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Hydrazine (N_2H_4) is weak base and its dissociation constant is 1.8xx10^(-6) . So, calculate pH of 0.25 M solution. |
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Answer» Solution :Suppose dissociation of hydrazine is `alpha` . `therefore 0.25 alpha` M Hydrazine dissociated So, [OH] and `[NH_2NH_3^+] = 0.25 alpha` `{:("EQUILIBRIUM" , NH_2NH_(2(aq)) + ,H_2O_((L)) hArr , NH_2NH_(3(aq))^(+) + , OH_((aq))^(-)),("Initial" , 0.25 M, -, 0.0,0.0),("Equilibrium change M" , -0.25 alpha, ,0.25 alpha, 0.25 alpha),("Equili. M" , 0.25-0.25 alpha, ,0.25 alpha, 0.25alpha):}` `=0.25(1-alpha)approx 0.25` `K_b=([NH_2NH_3^+ ][OH^-])/([NH_2NH_2])` `therefore 1.8xx10^(-6) = ((0.25alpha)(0.25alpha))/(0.25(1-alpha))` `=((alpha^2)0.25)/(1-alpha)` `therefore 1.8xx10^(-6) = 0.25 alpha^2` `prop= sqrt((1.8xx10^(-6))/0.25)=2.6833xx10^(-3)` `[OH^-]= C prop = 0.25 alpha= 0.25 (2.6833xx10^(-3))` `=6.708xx10^(-4)` pOH=-log `(6.708xx10^(-4))`= 3.1734 pH=14.0-pH =14.0 - 3.1734 = 10.827 |
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