Question

What mass of sodium hydroxide must be added to 75.0 mL of 0.205 M acetic acid in order to create a buffer with a pH of 4.74? acetic acid Ka=1.8x10^-5

(Answer is 0.31g how do we arrive there?)

Answer 1

Ans. Step 1: Calculate pKa of acid

            pKa = -log Ka = -log (1.8 x 10^-5) = 4.74

Step 2: Calculate relative concertation of weak acid and its conjugate base

Using Henderson- Hasselbalch equation –

            pH = pKa + log ([A^-] / [AH])                    - equation 1

            where, A^- = conjugate base (CH3COONa or simply CH3COO^-)

                        AH = Weak acid (CH3COOH)

Putting the values in equation 1-

            4.74 = 4.74 + log ([A^-] / [AH])

            Or, 4.74 – 74.74 = log ([A^-] / 0.100)

            Or, [A^-] / [AH] = antilog 0 = 1

            Or, [A^-] / [AH] = 1

            Hence, [A^-] = [AH] = 1

That is, at pH 4.74 both the acid and its conjugate base are at same concentration.

Now,

Let [A^-] = [AH] = X

So, total concentration of [A^-] and [AH] at equilibrium is-

[A^-] + [AH] = 2X

Given, total initial [AH] = 0.205 M

It is the initial [AH] that dissociates to form equilibrium [A^-]. The remaining un-dissociated initial [AH] (i.e. total initial [AH] – dissociated [AH] = [AH] at equilibrium)] is equal to [AH at equilibrium.

So,

            ([A^-] + [AH]) at equilibrium = Total initial [AH]

            Or, 2X = 0.205 M

            Hence, X = 0.205 M/ 2 = 0.1025 M

That is, at equilibrium, [A^-] = [CH3COONa] = 0.1025 M

Step 3: Calculating required amount of NaOH.

Balanced reaction:    CH₃COOH + NaOH ---------> CH₃COONa + H₃O⁺

Stoichiometry: 1 mol NaOH reacts with 1 mol CH3COOH to form 1 mol CH3COONa.

Moles of CH3COONa at equilibrium = Molarity x Volume of solution in liters

                                                = 0.1025 M x 0.075 L                       ; [1 L = 10³ mL]

                                                = (0.1025 mol/ L) x 0.075 L             ; [1 M = 1 mol/ L]’

                                                = 0.0076875 mol

Note the stoichiometry of balanced reaction, 1 mol NAOH produces 1 mol CH3COONa.

So,

Moles of NaOH required = 0.0076875 mol = Moles of CH3COONa at equilibrium

Mass of NaOH required = Moles x Molar mass

                                                = 0.0076875 mol x (40.00 g/mol)

                                                = 0.3075 g

                                                = 0.31 g