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A solution of 0.01 MCd^(2+) contains 0....

A solution of 0.01 `MCd^(2+) ` contains `0.01M NH_4OH.` What conc. Of `NH_4^(+) ` from `NH_4Cl` is necessary to prevent precipitation of `Cd( OH)_2? K_(sp) " of " (OH)_2 =2.0 xx 10^(-14) ,K_b " of " NH_4 OH = 1.8 xx 10^(-5) ` if answer is `1.272 xx 10^(-x) ` mol/litre then x=________?

Text Solution

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The correct Answer is:
1

` K_(sp) =[Cd^(+2) ][OH^(-) ] ^(2) , 2 xx 10 ^(-14) =10 ^(-2) [OH^(-) ]^(2) `
` [OH^(-) ] =1.414 xx 10 ^(-6) rArr K_b =([NH_4^(+) ][OH^(-) ])/( [NH_4OH]) `
` 1.8 xx 10 ^(-5) =([ NH_4^(+) ][1.4 xx 10 ^(-6)])/( 10^(-2))`
` [NH_4^(+) ]=1.2 xx 10 ^(-1) `
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