100 ml aqueous 0.1 molar M(CN)_(2), (80% ionized) solution is mixed with 100 ml of 0.05 molar H_(2),SO_(2) solution (80% ionized). (K_(6), of CN" = 10^(-6))

100 ml aqueous 0.1 molar M(CN)_(2), (80% ionized) solution is mixed wi

1.	100 ml aqueous 0.1 molar M(CN)_(2), (80% ionized) solution is mixed with 100 ml of 0.05 molar H_(2),SO_(2) solution (80% ionized). (K_(6), of CN" = 10^(-6))
Answer» <html><body><p>`6` <br/>`8` <br/>`9` <br/>`7` </p>Solution :`M(<a href="https://interviewquestions.tuteehub.com/tag/cn-907325" style="font-weight:bold;" target="_blank" title="Click to know more about CN">CN</a>)_2 hArr M^(+2)+ 2CN^(-) ` <br/>`10m " moles" ` <br/> ` {:( 10(1- alpha ) , 10 alpha =8 , 2 xx 10 alpha = 16 ),( H_2SO_4 hArr, 2H^(+) , SO_4^(2-)),( 5 " moles" ,,),(5(1- beta ), 2 xx 5 beta =8, 5 beta =4 ), (<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+) + , CN^(-) hArr, HCN),( 8,16,):}` <br/> `8,8 ` <br/>i.e., <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a> of HCN & `CN^(-) `<br/> ` therefore <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =pKa +<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> ""( S)/(A)=pKa =14 -6=8`</body></html>