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Question
Answer the following in one sentence :
Calculate the pH of 0.01 M sulphuric acid.
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Solution
Given: Concentration of sulphuric acid = 0.01 M
To find: pH
Formula: pH = `-"log"_10["H"_3"O"^+]`
Calculation:
Sulphuric acid (H2SO4) is a strong acid. It dissociates almost completely in the water as:
\[\ce{H2SO4_{(aq)} + 2H2O_{(l)} -> 2H3O^+_{ (aq)} + SO^{2-}_{4(aq)}}\]
Hence, [H3O+] = 2 × c = 2 × 0.01 M = 2 × 10-2 M
From formula (i),
pH = -log10[H3O+] = -log10[2 × 10-2] = `-"log"_10"2" - "log"_10"10"^-2`
= `-"log"_10"2" + 2 = 2 - 0.3010`
pH = 1.699
The pH of 0.01 M sulphuric acid is 1.699.
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