Question

$pH$ of saturated solution of silver salt of monobasic acid $HA$ is found ton be 9. Find the $K_{sp}$ of sparingly soluble salt $AgA$ (s).
Given : $K_a\left(HA\right)={10}^{10}$

• A. $1.1\times {10}^{-9}$
• B. $1.1\times {10}^{-10}$
• C. ${10}^{-12}$
• D. None of these

Answer 1

Answer: (A)

$1.1\times {10}^{-9}$

Let solutately of AgA be x moles/

$x=\left[A{g}^{+}\right]=\left[A^{-}\right]+\left[4A\right]$...(1)

from $\frac{\left[4A\right]}{\left[A\right]}=10$

$x=\left[A{g}^{+}\right]=\left[A^{-}\right]+10\left[A^{-}\right]=11\left[A^{-}\right]$

$x=\left[A{g}^{+}\right]=11\left[A^{+}\right]=11\left[A^{-}\right]\times \left[A^{-}\right]$...(2)

equations (2) can be written as

$A^{-}+4_{2}O\rightleftharpoons HA+O4$

$kw=\left[O{4}^{-}\right]\left[4^{+}\right]$

$\left[O{4}^{-}\right]=\frac{kw}{\left[4^{+}\right]}=\frac{{10}^{-14}}{1{-}^{-4}}={10}^{-5}$

At eq $\left[A^{-}\right]=\left[O{4}^{-}\right]={10}^{-5}$

from eq. (1) $\left[A{g}^{+}\right]=\left[A^{-}\right]={10}^{-}5$

ksp $=\left[A{g}^{+}\right]\left[\right(a^{-}\left)\right]+\left(4A\right)$

$={10}^{-5}\times 11]\times [A^{-}]$

$={10}^{-5}\times 11]\times {10}^{-5}$

$={10}^{-10}\times 11\times {10}^{-5}$

$={10}^{-10}\times 11=\frac{11}{10}\times {10}^{-10}\times 10$

$=1.1\times {10}^{-9}$