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NCERT Exemplar · Q30

Q.pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution a 100 times?

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Diluting a strong acid solution 100 times decreases its hydrogen ion concentration by a factor of 100. When the resulting concentration is 10−7 M10^{-7} \text{ M}, the autoionization of water must be considered, leading to a pH slightly less than 7. The final pH of the solution is approximately 6.79\boxed{6.79}.

The pH scale is a convenient way to express the hydrogen ion concentration ([H+][H^+]) in a solution. It's defined as the negative logarithm (base 10) of the hydrogen ion concentration.

pH=−log⁡[H+]pH = -\log[H^+]

A strong acid, like HCl or H2SO4H_2SO_4, completely dissociates in water. This means that if you have a 10−5 M10^{-5} \text{ M} solution of a strong monoprotic acid, the concentration of H+H^+ ions it contributes to the solution is also 10−5 M10^{-5} \text{ M}.

When a solution is diluted, the amount of solute (in this case, the acid) remains constant, but the volume of the solvent increases. This leads to a decrease in the concentration of the solute. If a solution is diluted 100 times, its concentration becomes 1/1001/100th of the original concentration.

A crucial point arises when diluting strong acids or bases to very low concentrations. Water itself undergoes autoionization, producing H+H^+ and OH−OH^- ions:

H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-

At 25∘C25^\circ C, the ion product of water, Kw=[H+][OH−]K_w = [H^+][OH^-], is 1.0×10−141.0 \times 10^{-14}. In pure water, [H+]=[OH−]=1.0×10−7 M[H^+] = [OH^-] = 1.0 \times 10^{-7} \text{ M}, resulting in a neutral pH of 7. When the concentration of H+H^+ from the added acid becomes comparable to or less than 10−7 M10^{-7} \text{ M}, the H+H^+ ions contributed by water's autoionization can no longer be ignored. The total [H+][H^+] in the solution will be the sum of H+H^+ from the acid and H+H^+ from water.

Let's work through the problem step-by-step.

  1. Determine the initial hydrogen ion concentration:

    The initial pH of the strong acid solution is given as 5.0.

    Using the pH formula:

    pH=−log⁡[H+]pH = -\log[H^+]

    5.0=−log⁡[H+]5.0 = -\log[H^+]

    [H+]=10−5.0 M[H^+] = 10^{-5.0} \text{ M}

    So, the initial hydrogen ion concentration is 10−5 M10^{-5} \text{ M}. Since it's a strong acid, this is also the initial concentration of the acid.

  2. Calculate the hydrogen ion concentration after dilution:

    The solution is diluted 100 times. This means the new volume is 100 times the original volume. Consequently, the concentration of the acid (and thus the H+H^+ ions from the acid) will decrease by a factor of 100.

    New [H+]acid=Initial [H+]Dilution factor[H^+]_{acid} = \frac{\text{Initial } [H^+]}{\text{Dilution factor}}

    New [H+]acid=10−5 M100[H^+]_{acid} = \frac{10^{-5} \text{ M}}{100}

    New [H+]acid=10−5 M102[H^+]_{acid} = \frac{10^{-5} \text{ M}}{10^2}

    New [H+]acid=10−5−2 M[H^+]_{acid} = 10^{-5-2} \text{ M}

    New [H+]acid=10−7 M[H^+]_{acid} = 10^{-7} \text{ M}

  3. Account for the autoionization of water:

    The calculated [H+]acid[H^+]_{acid} is 10−7 M10^{-7} \text{ M}. This concentration is exactly equal to the [H+][H^+] contributed by water in a neutral solution. At such low concentrations, the H+H^+ ions from the autoionization of water cannot be ignored. The total hydrogen ion concentration in the solution will be the sum of H+H^+ from the acid and H+H^+ from water.

    Let the total hydrogen ion concentration be [H+]total[H^+]_{total}.

    [H+]total=[H+]acid+[H+]water[H^+]_{total} = [H^+]_{acid} + [H^+]_{water}

    We know that for water, [H+][OH−]=Kw=10−14[H^+][OH^-] = K_w = 10^{-14} (at 25∘C25^\circ C).

    Also, the OH−OH^- ions in the solution come solely from the autoionization of water, so [OH−]=[H+]water[OH^-] = [H^+]_{water}.

    Substituting [H+]water=Kw[H+]total[H^+]_{water} = \frac{K_w}{[H^+]_{total}} into the equation for [H+]total[H^+]_{total}:

    [H+]total=[H+]acid+Kw[H+]total[H^+]_{total} = [H^+]_{acid} + \frac{K_w}{[H^+]_{total}}

    Let x=[H+]totalx = [H^+]_{total}. We have [H+]acid=10−7 M[H^+]_{acid} = 10^{-7} \text{ M} and Kw=10−14K_w = 10^{-14}.

    x=10−7+10−14xx = 10^{-7} + \frac{10^{-14}}{x}

    Multiply by xx to clear the denominator:

    x2=10−7x+10−14x^2 = 10^{-7}x + 10^{-14}

    Rearrange into a quadratic equation:

    x2−10−7x−10−14=0x^2 - 10^{-7}x - 10^{-14} = 0

  4. Solve the quadratic equation for [H+]total[H^+]_{total}:

    Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:

    Here, a=1a=1, b=−10−7b=-10^{-7}, c=−10−14c=-10^{-14}. …

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