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Exercises · 6.65

Q.Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?

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The pH of neutral water is defined by −log⁡[H+]-\log[\mathrm{H}^+], and at any temperature, neutrality means [H+]=[OH−][\mathrm{H}^+] = [\mathrm{OH}^-]. Using the ionic product Kw=[H+][OH−]=2.7×10−14K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 2.7 \times 10^{-14} at 310 K, we get [H+]=2.7×10−14≈1.643×10−7[\mathrm{H}^+] = \sqrt{2.7 \times 10^{-14}} \approx 1.643 \times 10^{-7} M, so pH ≈6.78\approx 6.78.

The key idea here is that neutral water is not defined by pH = 7 — that’s only true at 298 K (25 °C), where Kw=1.0×10−14K_w = 1.0 \times 10^{-14}. Neutrality means the concentrations of hydrogen ions and hydroxide ions are equal. At any temperature, if [H+]=[OH−][\mathrm{H}^+] = [\mathrm{OH}^-], the water is neutral. So the pH changes with temperature because KwK_w changes.

Let’s work through it.

  1. Recall the definition of KwK_w. The ionic product of water is

Kw=[H+][OH−]K_w = [\mathrm{H}^+][\mathrm{OH}^-]

At 310 K, we are given Kw=2.7×10−14K_w = 2.7 \times 10^{-14}.

  1. Neutral water condition. For pure water, every H+\mathrm{H}^+ comes from the dissociation of a water molecule, and an equal number of OH−\mathrm{OH}^- ions are produced. So

[H+]=[OH−][\mathrm{H}^+] = [\mathrm{OH}^-]

Let this common concentration be xx M.

  1. Substitute into KwK_w.

x⋅x=x2=2.7×10−14x \cdot x = x^2 = 2.7 \times 10^{-14}

Therefore

x=2.7×10−14x = \sqrt{2.7 \times 10^{-14}}

  1. Calculate xx carefully.

x=2.7×10−14=2.7×10−7x = \sqrt{2.7} \times \sqrt{10^{-14}} = \sqrt{2.7} \times 10^{-7}

Now 2.7≈1.643\sqrt{2.7} \approx 1.643 (since 1.642=2.68961.64^2 = 2.6896 and 1.652=2.72251.65^2 = 2.7225, so 1.643 is a good approximation).

Hence

[H+]≈1.643×10−7 M[\mathrm{H}^+] \approx 1.643 \times 10^{-7} \ \text{M}

  1. Find the pH.

pH=−log⁡10[H+]=−log⁡10(1.643×10−7)\text{pH} = -\log_{10}[\mathrm{H}^+] = -\log_{10}(1.643 \times 10^{-7})

Using logarithm rules:

pH=−(log⁡101.643+log⁡1010−7)=−(log⁡101.643−7)=7−log⁡101.643\text{pH} = -(\log_{10} 1.643 + \log_{10} 10^{-7}) = -(\log_{10} 1.643 - 7) = 7 - \log_{10} 1.643 …

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