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Calculate the degree of ionization and pH of 0.05 M ammonia solution. The ionization constant of ammonia (K_(b)) is 1.77xx10^(-5). Also calculate the ionic constant of the conjugate acid of ammonia . |
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Answer» Solution :`{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.05M,,,,,,),("Eqm. conc. ",0.05(1 -alpha),,,,0.05 alpha,,0.05 alpha):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` `:. 1.77xx10^(-5)=((0.5 alpha)^(2))/(0.05(1-alpha))~=0.05 alpha^(2) or alpha^(2) = (1.77xx10^(-5))/(0.05) = 3.54xx10^(-4) or alpha = 1.88xx10^(-2)` Alternatively, applydirectly theformula : `alpha =sqrt(K_(b)//C)` (II) `[OH^(-)]=0.05 alpha = 0.05 xx 1.88xx10^(-2) = 9.4 xx 10^(-4) M` `pOH = log (9.04xx10^(-4)) =4-0.9562=3.04 :. pH = 14-3.04 = 10.96` (iii) For conjugate acid BASE pair, `K_(a) = (K_(w))/(K_(b))=(10^(-14))/(1.77xx10^(-5))=5.64xx10^(-10)` |
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