Question

Discussion. Write out the answers to the following calculations neatly on a separate piece of paper. Staple the paper to the back of your lab report. (You can type it out if you feel like it.) (2 points each)

Assume that you dissolve 0.235 g of the weak acid benzoic acid, C₆H₅COOH, in enough water to make 1.00×10² mL of solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.

1. What is the pH of the original benzoic acid solution before the titration is started?

2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)

3. What is the pH at the equivalence point?

Answer 1

m = 0.235 g of benzoic acid

V = 100 mL = 0.1 L

titration with 0.108M of bacE:

Q1

pH original

First, assume the acid:

HBenzoic

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = mol/V = (mass/MW) / V= (0.235/122.1213) / 0.1 = 0.019243

x^2 + (10^-4.2)x - 0.019243*(10^-4.2) = 0

solve for x

x =0.00107

substitute

[H+] = 0 + 0.00107= 0.00107M

pH = -log(H+) = -log(0.00107) = 2.97

Q2

mmol of base = MV = 0.108*7 = 0.756

mmol of acid = (0.235/122.1213) * 10^3 = 1.9243

after reaction

acid left = 1.9243-0.756 = 1.1683

conjugate formed = 0.756

pH = pKa+ log(conjugate/acid)

pH = 4.20 + log(0.756/1.1683) = 4.0109

Q3

in equivalence

Vacid = mmol/M = 1.9243 / 0.108 = 17.81 mL

Vtotal = V1+V2 = 17.81 + 100 = 117.81

[Benzoate] = mmol/Vtotal = (1.9243)/ 117.81 = 0.01633

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(10^-4.20) = 1.584*10^-10

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

1.584*10^-10 = x*x/(0.01633-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =1.608*10^-6

[OH-]  =1.608*10^-6

pOH = -log(OH-) = -log(1.608*10^-6) = 5.7937

pH = 14-5.7937= 8.2063

pH = 8.2063