The pH of a 0.01 M weak acid HX (Ka = 4 10-10) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as x 10-4 M. The value of x is (nearest integer)
Answer
25
Solution
For a weak acid HX of concentration C (small degree of ionisation):
[H+] = Ka C
⇒ C = [H+]2Ka
After dilution, pH = 6 ⇒ [H+] = 10-6 M
⇒ C = (10-6)24 10-10
= 10-124 10-10 = 2.5 10-3 M
= 25 10-4 M
∴ x = 25
Answer: 25