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  • Calculate the degree of hydrolysis of 0.005 M K2CrO4. K2 = 5.0  10<

Calculate the degree of hydrolysis of 0.005 M K2CrO4. K2 = 5.0 \times 10-7 for H2CrO4. (It is essentially strong for first ionization).

Option: 1

2 \times 10^{-7}


Option: 2

4 \times 10^{-5}


Option: 3

2 \times 10^{-3}


Option: 4

2 \times 10^{3}


Answers (1)

best_answer

According to the question, we have to only assume the first hydrolysis of the Chromate ion. So the reaction will be 

\mathrm{CrO _{4}^{2-}+ H _{2} O ( l ) \rightleftharpoons HCrO _{4}^{-}+ OH ^{-}}

\mathrm{K_h = \frac{K_w}{K_{a_2}} = \frac{10^{-14}}{5\times 10^{-7}}=2 \times 10^{-8}}

Now,

\mathrm{K_h = c \times h^2 \Rightarrow h= \sqrt{\frac{K_h}{c}}}

\therefore \mathrm{h= \sqrt{\frac{2 \times 10^{-8}}{0.005}}=2 \times 10^{-3}}

Therefore, option(3) is correct

Posted by

avinash.dongre

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