Question

1. Calculate the pH of a 0.10 M ammonia solution.

2. Calculate pH when the following volumes of 0.10 M HCl are added to 20.0 mL of 0.10 M ammonia solution. a. 5.0 mL b. 10.0 mL c. 15.0 mL d. 20.0 mL e. 25.0 mL

3. What are the major species present under the conditions in 2.

4. Sketch the titration Curve

.5. What was the pH at the equivalence point ( where equal amounts of acid and base were present)?

Step by step please and thank you!

Answer 1

1. Calculate the pH of a 0.10 M ammonia solution.

pKb of NH3 = 4.74

pOH = 1/2 [pKb - log C]

pOH = 1/2 [4.74 - log 0.1]

POH = 2.87

pH + pOH = 14

pH = 11.13

2.

a) 5.0 mL HCl

millimoles of NH3 = 20 x 0.1 = 2

millimoles of HCl = 5 x 0.1 = 0.5

NH3 + HCl -----------------> NH4Cl

2           0.5                           0

1.5        0                              0.5

pOH = pKb + log [NH4Cl / NH3]

pOH = 4.74 + log (0.5 / 1.5)

pOH = 4.26

pH + pOH = 14

pH = 9.74

b) 10 mL HCl

it is half equivalence point

here pOH = pKb

pOH = 4.74

pH = 9.26

c) 15 mL

millimoles of NH3 = 20 x 0.1 = 2

millimoles of HCl = 15 x 0.1 = 1.5

NH3 + HCl -----------------> NH4Cl

2           1.5                           0

0.5        0                              1.5

pOH = pKb + log [NH4Cl / NH3]

pOH = 4.74 + log (1.5 / 0.5)

pOH = 4.22

pH + pOH = 14

pH = 8.78

d) 20 mL HCl

here only salt NH4Cl remains

salt concentration = 20 x 0.1 / (20 + 20) = 0.05 M

pH = 7 - 1/2 [pKb + log C]

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

pH = 5.28

e) 25 mL of HCl

[H+] = 25 x 0.1 - (20 x 0.1) / (25 + 20)

         = 0.011 M

pH = -log [H+]

pH = 1.95