7.00 5 x 4,4 x10 19, Buff and Titrations ers 19. A 0 mL solution of 0.40 MCHacoo

Question

7.00 5 x 4,4 x10 19, Buff and Titrations ers 19. A 0 mL solution of 0.40 MCHacooH(aq) and 0.40 M Ni has 20.0 mL of 1.0 M HCIOaq) added to it. What is the pH after the HCI has been added? (Ka 1.8x 10 for CHscooH) a) 4.7 4.5 4.9 d) 7.0 e) 9.5 20. Consider the titration of 40.0 mL of 0.100 M acetic acid (CH3CooH, 1.8 x 10) with 0.150 M NaoH. What is the pH after the addition of 15.0 mL of NaOH? a) 4.74 b) 4.92 c) 4.49 4.64 e) 4.85 21. Consider the titration of 50.00 mL of 0.100 M hydrofluoric acid (HF, Ka 6.76 x 104) with o 125 M NaoH. What is the pH at the equivalence point? 7.00 c) 3.17 d) 10.83 e) 12.74 AG" 4.73 ku/mol at 25 C

Explanation / Answer

pH = pKa + log( Base/acid) ....(Henderson-Hasselbalch equation)
in initial 200 ml of 0.4 M solution of CH3COOH and same 0.4 M CH3COO are present mean acid base ratio is 1:1

but after addition of 20 ml of 1.0 M HCl acid will increse and base will decrease

moles of base = ( 0.4 M x 0.200 L ) = 0.08 moles

moles of acid (CH3COOH) = (0.4 M x0.2 L) = 0.08 moles

20 ml 1.0 M HCl moles = (0.02 L x 1.0) = 0.02 moles

base = 0.08 - 0.02 = 0.06 moles

acid = 0.08 + 0.02 = 0.1 moles

pH = -log(1.8 x10-5 ) - log(base / acid)

pH = 4.74 - log(0.06 / 0.1)

pH = 4.51

20) Calculate moles of acid and base in solution before reaction:

CH3COOH: (0.04 L x 0.1 M) = 0.004 mol
NaOH: (0.015 L) (0.150 mol/L) = 0.00225 mol

Determine amounts of acid and acetate ion after reaction:

CH3COOH: 0.004 mol - 0.00225 mol = 0.00175 mol
CH3COONa: 0.00225 mol

Use Henderson-Hasselbalch equation to determine pH of (now) buffered solution:

pH = pKa + log (base/acid)

pH = 4.74 + log (0.00225/0.00175)

pH = 4.85

21)

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

molarity of NaOH: 0.125 M

Calculate the Kb of sodium acetate:

Kw = KaKb

1.00 x 10-14 = (6.76 x 10-4 ) (x)

x = 1.48 x 10-9

Calculate pH of the solution:

1.48 x 10-9 = [(x) (x)] / 0.125

x = 1.3 x 10-6 M (this is the hydroxide ion concentration)

pOH = -log(OH-)

pOH = 5.9

pH = 14 - pOH

pH = 8.1 ~ 7.96

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