What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 72.5 mL?

Question

to 0.20 M NH3/0.17 M NH4Cl buffer solution

Answer 1

Initial solution has 0.20 M NH3 and 0.17 M NH4+
Let x=moles/L NH3 that react NH3+H2O>>NH4+ + OH-
This will give at equilibrium
NH3= 0.20 - x NH4+=0.17+x OH-=x
K=1.81 10^-5= (0.17+x)(x)/0.20-x
x=0.0000213 = OH- pOH=4.67 pH=9.33

Now we add 10 mL of 0.1M HCl which gives 0.001 mole H+.
This H+ willi be neutralized by the base NH3. This convert the 0.001 mole H+ into 0.001 mole NH4+ and uses up 0.001 mole NH3.
The initial solution contained
(0.0725)(0.2)=0.0145 mole NH3
(0.0725)(0.17)=0.0123 mole NH4+
These will be decreased and incraesed respectively to give
0.0145-0.001=0.0135 mole NH3
0.0123+0.001=0.0133 mole NH4+
The total volume has gone from 0.0725 L to 0.0825 L
The concentration are
NH3=0.0135/0.0825=0.164
NH4+=0.0133/0.0825=0.161
At equilibrium
NH3=0.164-x
NH4+=0.161+x
OH-=x
1.81 10^-5=(0.161+x)(x)/0.164-x
x=0.0000184=OH- pOH=4.73 pH=9.26