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The hydrogen ion concentration of a $10^{-8}$ M HCl aqueous solution at 298 K ($K_w = 10^{-14}$) is :-
A
$1.0 \times 10^{-6}$ M
B
$1.0525 \times 10^{-7}$ M
C
$9.525 \times 10^{-8}$ M
D
$1.0 \times 10^{-8}$ M
Detailed Solution
The acid is extremely dilute, so the $H^+$ ions from the ionisation of water cannot be neglected.
$[H^+]$ from HCl $= 10^{-8}$ M
Approximate treatment: $[H^+]$ from water $\approx 10^{-7}$ M, so $[H^+] = 10^{-8} + 10^{-7} = 10^{-7}(0.1 + 1) = 1.1 \times 10^{-7}$ M
More exactly, the $H^+$ from HCl suppresses the ionisation of water (common ion effect). Let water give $x$ M each of $H^+$ and $OH^-$.
$K_w = [H^+][OH^-] = (10^{-8} + x)(x) = 10^{-14}$
$x^2 + 10^{-8}x - 10^{-14} = 0 \Rightarrow x = 9.525 \times 10^{-8}$ M
$[H^+] = 10^{-8} + 9.525 \times 10^{-8} = 1.0525 \times 10^{-7}$ M
The solution is acidic, so $[H^+]$ must be slightly greater than $10^{-7}$ M; it cannot be $10^{-8}$ M.
$[H^+]$ from HCl $= 10^{-8}$ M
Approximate treatment: $[H^+]$ from water $\approx 10^{-7}$ M, so $[H^+] = 10^{-8} + 10^{-7} = 10^{-7}(0.1 + 1) = 1.1 \times 10^{-7}$ M
More exactly, the $H^+$ from HCl suppresses the ionisation of water (common ion effect). Let water give $x$ M each of $H^+$ and $OH^-$.
$K_w = [H^+][OH^-] = (10^{-8} + x)(x) = 10^{-14}$
$x^2 + 10^{-8}x - 10^{-14} = 0 \Rightarrow x = 9.525 \times 10^{-8}$ M
$[H^+] = 10^{-8} + 9.525 \times 10^{-8} = 1.0525 \times 10^{-7}$ M
The solution is acidic, so $[H^+]$ must be slightly greater than $10^{-7}$ M; it cannot be $10^{-8}$ M.