In aqueous solution, NH_3NH3 reacts with water according to the following reaction:
" " " " " " " "NH_3(aq)+H_2O(l)->NH_4^+(aq)+OH^(-)(aq) NH3(aq)+H2O(l)→NH+4(aq)+OH−(aq)
Initial" " " "15M" " " " " " " " " " " "0M" " " " " "0MInitial 15M 0M 0M
"Change" " " " "-x" " " " " " " " " " " "+x" " " " " "+xChange −x +x +x
Equilibrium" "(15-x)M " " " " " " ""+x" " " " " "+xEquilibrium (15−x)M +x +x
The equilibrium constant is written as:
K_b=([NH_4^+][OH^(-)])/[NH_3]=1.8xx10^(-5)Kb=[NH+4][OH−]NH3=1.8×10−5
Replacing the equilibrium concentrations by their values in the expression of K_bKb:
K_b=((x)(x))/((15-x))=1.8xx10^(-5)Kb=(x)(x)(15−x)=1.8×10−5
since the value of K_bKb value is small, we consider x"<<"15x<<15
Solving for xx, x=1.64xx10^(-2)Mx=1.64×10−2M
xx represents the concentration of OH^-OH−.
Using the expression of K_w=[H^+][OH^-] =1.0xx10^(-14)Kw=[H+][OH−]=1.0×10−14
[H^+]=(K_w)/([OH^-])=(1.0xx10^(-14))/(1.64xx10^(-2))=6.09xx10^(-13)M[H+]=Kw[OH−]=1.0×10−141.64×10−2=6.09×10−13M
Then, the pH is calculated by:
pH=-log([H^+])pH=−log([H+]) => pH=-log(6.09xx10^(-13))⇒pH=−log(6.09×10−13)
color(blue)(pH=12.22)pH=12.22
