Question

A weak electrolyte $A_2{H}_4$ is dissolved in water to form aqueous solution. If its base dissociation constants are $K_{b_1}=4\times {10}^{-5}$ and $K_{b_2}={10}^{-8}$ respectively then which of the options are correct regarding approximate concentration of species when 0.1 M aqueous solution of $A_2{H}_4$ is taken.

• A. $\left[A_2{H}_6^{2+}\right]={10}^{-8}$ M
  
• B. $\left[H^{+}\right]=5\times {10}^{-12}$ M
  
• C. $\left[A_2{H}_5^{+}\right]=2\times {10}^{-3}$ M
  
• D. $\left[O{H}^{-}\right]=2\times {10}^{-3}$ M

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\left[A_2{H}_6^{2+}\right]={10}^{-8}$ M

B $\left[H^{+}\right]=5\times {10}^{-12}$ M

C $\left[A_2{H}_5^{+}\right]=2\times {10}^{-3}$ M

D $\left[O{H}^{-}\right]=2\times {10}^{-3}$ M

A substrance is dissolved in water only if it forms ions after dissolving.
As base disosiation constant is given then after dissolve in water the solution will have $O{H}^{-}$.
$\begin{array}{ccccc}& A_2{H}_4+H_{2}O& \to & A_2{H}_5^{+}+& O{H}^{-}\\ t=0& 0.1& & 0& 0\\ t=t& (0.1-x)& & \left(x\right)& \left(x\right)\\ & & & & K{b}_1=4\times {10}^{-5}\end{array}$
$\begin{array}{ccccc}& A_2{H}_5^{+}+H_{2}O& \to & A_2{H}_6^{+}+& O{H}^{-}\\ t=t& x& & 0& x\\ t=t^{\prime }& (x-y)& & \left(y\right)& (x+y)\\ & & & & K{b}_2={10}^{-8}\end{array}$
Now total concentration of $[OH{]}^{-}=x+y$
$\therefore y<<<<x;K{b}_1>K{b}_2$
$\frac{\left(x\right)(x+y)}{(0.1-x)}\approx \frac{x.x}{0.1}=4\times {10}^{-5}$
(As it is a weak electrolyte then $(0.1-x)\approx 0.1$)
$x=2\times {10}^{-3}$
$\left[A_2{H}_5^{+}\right]=\left[O{H}^{-}\right]=2\times {10}^{-3}$
$\frac{y(y+x)}{(x-y)}\approx \frac{y.x}{x}={10}^{-8}$
$y={10}^{-8}$
$\left[A_2{H}_6^{+}\right]={10}^{-8}$
$(2\times {10}^{-3})\times \left[H^{+}\right]={10}^{-14};$
$\left[H^{+}\right]=\frac{{10}^{-14}}{2\times {10}^{-3}}=5\times {10}^{-12}$