Question

Calculate the extent of hydrolysis of $0.005M$ $K_{2}Cr{O}_4$.

$[K_2=3.1\times {10}^{-7}$ for $K_{2}Cr{O}_4;$ It is essentially strong for first ionization].

• A. $0.26$%
• B. $12.50$%
• C. $88.5$%
• D. $99.74$%

Answer 1

Answer: (A)

$0.26$%

$K_{2}Cr{O}_4$ is a Salt of Weak Acid ($H_{2}Cr{O}_4$) and Strong Base ($KOH$).
The given salt is essentially strong for first ionisation. So,second ionisation would be reversible. Hence,
$K_2=Ka$
Hence,
$K_h$=$\frac{K_w}{Ka}$=$\frac{{10}^{-14}}{3.1\times {10}^{-7}}$$=3.22\times {10}^{-8}$
Also,
$K_h=$$h^{2}C$
$h^2=$$\frac{3.22\times {10}^{-8}}{0.005}=6.45\times {10}^{-6}$

$h$=$\sqrt{6.45\times {10}^{-6}}=2.54\times {10}^{-3}$
Percent of hydrolysis$=h\times 100=$$0.25\approx 0.26\%$