Equilibrium Question 895

Question: Which relation is wrong

Options:

A) $ {10^{-pH}}+{10^{-pOH}}={10^{-14}} $

B) $ pH\alpha \frac{1}{[{H^{+}}]} $

C) $ K_{w}\alpha ,T $

D) dissociation constant of water $ K=1.8\times {10^{-16}} $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $ {10^{-pH}}\times {10^{-pOH}}={10^{-Kw}} $

[b] as $ [{H^{+}}] $ increases, pH decreases, [c] since ionization of water is endothermic in nature hence increase in temperature increases $ K_{w} $

[d] $ K_{a} $ of $ H_2O=\frac{[{H^{+}}][O{H^{-}}]}{[H_2O]}=\frac{{10^{-14}}}{55.55}=1.8\times {10^{-16}} $



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