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InterviewSolution
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Calculate the hydrolysis constant, degree of hydrolysis and pH of 0.10 M KCN solution at `15^(@)`C . For HCN, `K_(a)=6.2xx10^(-10)`. |
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Answer» As KCN is a salt of strong base and weak acid, Hydrolysis constant, `K_(h) = (K_(w))/(K_(a))=(10^(-14))/(6.2xx10^(-10))=1.6xx10^(-5)` Degree of hydrolysis, `h=sqrt((K_(h))/(c))=sqrt((1.6xx10^(-5))/(0.1))=1.26xx10^(-2)` The hydrolysis reaction will be : `{:(,CN^(-),+,H_(2)O,hArr,HCN,+,OH^(-)" ...(i)"),("Initial conc.",cM,,,,0,,0),("At eqm.",c-x,,,,x,,x " (x-No. of moles of" CN^(-) "reacted)"),(,,,,,,,):}` `K(h)=([HCN][OH^(-)])/([CN^(-)])=(x xx x)/(c-x)=(x^(2))/(c) or x=sqrt(K_(h)xxc)=sqrt((1.6xx10^(-5))(0.1))=1.26xx10^(-3)` i.e., `[OH^(-)]=1.26xx10^(-3)` `:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.26xx10^(-3))=7.94xx10^(-12)` `pH = - log [H^(+)]=-log (7.94xx10^(-12))=12-0.90=11.1` Alternatively, reaction (i) is the hydrolysis of the base `(CN^(-))`. We need `K_(b)` for this reaction. We are given `K_(a)` of the conjugate acid (HCN) . Hence, `K_(b)=(K_(w))/(K_(a))=(10^(-14))/(6.2xx10^(-10))=1.61xx10^(-5)` For equilibrium (i), `K_(b)=(x xx x)/(c-x) ~= (x^(2))/(c)` `:. 1.61xx10^(-5)=(x^(2))/(0.1) or x^(2)xx1.61xx10^(-6) or x = 1.26xx10^(-3)` i.e., `[OH^(-)]=1.26xx10^(-3)` `pOH = log (1.26xx10^(-3))=2.9` `:. pH = 14-2.9 = 11.1` Alternatively, pH can be calculated directly by applying the formula, `pH = 7 +(1)/(2) [pK_(z)+log c]` |
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