Question

One litre of $1\,M$ solution of an acid $HA(K_{a} = (10)^{- 4}$ at $25 ^\circ C)$ has $pH=2.$ It is diluted by water so the new $pH$ becomes double. The solution was diluted to $y\times 10^{z} \, ml$ . The value of $y+z$ is:

Answer 1

Ostwald's dilution law describes the dissociation constant of the weak electrolyte with the degree of dissociation $(\alpha)$ and the concentration of the weak electrolyte.
Dissociation of weak acid:

$pH=2\Rightarrow \left[H^{+}\right]_{1}=10^{- 2}=c_{1}\alpha _{1}=1\times \alpha _{1}$
$\alpha _{1}=10^{- 2}$
$pH=4\Rightarrow \left[H^{+}\right]_{2}=10^{- 4}=\frac{1 \times 1000}{V}\times \alpha _{2}$
$k_{a}=10^{- 4}=\frac{\alpha _{2}^{2} C_{2}}{1 - \alpha _{2}}=\frac{\alpha _{2} \times \alpha _{2} \times C_{2}}{1 - \alpha _{2}}$
$10^{- 4}=\frac{\alpha _{2} \times 10^{- 4}}{1 - \alpha _{2}}\Rightarrow \alpha _{2}=0.5$
$\left[H^{+}\right]_{2}=\frac{1 \times 1000}{V}\times \alpha _{2}=10^{- 4}$
$\frac{1 \times 1000}{V}\times 0.5=10^{- 4}$
$V=5\times 10^{6} \, ml$
$y+z=5+6=11$