100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then(K_a= 1.2 x

1.	100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then(K_a= 1.2 xx 10^(-6))
Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/ocn-2202269" style="font-weight:bold;" target="_blank" title="Click to know more about OCN">OCN</a>^(-) `concentration in the solution is `1.2 xx10^(-6) ` <br/>pH of the solution is 1.3<br/>solution is a buffer <br/>`H^(+) ` in the solution is `10^(6) ` </p>Solution :`HOCNhArr H^(+) +OCN^(-) ` <br/> ` K_a <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> ([H^(+)][OCN^(-)])/( [HOCN]) `<br/> ` 1.2 xx 10 ^(-6)= (5xx 10 ^(_2)[OCN^(-) ])/( 5 xx10 ^(-2)) ` <br/>`[ OCN^(-)]= 1.2 xx 10 ^(-6) ` <br/> Approximation method, <a href="https://interviewquestions.tuteehub.com/tag/weak-729638" style="font-weight:bold;" target="_blank" title="Click to know more about WEAK">WEAK</a> acid conribution is neglected. <br/>` [H^(+) ] =(100 xx <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.1 +0)/( <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>) = 0.05 therefore pH =1.3 `</body></html>

100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then(K_a= 1.2 xx 10^(-6))