Question

NAME _____________________________                                                                               Quiz2 Ch 16

1) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution before the addition of any KOH. The Ka of HF is 3.5 x 10-4.

A) 4.15

B) 0.70

C) 2.08

D) 3.46

E) 1.00

2) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 × 10-4.

A) 2.08

B) 3.15

C) 4.33

D) 3.46

E) 4.15

3) A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 100.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

A) 6.58     B) 10.56

C) 8.72

D) 3.44

E) 5.28

4) A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

A) 4.74

B) 7.78

C) 7.05

D) 9.26

E) 10.34

5) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 300.0 mL of KOH. The Ka of HF is 3.5 × 10-4.

A) 12.40   

B) 9.33  

C) 5.06

D) 8.94

E) 12.00

Answer 1

1) [H+] = (Ka*C)^1/2
   [H+] = (3.5*10^-4 * 0.20)^1/2
   [H+] = 8.4*10^-3
Ph = -log[H+]
Ph = -log[8.4*10^-3]
Ph = 2.08
option C is correct

2)
no.of moles of HF = 0.20*0.1 lt = 0.02 moles
no. of moles of KOH = 0.1*0.1lt = 0.01 moles

0.01 moles of OH- will react with 0.01 moles of HF generating 0.01 moles of F- and leaving 0.01(0.02-0.01) moles of HF

Ph = Pka + log(0.01/0.01)
Pka = -log(Ka) = -log(3.5*10^-4)
Pka = 3.46

   Ph = 3.46 + 0
   Ph = 3.46
so option D is correct