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NCERT Exemplar · Q29

Q.On the basis of the equation pH = – log [H^+], the pH of 10^-8 mol dm^-3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason.

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The pH of a 10−810^{-8} M HCl solution is less than 7 because the autoionization of water contributes a significant amount of H+H^+ ions at such low acid concentrations, making the total [H+][H^+] greater than 10−710^{-7} M.

The equation pH=−log⁡[H+]\text{pH} = -\log [H^+] works perfectly for strong acids at moderate concentrations. But when the acid is extremely dilute — like 10−810^{-8} M HCl — a naive calculation gives pH = 8, which would imply a basic solution. That cannot be right, because adding an acid to pure water should never make it basic. The flaw is in forgetting that water itself is a source of H+H^+ ions.

Pure water at 25°C has [H+]=10−7[H^+] = 10^{-7} M from its own autoionization:

H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-, with Kw=[H+][OH−]=10−14K_w = [H^+][OH^-] = 10^{-14}.

When you add a tiny amount of HCl, the H+H^+ from the acid and the H+H^+ from water both contribute to the total [H+][H^+]. At 10−810^{-8} M HCl, the acid's contribution is ten times smaller than water's own 10−710^{-7} M — so water's contribution cannot be ignored. The total [H+][H^+] will be slightly above 10−710^{-7} M, giving a pH just below 7.

Let's work through it properly.

  1. Set up the charge balance. In the solution, the positive ions are H+H^+ (from both HCl and water) and the negative ions are Cl−Cl^- (from HCl) and OH−OH^- (from water). Charge neutrality demands:

[H+]=[Cl−]+[OH−][H^+] = [Cl^-] + [OH^-]

  1. Express known concentrations.

    HCl is a strong acid, so [Cl−]=10−8[Cl^-] = 10^{-8} M.

    From water's autoionization, [OH−]=Kw[H+]=10−14[H+][OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{[H^+]}.

  2. Substitute into the charge balance equation.

[H+]=10−8+10−14[H+][H^+] = 10^{-8} + \frac{10^{-14}}{[H^+]}

  1. Solve for [H+][H^+]. Multiply through by [H+][H^+]:

[H+]2=10−8[H+]+10−14[H^+]^2 = 10^{-8}[H^+] + 10^{-14}

Rearrange into a quadratic:

[H+]2−10−8[H+]−10−14=0[H^+]^2 - 10^{-8}[H^+] - 10^{-14} = 0

Solve using the quadratic formula (taking the positive root):

[H+]=10−8+(10−8)2+4×10−142[H^+] = \frac{10^{-8} + \sqrt{(10^{-8})^2 + 4 \times 10^{-14}}}{2}

[H+]=10−8+10−16+4×10−142[H^+] = \frac{10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2}

[H+]=10−8+4.01×10−142[H^+] = \frac{10^{-8} + \sqrt{4.01 \times 10^{-14}}}{2}

[H+]≈10−8+2.0025×10−72[H^+] \approx \frac{10^{-8} + 2.0025 \times 10^{-7}}{2} …

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