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" " " "15M" " " " " " " " " " " "0M" " " " " "0M
"Change" " " " "-x" " " " " " " " " " " "+x" " " " " "+x
Equilibrium" "(15-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))/((15-x))=1.8xx10^(-5)
since the value of K_b value is small, we consider x"<<"15
Solving for x, x=1.64xx10^(-2)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))/(1.64xx10^(-2))=6.09xx10^(-13)M
Then, the pH is calculated by:
pH=-log([H^+]) => pH=-log(6.09xx10^(-13))
color(blue)(pH=12.22)
