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Chemistry · EQUILIBRIUM
हिन्दीConsidering the self-ionization of water, what is the exact \( [\text{OH}^-] \) in an extremely dilute \( 1.0 \times 10^{-8}\text{ M}\text{ HCl} \) solution at \( 298\text{ K} \)?
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Correct answer · Option C
\( 9.5 \times 10^{-8}\text{ M} \)
Explanation
Let \( [\text{OH}^-] = x \), then \( [\text{H}^+] = 10^{-8} + x \). \( x(10^{-8} + x) = 10^{-14} \implies x^2 + 10^{-8}x - 10^{-14} = 0 \), which yields \( x = [\text{OH}^-] = 9.5 \times 10^{-8}\text{ M} \).
