Question

1. A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Kb, NH3 = 1.8 × 10−5. Determine the pH of the solution at each of the following points in the titration:

(a) before addition of any HNO3

(b) after the addition of 50.0 mL HNO3

(c) after the addition of 75.0 mL HNO3

(d) at the equivalence point

(e) after the addition of 150.0 mL HNO3

2. (12 points) Consider the titration of 37.0 mL of 0.200 M ammonia, NH₃ (K_b = 1.8 × 10−5) with 0.130 M HCl

(a) Calculate the volume (mL) of HCl required to reach the equivalence point.

(b) Calculate the Ka of the conjugate acid, NH₃+

(c) Calculate the pH at each of the following points:

i. initial pH

ii. halfway to the equivalence point

iii. at the equivalence point

iv. after the addition of 20.0 mL excess HCl

Plot the points from part (c) on a graph. Please label each point with volume and pH. Then sketch the shape of the titration curve. ​

Answer 1

1. :

(a) before addition of any HNO3

pOH = 1/2 [pKb - logC]

pOH = 1/2 [4.74 -log0.10]

pOH   = 2.87

pH + pOH = 14

pH = 11.13

(b) after the addition of 50.0 mL HNO3

it is half equivalence point here pOH = pKb

pOH = 4.74

pH = 9.26

(c) after the addition of 75.0 mL HNO3

millimoles of NH3 = 100 x 0.1 = 10

millimoles of HNO3 = 75 x 0.1 = 7.5

NH3    + HNO3 -----------------------> NH4+

10            7.5                              0

2.5              0                             7.5

pOH = 4.74 + log (7.5 / 2.5)

pOH = 5.22

pH = 8.78

(d) at the equivalence point

here only salt remains

salt concentraiton = 10 / 200 = 0.05 M

pH = 7 - 1/2 [pKb + logC]

pH = 7 - 1/2 [4.74 + log 0.05]

pH = 5.28

(e) after the addition of 150.0 mL HNO3

millimoles of acid remains = 15 - 10 = 5

[H+] = 5 / 150 + 100 = 0.02 M

pH = -log [H+] = -log (0.02)

pH = 1.70