All three questions appear to be almost identical, but I can\'t figure out how t

Question

All three questions appear to be almost identical, but I can't figure out how to do them.

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 200.0 mL of KOH. The K_a of HF is 3.5 times 10^-4. A) 7.00 B) 3.46 C) 10.54 D) 9.62 E) 8.14 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 K_a of HF is 3.5 times 10^-4. A) 5.06 B) 12.00 C) 8.94 D) 9.33 E) 12.40 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 400.0 mL of KOH. The K_a of HF is 3.5 times 10^-4. A) 12.60 B) 13.85 C) 12.78 D) 12.30 E) 13.08

Explanation / Answer

28)Given:

M(HF) = 0.20 M

V(HF) = 100.00 mL

M(KOH) = 0.10 M

V(KOH) = 200.00 mL

mol(HF) = 0.20 * 100.00 = 20.00 mmol

mol(KOH) = 0.10 * 200.00 = 20.00 mmol

20.00 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 20.00 mmol

Volume of Solution = 100.00 + 200.00 = 300.00 mL

Kb of A- = Kw/Ka = 1.0E-14/3.5E-4 = 2.86E-11

concentration ofA-,c = 20.00/300.00 = 0.0667 M

for simplicity lets write weak base as A-

A- + H2O -----> AH + OH-

0.0667 0 0

0.0667-x x x

Kb = [AH][OH-]/[B-]

Kb = x*x/(c-x)

since kb is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.86E-11)*0.0667) = 1.38E-6

[OH-] = x = 1.38E-6 M

pOH = -log [OH-] = -log (1.38E-6) = 5.86

PH = 14 - pOH = 14 - 5.86 = 8.14

PH = 8.14

Answer: E

i am allowed to answer only 1 question at a time

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