(1.2-x)/x =3.981
1.2 - x = 3.981 x
x= 0.24 mmol

when 0.2 mmoles of H+ is fomed, 0.2 moles of citric acid must have converted into 0.2 mmoles of citrate and 0.2 mmoles of H+

[citric acid] =(x-0.2)/10 = (0.24-0.2) / 10 = 0.004 M
[citrate] = (1.2-x+0.2)/10 = (1.2-0.24+0.2) /10 = 1.24/10 = 0.124 M
pH = pKa + log [citrate] / [citric acid]
    =6.4 + log (0.124/0.004)
   = 7.9

B)
adding H+ in neutral water,
moles of H+ addec= 0.2 mmol
volume = 10 mL
[H+] = 0.2/10 =0.02
pH = -log [H+]
= -log (0.02)
   = 1.7