Calculate the ph of a solution formed by mixing 250.0 ml of 0.15 m nh4cl with 100.0 ml of 0.20 m nh3. the kb for nh3 is 1.8 x 10-5.

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

Answer is to avoid experimental error

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