InterviewSolution
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InterviewSolution
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Calculate the pH of a solution formed on mixing 0.2 M `NH_(4)Cl` and 0.1 M `NH_(3)`. The `pK_(b)` of ammonia solution is 4.75. |
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Answer» `NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-), pK_(b)=4.75` On mixing `NH_(3) ` with `NH_(4)Cl` solution, we have `{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.1M,,,,0.2M,,0),("At eqm.",0.1 -x,,,,0.2+x,,x):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` As `pK_(b) = 4.75 , i.e., -log K_(b) = 4.75 or log K_(b) = -4.75 = bar(5) .25 :. K_(b) = 1.77xx10^(-5)` `:. 1.77xx10^(-5) = ((0.2+x)(x))/(0.1-x)~=((0.2)x)/(0.1)` (neglecting x in comparison to 0.1 and 0.2) or, `x=010^(-5),i..e., [OH^(-)]=0.885xx10^(-5)=8.85xx10^(-6)M` or `[H^(+)]=(10^(-14))/(8.85xx10^(-6))=1.13xx10^(9) :. pH = - log [H^(+)]=-log (1.13xx10^(-9))=9-0.053=8.95` Alternatively, the given solution is a buffer solution of a weak base and is salt with a strong acid (discussed in sec. 7.31). For such a buffer, `pOH=pK_(b)+log.(["Salt"])/(["Base"])=4.75+log. (0.2)/(0.1) = 4.75+log 2 = 4.75 + 0.30 = 5.05 `[ Eqn . (vi)] `pH = 14 -pOH = 14 - 5.05=8.95` |
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