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If the equilibrium for reaction of HCN with NaOH is 10^(10) then calculate pH of 10^(-3)MNaCN solution of 25^(@)C |
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Answer» Solution :`{:(,HCN+NaOH,HARR,H_(2)O,+,NaCN,K=10^(10)),(rArr,CN^(-)+ H_(2)O,hArr,HCN,+,OH^(-),K=10^(10)),(t=0,10^(-3)M,,0,,0,),(at eq.,10^(-3)(1-h),,10^(-3)h,,10^(-3)h,):}` `K_(h)=10^(-10)=(10^(-3)hxx10^(-3)h)/(10^(-3)(1-h))rArrsqrt(K_(h)/(c ))=sqrt(10^(-7))( lt0.1)` `pH=7-(1)/(2)log10^(-10)+(1)/(2)log10^(-10)=7+5-(3)/(2)=10.5` |
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