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» <html><body><p><br/></p>Solution :` K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)=[<a href="https://interviewquestions.tuteehub.com/tag/cd-407381" style="font-weight:bold;" target="_blank" title="Click to know more about CD">CD</a>^(+2)][OH^(-) ] ^(2), 2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10 ^(-14)=10 ^(-2)[OH^(-)]^(2) ` <br/> ` [OH^(-) ] =<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.414 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)rArr K_b =([NH_4^(+) ][OH^(-) ])/( [NH_4OH]) ` <br/> ` 1.8 xx 10 ^(-5)=([ NH_4^(+)][1.4 xx 10 ^(-6)])/( 10^(-2))` <br/> ` [NH_4^(+) ]=1.2 xx 10 ^(-1) `</body></html>