22. (15 points) You are Walter White. You just dissolved the corpse of a local d

Question

22. (15 points) You are Walter White. You just dissolved the corpse of a local drug lord in a polyethylene tub using hydrofluoric acid. Because you are an environmentally-conscious baddie, you want to neutralize the remaining acid solution before disposing of it. To this end, consider the titration of 2.5 L of 0.300 M HF with 0.100 M KOH. (KHF = 6.6 × 104) a) (2 points) How much solid KOH (in grams) do you need to buy to prepare a 0.100 M solution that would exactly neutralize all HF? b) (1 point) Based on the titrant and titrand volumes, what size polyethylene tub do you need for this work?

Explanation / Answer

a)

Initial moles of HF present = 2.5 L * 0.3 M

= 0.75 moles

Moles of strong base KOH required to completely neutralize HF = 0.75

Molecular mass of KOH = 56.1 g/mol

Mass of KOH required = 0.75 * MWKOH

= 0.75 * 56.1 = 42.075 g

b)

M1 * V1 = M2 * V2

0.3 M * 2.5 L = 0.1 M * Volume of KOH

Volume of KOH required = 7.5 L

c)

(i)

HF --> H+ + Cl-

0.3 –x --> x + x

KHF = [H+] [Cl-] / [HF] = 6.6 x 10-4

x2 / (0.3 – x) = 6.6 x 10-4

Solving we get, x = 0.014 M

[H+] = x = 0.014 M

pH = - log([H+]) = -log(0.014)

= 1.86

(ii)

4 L of 0.1 M KOH addition

Moles of H+ neutralized = 4 * 0.1 = 0.4 moles

Initial moles of HF = 0.75 moles

Total Volume, V1 = 2.5 L + 4 L = 6.5 L

HF --> H+ + Cl-

0.75 – x --> (x – 0.4) + x

KHF = [H+] [Cl-] / [HF] = 6.6 x 10-4

(x – 0.4)/V1 * x /V1 /((0.75 – x)/V1) = 6.6 x 10-4

(x – 0.4)/6.5 * x /6.5 /((0.75 – x)/6.5) = 6.6 x 10-4

Solving we get, x = 0.4037 moles

[H+] = (x – 0.4) / 6.5 L = 5.66 x 10-4 M

pH = - log([H+]) = -log(5.66 x 10-4)

= 3.25

iii)

7.5 L of 0.1 M KOH addition

Moles of KOH added = 7.5 * 0.1 = 0.75 moles

Initial moles of HF = 0.75 moles

Total Volume, V1 = 2.5 L + 7.5 L = 10 L

HF + KOH --> KF + H2O

Moles of KF formed = 0.75 moles

[KF] = 0.75 / 10 L = 0.075 M

As KF is a strong electrolyte, initial [F-] = 0.075 M

F- + H2O --> HF + OH-

(0.075 – y) --> y + y

Kw/KHF = y2 / (0.075 – y) = 10-14 / 6.6 x 10-4

y2 / (0.075 – y) = 1.52 x 10-11

solving we get, y = 1.07 x 10-6 M

pOH = -log([OH-]) = - log(y) = 5.97

pH = 14 – 5.97 = 8.03

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