Calculate pH of 0.01M HCl solution

hard • 2 min read
Tags Ionic Equilibrium

Question

Calculate the pH of a 0.01 M HCl solution.

Solution — Step by Step

HCl is a strong acid — it dissociates completely in water. This is the critical fact.

HCl→H++Cl−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-

Since dissociation is 100%, every mole of HCl produces exactly 1 mole of H⁺ ions.

Since HCl is completely dissociated:

[H+]=[HCl]=0.01 M=10−2 M[\text{H}^+] = [\text{HCl}] = 0.01\ \text{M} = 10^{-2}\ \text{M} pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+] pH=−log⁡(10−2)=−(−2)=2\text{pH} = -\log(10^{-2}) = -(-2) = 2 pH=2\boxed{\text{pH} = 2}

At 25°C: pH+pOH=14\text{pH} + \text{pOH} = 14

pOH=14−2=12\text{pOH} = 14 - 2 = 12

[OH−]=10−12 M[\text{OH}^-] = 10^{-12}\ \text{M}, which is very small compared to [H+]=0.01 M[\text{H}^+] = 0.01\ \text{M} ✓ (acidic solution confirmed).

Why This Works

The pH scale is a logarithmic measure of hydrogen ion concentration. Because hydrogen ion concentrations span many orders of magnitude (from 10 M in concentrated acid to 10⁻¹⁴ M in concentrated base), the logarithmic scale compresses this into a manageable range of 0 to 14.

For strong acids, the calculation is direct: concentration of acid = concentration of H⁺. For weak acids, we’d need to use the dissociation constant KaK_a and solve a quadratic, since only partial dissociation occurs.

Alternative Method — Mental Shortcut

[H+]=10−n⇒pH=n[H^+] = 10^{-n} \Rightarrow \text{pH} = n

Here 0.01=10−20.01 = 10^{-2}, so pH = 2. For any strong acid of the form 10−n10^{-n} M, the pH equals nn directly.

Quick pH values to know: 0.1 M HCl → pH 1; 0.01 M HCl → pH 2; 0.001 M HCl → pH 3. Each 10-fold dilution increases pH by 1. For JEE, problems often ask what happens to pH when you dilute a strong acid 10× or 100×.

Common Mistake

The most common error is applying the strong acid formula to a weak acid like acetic acid. If the question says “0.01 M CH₃COOH,” you CANNOT write [H⁺] = 0.01 M. Acetic acid only partially dissociates; you must use Ka=1.8×10−5K_a = 1.8 \times 10^{-5} and solve: [H+]=Ka×C[\text{H}^+] = \sqrt{K_a \times C}. Only use [H⁺] = C for fully dissociating strong acids (HCl, H₂SO₄ for first dissociation, HNO₃, HBr, HI, HClO₄).

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