In aqueous solution, #NH_3# reacts with water according to the following reaction:
#" " " " " " " "NH_3(aq)+H_2O(l)->NH_4^+(aq)+OH^(-)(aq)#
#Initial" " " "0.10M" " " " " " " " " " " "0M" " " " " "0M#
#"Change" " " " "-x" " " " " " " " " " " "+x" " " " " "+x#
#Equilibrium" "(0.10-x)M " " " " " " ""+x" " " " " "+x#
The equilibrium constant is written as:
#K_b=([NH_4^+][OH^(-)])/[NH_3]=1.8xx10^(-5)#
Replacing the equilibrium concentrations by their values in the expression of #K_b#:
#K_b=((x)(x))/((0.10-x))=1.8xx10^(-5)#
since the value of #K_b# value is small, we consider #x"<<"0.10 #
Solving for #x#, #x=1.34xx10^(-3)M#
#x# represents the concentration of #OH^-#.
Using the expression of #K_w=[H^+][OH^-] =1.0xx10^(-14)#
#[H^+]=(K_w)/([OH^-])=(1.0xx10^(-14))/(1.34xx10^(-3))=7.46xx10^(-12)M#
Then, the pH is calculated by:
#pH=-log([H^+])# #=> pH=-log(7.46xx10^(-12))#
#color(blue)(pH=11.13)#
