1.

The pH of a 0.1 N solution of NH_4 Cl is 5.4. What will the hydrolysis constant? (supposing degree of hydrolysis as very small)

Answer»

`2.42 xx 10^(-10)`
`1.42 xx 10^(-10)`
`1.58 xx 10^(-9)`
`2.82 xx 10^(-9)`

SOLUTION :`NH_4CI `is a saltofstrong acidand WEAKBASE
`[H^+] =CH`
`pH =-log[H^+] =-logCh = 5.4`
`thereforeCh= 3.98 x 10^(-6) `
` thereforeh=( 3.98 xx 10^(-6) )/(0.1 ) = 3.98xx 10^(-5)`
Now `K_h = (h^2)/1-h^(2)= h^2 [ :. h lt LT1,therefore1- h^2~~1]`
`K_h =h^2= ( 3.98 xx 10^(-5))^2 =15.8 xx 10^(10) = 1.58 xx 10^(-9)`


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