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JEE Advanced 2025 · Paper 1, Chemistry question 9

At 25 °C, the concentration of HX+\ce{H+} ions in 1.00×10−31.00\times10^{-3} M aqueous solution of a weak monobasic acid having acid dissociation constant (KaK_a) of 4.00×10−114.00\times10^{-11} is X×10−7\boldsymbol{X}\times10^{-7} M. The value of X\boldsymbol{X} is _____.

Use: Ionic product of water (Kw)=1.00×10−14(K_w) = 1.00\times10^{-14} at 25 °C

Answer and worked solution

Official answer: 2.2

The acid is so weak that KaC=2×10−7\sqrt{K_aC}=2\times10^{-7} M is comparable with water's own HX+\ce{H+}, so water cannot be ignored.

Charge balance. Since the acid barely dissociates ([HA]≈C[\ce{HA}]\approx C),
[HX+]2=KaC+Kw=4×10−14+1×10−14=5×10−14.[\ce{H+}]^2=K_aC+K_w=4\times10^{-14}+1\times10^{-14}=5\times10^{-14}.

[HX+]=2.24×10−7[\ce{H+}]=2.24\times10^{-7} M, so X≈2.24X\approx2.24.

Topics:pH and ionisation
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