Equation is as under:
pOH = pKb + log ([salt]/[base])
Now:
We have to find pKb for NH3:
As,
pKb = - log [Kb]
pKb of NH3 = -log Kb = -log (1.8*10^-5) = 4.74
Volume of final solution = 200+250 = 450mL
As the equation of molarity is M1V1 = M2V2
So,
M1V1 = M2V2
V1 = 450mL
V2 = 250mL
M2 = 0.15 M
Putting all values into equation:
M1*450 = 0.15*250
M1 = 0.15*250/450
M1 = 0.0833M
Now we have to find molarity of [NH3]:
[NH3]
M1V1 = M2V2
V1 = 50 mL
V2 = 200mL
M2 = 0.12M
Putting all the values again in equation:
M1*50 = 0.12*200
M1 = 0.12*200/450
M1 = 0.05333
Using the equation pOH = pKb + log ([salt]/[base])
pkb = 4.74
salt =0.0833 M
base = 0.0533M
Substitute:
pOH = 4.74 + log(0.0833/0.0533)
pOH = 4.74 + log 1.563
pOH = 4.74 + 0.19
pOH = 4.93
pH = 14.00-pOH
pH = 14.00-4.93
pH = 9.07
