Answer:
[tex]pH=11.12[/tex]
Explanation:
Hello,
In this case, ammonia dissociation is:
[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]
So the equilibrium expression:
[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
That in terms of the reaction extent and the initial concentration of ammonia is written as:
[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]
Thus, solving by using solver or quadratic equation we find:
[tex]x=0.00133M[/tex]
Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:
[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]
And the pH from the pOH is:
[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]
Best regards.
