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6-2.Equilibrium-II (Ionic Equilibrium)
normal
The solubility of a salt of weak acid $( A B )$ at $pH 3$ is $Y \times 10^{-3} mol L ^{-1}$. The value of $Y$ is
. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
A
$4.47$
B
$4.48$
C
$4.49$
D
$4.50$
(IIT-2018)
Solution
Solubility $=\sqrt{ K _{ sp }\left(1+\frac{\left[ n ^{+}\right]}{ Ka }\right)}$
$Y \times 10^{-3} =\sqrt{2 \times 10^{-10}\left(1+\frac{10^{-3}}{10^{-8}}\right)}=\sqrt{2 \times 10^{-5}}$
$Y \times 10^{-3} =\sqrt{20 \times 10^{-6}}=\sqrt{20} \times 10^{-3}$
$Y =\sqrt{20}=4.47$
Standard 11
Chemistry
