Question

A 25.0 mL sample of .100 M NH3 (Kb = 1.8 x 10^-5) was titrated with .200 M HCl. Calculate the pH after addition of: a. 0.00 b. 6.25 c. 9.50 d. 12.5 e. 15.0 If I asked to many questions, please just post for c.

Answer 1

c)

Given:

M(HCl) = 0.2 M

V(HCl) = 9.5 mL

M(NH3) = 0.1 M

V(NH3) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.2 M * 9.5 mL = 1.9 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCl) = 1.9 mmol

mol(NH3) = 2.5 mmol

1.9 mmol of both will react

excess NH3 remaining = 0.6 mmol

Volume of Solution = 9.5 + 25 = 34.5 mL

[NH3] = 0.6 mmol/34.5 mL = 0.0174 M

[NH4+] = 1.9 mmol/34.5 mL = 0.0551 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.7447+ log {0.0551/0.0174}

= 5.2453

use:

PH = 14 - pOH

= 14 - 5.2453

= 8.75

PH = 8.75