Question

Part C

Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C.

Part D

A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 ∘C.

Answer 1

C)

we have:

Molarity of HNO3 = 0.2 M

Volume of HNO3 = 50 mL

Molarity of NH3 = 0.2 M

Volume of NH3 = 50 mL

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.2 M * 50 mL = 10 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.2 M * 50 mL = 10 mmol

We have:

mol of HNO3 = 10 mmol

mol of NH3 = 10 mmol

10 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 10 mmol

Volume of Solution = 50 + 50 = 100 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 10 mmol/100 mL = 0.1 M

NH4+ + H2O -----> NH3 + H+

0.1 0 0

0.1-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.1) = 7.454*10^-6

since c is much greater than x, our assumption is correct

so, x = 7.454*10^-6 M

[H+] = x = 7.454*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (7.454*10^-6)

= 5.13

Answer: 5.13

D)

we have:

Molarity of CH3COOH = 0.5 M

Volume of CH3COOH = 30 mL

Molarity of NaOH = 0.5 M

Volume of NaOH = 30 mL

mol of CH3COOH = Molarity of CH3COOH * Volume of CH3COOH

mol of CH3COOH = 0.5 M * 30 mL = 15 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.5 M * 30 mL = 15 mmol

We have:

mol of CH3COOH = 15 mmol

mol of NaOH = 15 mmol

15 mmol of both will react to form F- and H2O

F- here is strong base

F- formed = 15 mmol

Volume of Solution = 30 + 30 = 60 mL

Kb of F- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofF-,c = 15 mmol/60 mL = 0.25M

F- dissociates as

F- + H2O -----> CH3COOH + OH-

0.25 0 0

0.25-x x x

Kb = [CH3COOH][OH-]/[F-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*0.25) = 1.179*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.179*10^-5 M

[OH-] = x = 1.179*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.179*10^-5)

= 4.9287

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.9287

= 9.07

Answer: 9.07