Question

What is the concentration of `[OH^-]` in a solution of 0,025 M NH3?

Answer 1

You have not quoted `pK_b` for ammonia....

Explanation:

We interrogate the equilibrium...

`NH_3(aq) +H_2O(l)rightleftharpoonsNH_4^+ + HO^(-)`..

for which...`K_b=10^(-4.76)=([NH_4^+][HO^-])/([NH_3])`

And so at equilibrium...if `x*mol*L^-1` ammonia associates...we gots...

`10^(-4.76)=(x^2)/(0.025-x)`

And so `x=sqrt(10^(-4.76)xx(0.025-x))`

And so `x~=sqrt(10^(-4.76)xx0.025)`...if `0.025">>"x`

Well, we solve for `x_1=6.59xx10^-4*mol*L^-1`

And now we have an approx. for `x`, we can plug this value back into the expression, and see how the value evolves...

`x_2=6.50xx10^-4*mol*L^-1`

And on the third approx., the values converge...

`x_3=6.50xx10^-4*mol*L^-1=[HO^-]`...`pH=10.81`