20.00 mL of an unknown concentration of Acetate (C2H3O2-) is titrated to its equ

Question

20.00 mL of an unknown concentration of Acetate (C2H3O2-) is titrated to its equivalence point with 20.00 mL of 0.100 M HCl. The pka of acetic acid = 4.76. Draw a titration curve and answer the following questions:

g. What is the concentration of Acetate in the solution?

h. What is the pH of the acetate solution before the addition of any HCl?

i. What is the pH of the solution after the addition of 5.00 mL of HCl?

j. What is the pH of the solution after the addition of 15.00 mL of HCl?

k. What is the pH at the equivalence point?

l. What is the pH after the addition of 30.00 mL of HCl?

Explanation / Answer

(g) M1 V1 = M2V2

where M1 and V1 is the concentration od volume of acetate. m2 and V2 is the concentration and volume of acetate.

Concentration of acetate = 20mL * 0.1M/20mL = 0.1M

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(h) CH3COO- + H2O <==> CH3COOH + OH-

pKb of this reaction = pKw-pKa = 14-4.76 = 9.24

Kb = 10^-9.24 = 5.75*10^-10

Kb = [CH3COOH][OH-]/[CH3COO-]

5.75*10^-10 = x*x/0.1-x

as x is very small, 0.1-x~0.1

So, x = 7.59*10^-6 M

[OH-] = 7.59*10^-6 M

pOH = 5.12, pH = 8.88

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when HCl is added it will react with acetate to convert it to acetic acid. Acetic and acetate together will form a buffer. pH of this buffer can be calculated by using hinderson Hasselbalch equation :

pH = pKa + log [acetate/acetic acid]

moles of acetate innitially present = 0.02 L * 0.1 M = 0.002 moles

Moles of acetic acid formed = moles of acid added = 0.005 L * 0.1 = 0.0005 moles

Moles of acetate present after adding HCl = 0.002-0.0005 = 0.0015 moles

pH = 4.76 + log [0.0015/0.0005] = 5.24

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moles of acetate innitially present = 0.02 L * 0.1 M = 0.002 moles

Moles of acetic acid formed = moles of acid added = 0.015 L * 0.1 = 0.0015 moles

Moles of acetate present after adding HCl = 0.002-0.0015 = 0.0005 moles

pH = 4.76 + log [0.0005/0.0015] = 4.28

CH3COO- CH3COOH OH- initial 0.1 0 0 change -x +x +x equilibrium 0.1-x x x

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