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Sum

**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 (H_{2}SO_{4}) is a strong acid. It dissociates almost completely in the water as:

\[\ce{H2SO4_{(aq)} + 2H2O_{(l)} -> 2H3O^+_{ (aq)} + SO^{2-}_{4(aq)}}\]

Hence, [H_{3}O^{+}] = 2 × c = 2 × 0.01 M = 2 × 10^{-2 }M

From formula (i),

pH = -log_{10}[H_{3}O^{+}] = -log_{10}[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.

Concept: The pH Scale

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