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The ionization constant of nitrous acid is `4.5xx10^(-4)`. Calculate the `pH` of `0.04 M` sodium nitrite solution and also its degree of hydrolysis. |
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Answer» Correct Answer - `pH = 7.97`. Degree of hydrolysis `= 2.36 xx 10^(-5)` `"NaNO"_(2)` is the salt of a strong base (NaOH) and a weak acid `"HNO"_(2)`. `NO_(2)^(-)+H_(2)OleftrightarrowHNO_(2)+OH^(-)` `K_(h) = ([HNO_(2)][OH^(-)])/([NO_(2)^(-)])` `rArr (K_(w))/(K_(a)) = (10^(-14))/(4.5 xx 10^(-4)) = .22 xx 10^(10)` Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: `[NO_(2)^(-)] = .04-x , 0.04` `[HNO_(2)] = x` `[OH^(-)] = x` `K_(b) = (x^(2))/(0.04) = 0.22 xx 10^(-10)` `x^(2) = .0088 xx 10^(-10)` `x = .093 xx 10^(-5)` `:. [OH^(-)] = 0.093 xx 10^(-5) M` `[HO^(+)] = (10^(-14))/(.093 xx 10^(-5)) = 10.75 xx 10^(-9) M` `rArr pH = - "log" (10.75 xx 10^(-19))` `= 7.96` Therefore, degree of hydrolysis `= (x)/(0.04) = (.093 xx 10^(-5))/(.04) = 2.325 xx 10^(-5)` |
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