pH calculation upon dilution of a strong acid solution is generally done by equating 
in original solution & diluted solution. However, if strong acid solution is very dilute, then H + from water are also to be considered.
Take log 3.7 = 0.568 and answer the following questions.
(i) A 1 litre solution of pH = 4 (solution of a strong acid) is added to the 7/3 litre of water. What is the pH of resulting solution ?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) Initial pH = 4
[H + ] = 10 –4
N 1 V 1 = N 2 V 2 ⇒ 10 –4 = N 2 × 
10 –4 = N 2 × 
⇒ N 2 = 3 × 10 –5 > 10 –6
So [H + ] of water is not consider
[H + ] = 3 × 10 –5 So pH = 5 – log3 = 5 – 0.48 = 4.52
(ii) pH = 6
[H + ] = 10 –6
N 1 V 1 = N 1 V 2
⇒ 10 –6 × 1 = N 2 
⇒ 10 –6 = N 2 × 
N 2 = 
× 10 –6 ⇒ N 2 = 3 × 10 –7
[H + ] < 10 –6
So [H + ] of water is also added. as common ion effect on H 2 O is neglected so
[H + ] = 3 × 10 –7 + 10 –7 = 4 × 10 –7 M
⇒ pH = 7 – log 4 = 7 – 0.60 = 6.4
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