Start with 40.0 mL of your acid. Use your acid and base to answer the following

Question

Start with 40.0 mL of your acid. Use your acid and base to answer the following questions.

phemol = .20M

NaOh = .20M

Calculate the pH of the acid initially. 3 points

Calculate the pH ½ way to the equivalence point. 3 points

Calculate the pH 1 mL before the equivalence point. 3 points

Calculate the pH at the equivalence point. 3 points

Calculate the pH 2 mL after the equivalence point. 3 points

Graph the titration curve. Label the axes, label where there is a weak acid, weak base, strong base and a buffer. 5 points.

Explanation / Answer

Ka of phenol (C6H5OH, denote as HA) = 1.6*10-10; pKa = -log (Ka ) = -log (1.6*10-10) = 9.80

a) Concentration of HA = 0.20 M

Write down the dissociation reaction:

HA (aq) <=====> H+ (aq) + A- (aq)

The acid dissociation constant is

Ka = [H+][A-]/[HA] = (x).(x)/(0.20 – x)

===> 1.6*10-10 = x2/(0.20 – x)

Use the small x approximation, i.e, x << 0.20 and write

1.6*10-10 = x2/0.20

===> x2 = 3.2*10-11

===> x = 5.65*10-6

Thus, [H+] = 5.65*10-6 M and pH = -log [H+] = -log (5.65*10-5) = 5.25 (ans)

b) Write down the neutralization reaction.

HA (aq) + NaOH (aq) -------> NaA (aq) + H2O (l)

As per the stoichiometric equation,

1 mole HA = 1 mole NaOH

Moles of HA added = (4.0 mL)*(0.20 mol/L) = 0.80 mmole.

Since the acid is half neutralized, mmole of NaOH added = 0.40 mmole and mmole of NaA formed = 0.40 mmole; mmole of unneutralized acid = (0.80 – 0.40) mmole = 0.40 mmole

Volume of NaOH added = (0.40 mmole)/(0.20 mol/L) = 2.0 mL.

Volume of solution = (4.0 + 2.0) mL = 6.0 mL.

[HA] = (0.40 mmole)/(6.0 mL) = (0.40/6.0) M

[A-] = (0.40 mmole)/(6.0 mL) = (0.40/6.0) M

Use the Henderson-Hasslebach equation:

pH = pKa + log [A-]/[HA] = 9.80 + log [(0.40/6.0) M/(0.40/6.0) M] = 9.80 + log (1) = 9.80 (ans).

c) Volume of NaOH added to reach the equivalence point = (mmoles of HA)/(concentration of NaOH) = (0.80 mmole)/(0.20 mol/L) = 4.0 mL.

Volume of NaOH added in this particular instance = 3.0 mL; mmoles of NaOH added = (3.0 mL)*(0.20 mol/L) = 0.60 mmole.

Mmoles of HA neutralized = 0.60; mmoles of HA unreacted = 0.20 mmole; mmoles of NaA formed = 0.60 mmole.

Volume of solution = (4.0 + 3.0) mL = 7.0 mL.

[HA] = (0.20 mmole)/(7.0 mL) = (0.20/7.0) M

[A-] = (0.60 mmole)/(7.0 mL) = (0.60/7.0) M

Use the Henderson-Hasslebach equation:

pH = pKa + log [A-]/[HA] = 9.80 + log [(0.60/7.0) M/(0.20/7.0) M] = 9.80 + log (3) = 10.277 10.28 (ans).

d) Volume of NaOH added = 4.0 mL; mmoles of NaOH added = 0.80 mmole.

HA is completely neutralized and NaA is formed; mmoles of NaA formed = mmoles of NaOH added = 0.80 mmole.

Volume of solution = (4.0 + 4.0) mL = 8.0 mL.

[NaA] = (0.80 mmole)/(8.0 mL) = 0.10 M

NaA (A-) is the conjugate base of the weak acid HA and establishes equilibrium as

A- (aq) + H2O (l) ------> HA (aq) + OH- (aq)

The base constant, Kb is given as

Kb = [HA][OH-]/[A-] = Kw/Ka = (1.0*10-14)/(1.6*10-10) = 6.25*10-5 = (x).(x)/(0.10 – x)

Use small x approximation and write

6.25*10-5 = x2/0.1

===> x2 = 6.25*10-6

===> x = 2.5*10-3

Therefore, [OH-] = 2.5*10-3 M and pOH = -log [OH-] = -log (2.5*10-3) = 2.60.

We know that pH + pOH = 14; thus pH = 14 – pOH = 14 – 2.60 = 11.40 (ans).

e) Volume of NaOH added = 2.0 mL; concentration of NaOH added = 0.20 M.

Total volume of solution = 10.0 mL; concentration of NaOH in the dilute solution = (2.0 mL)*(0.20 M)/(10.0 mL) = 0.04 M

Therefore, [OH-] = 0.04 M and pOH = -log [OH-] = -log (0.04) = 1.40

Thus, pH = 14 – pOH = 14 – 1.40 = 12.60 (ans).

Now plot the graph of pH vs volume of NaOH added as shown below:

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