How do you calculate the pH of 0.15 M aqueous solution of hydrazine?

1 Answer
Aug 11, 2017

Answer:

We interrogate the reaction.........
#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-# and finally get #pH=10.6........#

Explanation:

...............

#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#

#K_"eq"=([N_2H_5^+][HO^-])/([N_2H_4])=1.0xx10^-6#, and if #x*mol# of hydrazine associate, we can put in the numbers........

#K_"eq"=1.0xx10^-6=((x)xx(x))/(0.15-x)=x^2/(0.15-x)#

This is a quadratic in #x# that we could solve exactly. However, because #K_"eq"# is so small, we are justified in making the approx. #0.15">>"x#, and we write.........

#1.0xx10^-6~=x^2/0.15#

#x_1=sqrt(1.0xx10^-6xx0.15)=3.87xx10^-4#, this is indeed small compared to #0.15*mol*L^-1#, but now we have an approximation for #x# we can plug this value back into the equation to get #x^2#...

#x_2=3.87xx10^-4*mol*L^-1#, and given the convergence of #x_1# and #x_2# we are satisfied that these values are as good as if we used the quadratic equation............

But we are not finished there.....we know that ..............

#[N_2H_5^+]=[HO^-]=3.87xx10^-4*mol*L^-1#...

We DO NOT KNOW #pH#.

But in aqueous solution, #14=pH+pOH#

#pOH=-log_10[HO^-]=-log_10(3.87xx10^-4)=3.41#

#pH=14-3.41=10.6#