Question

Suppose you are titrating 50.00 mL of 0.1752 M acetic acid (CH3COOH; Ka = 1.76 × 10^-5) with 0.1998 M sodium hydroxide (NaOH). Calculate the pH of a titree solution when:

(a) 15.25 mL of NaOH is added. (3 pts)
(b) the equivalent volume (equivalence point) of NaOH is added. (4 pts)

(c) a half of the equivalent volume is added (3 pts)
(d) 46.00 mL of NaOH is added. (3 pts)

Answer 1

Henderson–Hasselbalch equation for weak acid/base is given by

pH= pKa +log [A-]/[HA], [HA] is the molarity of undissociated weak acid, [A⁻] is the molarity of this acid's conjugate base, pH = -log[H3O+]

Acetic acid exists in following equilibrium,

CH3COOH <-==> CH3COO- + H3O+,

Ka = [H3O+][CH3COO-] / [CH3COOH] = 1.76 x 10^-5.

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A) On adding 15.25 mL NaOH,

No. of moles of acetic acid in 50.0 mL, 0.1752 M solution = 0.1752 mol/L x 0.050 L = 0.00876 mol

No. of moles of NaOH in 15.25 mL, 0.1998 M Solution = 0.1998 mol/L x 0.01525 L = 0.00305 mol

On adding NaOH (strong base), following reaction takes place,

CH3COOH + NaOH ==> CH3COONa + H2O

[CH3COOH] = 0.00876 - 0.00305 = 0.00571 mol

[CH3COO-] = 0.00305 mol

pH of this solution can be calculated by Henderson–Hasselbalch equation

pH= pKa +log [A-]/[HA]

pKa = -log(Ka) = -log(1.76 x 10^-5) = 4.75

It is not required to convert mole to molarity as in both numerator and denominator we use same unit.

pH = 4.75 + log(0.00305/0.00571) = 4.4776 = 4.48

pH = 4.48

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B) the equivalent volume (equivalence point) of NaOH is added.

On adding equivalent volume of NaOH, i.e. 0.00876 mol of NaOH, neutralization takes place and all of the CH3COOH ( 0.00876 mol) will be converted to CH3COONa (0.00876 mol)

Then the pH of the solution is determined by the dissociation of the conjugate base, CH3COONa

CH3COO- + H2O <==> CH3COOH + OH-

Which will have Kb = [CH3COOH][OH-]/CH3COO-]

Since Ka.Kb = Kw = 1 x 10^-14

Kb = Kw/Ka = 1 x 10^-14/1.76 x 10^-5 = 5.68 x 10^-10

Volume of NaOH added = 50.0 mL x 0.1752 M / 0.1998 M = 43.84 mL

Total volume of solution = 50.0 mL + 43.84 mL = 93.84 mL = 0.09384 L

No. of moles of CH3COO- = 0.00876 mol

Molarity of [CH3COO-] ions = 0.00876 mol/ 0.09384 L = 0.09335 M

If x fraction of CH3COO- is reacted with water to form CH3COOH and OH-, then

[CH3COO-] = 0.09335 -x

[CH3COOH] = x

[OH-] = x

then, kb = (x) .(x) /(0.09335-x) = x^2/0.09335-x = 5.68 x 10^-10

Solving x = 7.28 × 10^-6

[OH-] = 7.28 × 10^-6 M

pOH = -log[OH-] = -log 7.28 × 10^-6 = 5.138

Since pH + pOH = 14

pH = 14 - 5.138 = 8.862

pH = 8.86

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C) half of the equivalent volume is added

No. of moles of acetic acid in 50.0 mL, 0.1752 M solution = 0.00876 mol

No. of moles of NaOH added at half of equivalence point = 0.00876/2 = 0.00438 mol

On adding NaOH (strong base), following reaction takes place,

CH3COOH + NaOH ==> CH3COONa + H2O

[CH3COOH] = 0.00876 - 0.00438 = 0.00438 mol

[CH3COO-] = 0.00438 mol

pH= pKa +log [A-]/[HA]

pH = 4.75 + log(0.00438/0.00438) = 4.75

pH = 4.75

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D) 46.00 mL of NaOH is added

No. of moles of acetic acid in 50.0 mL, 0.1752 M solution = 0.00876 mol

No. of moles of NaOH in 46.00 mL, 0.1998 M Solution = 0.1998 mol/L x 0.046 L = 0.009191 mol

Here the No. of moles of NaOH is more than that of CH3COOH, hence NaOH is added in excess.

Excess NaOH = 0.009191 - 0.00876 = 0.000431 mol.

Since kb of CH3COONa is   5.68 x 10^-10, the contribution from CH3COONa will be negligible

Total volume of solution = 50 mL + 46 mL = 96 mL = 0.096 L

Molarity of [OH-] = 0.000431 mol/0.096L = 4.4896 x 10^-3 M

pOH = -log 4.4896 x 10^-3 = 2.348

pH = 14 - 2.348 = 11.652

pH = 11.65