Question

Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine the pH of the solution after the addition of the following volumes of titrant.

a. 5.00 mL

b. 10.00 mL

c. 15.00 mL

d. 20.00 mL

e. 25.00 mL

Answer 1

millimoles of NH3 = 20 x 0.1 = 2

kb= 1.8x10^-5

pKb = -logKb = -log (1.8x10^-5) = 4.74

a) after the addition of 5.00 mL HI

millimoles of HI = 5 x0.1 = 0.5

NH3 + HI ----------------------> NH4+I-

2 0.5 0

1.5 0 0.5

pOH = pKb + log [salt / base]

= 4.74 + log [0.5 / 1.5]

= 4.26

pH = 9.73

b) after the addition of 10 mL HI

millimoles of acid = 10 x 0.1 = 1

it is half - equivalence point .so here

pOH = pKb

pOH = 4.74

pH = 9.26

c) after the addition of 15 mL HI

millimoles of HBr = 15 x0.1 = 1.5

NH3 + HI ----------------------> NH4+I-

2 1.5 0

0.5 0 1.5

pOH = pKb + log [salt / base]

= 4.74 + log [1.5 / 0.5]

= 5.22

pH = 8.78

d) after the addition of 20 mL HI

millimoles of HI = 20 x 0.1 = 2

it is equivalence point . so here salt only remained. so

salt concentration = millimoles / total volume = 2 / (20 + 20) = 0.05 M

salt is from strong acid weak base so pH <7

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

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

pH = 5.28

e) after the addition of 25 mL HI

millimoles of HBr = 25 x 0.1 = 2.5

NH3 + HI ----------------------> NH4+I-

2 2.5 0

0 0.5 2

strong acid remained in the solution

[H+] = 0.5 / (25+20) = 0.011 M

pH = -log(0.011)

pH = 1.95