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}} $
