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