InterviewSolution
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InterviewSolution
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a. Calculate the percentage hydrolysis of `0.003M` aqueous solution of `NaOH. K_(a)` for `HOCN = 3.3 xx 10^(-4)`. b. What is the `pH` and `[overset(Theta)OH]` of `0.02M` aqueous solution of sodium butyrate. `(K_(a) = 2.0 xx 10^(-5))`. |
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Answer» a. `NaOCN + H_(2)O hArr NaOH + HCN` `h = sqrt(((K_(h))/(c))) = sqrt(((K_(w))/(K_(a).C)))` `=sqrt((10^(-14))/(3.33xx10^(-4)xx0.003))` `h = 10^(-4)` `:. %` hydrolysis `= 10^(-4) xx 100 = 10^(-2)`. b. `{:(,NaBu+,H_(2)OhArr,NaOH+,BuH),("Conc before hydrolysis",1,,0,0),("Conce after hydrolysis",1-h,,h,h):}` `:. [OH^(Theta)] = C.h = sqrt(((K_(h))/(c)))` `= sqrt((K_(h).C)) = sqrt(((K_(w))/(K_(a)).C ))` `[OH^(Theta)] = (10^(-14)xx0.2)/(2xx106(-5)) = sqrt(10^(-10)) = 10^(-5)`, `pOH = 5, pH = 9` |
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