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=________?

A solution of 0.01 MCd^(2+)contains 0.01M NH_4OH. What conc. Of NH_4^(

1.	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=________?
Answer» Solution :` 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) `