Question

A 1.944 g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES-K+, MW = 220.29 g/mol) is dissolved in 52.38 mL of water. To the solution, 27.66 mL of HCl is added, resulting in a pH of 6.95. Calculate the concentration of the HCl solution. The pK, of ACES is 6.85. (HCI) = 3.80 x10-6

Answer 1

[HCl] = 0.1412 M

Explanation

mass ACES^-K⁺ = 1.944 g

moles ACES^-K⁺ = (mass ACES^-K⁺) / (molar mass ACES^-K⁺)

moles ACES^-K⁺ = (1.944 g) / (220.29 g/mol)

moles ACES^-K⁺ = 0.00882473 mol

Let moles HCl added = x

HCl will convert ACES^-K⁺ salt to ACES

moles ACES formed = moles HCl added = x

moles ACES^-K⁺ remaining = (initial moles ACES^-K⁺) - (moles HCl added)

moles ACES^-K⁺ remaining = (0.00882473 mol) - (x)

moles ACES^-K⁺ remaining = 0.00882473 mol - x

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles ACES^-K⁺ remaining / moles ACES formed)

6.95 = 6.85 + log((0.00882473 mol - x) / x)

log((0.00882473 mol - x) / x) = 6.95 - 6.85

log((0.00882473 mol - x) / x) = 0.10

(0.00882473 mol - x) / x = 10^0.10

(0.00882473 mol - x) / x = 1.26

(0.00882473 mol / x) - 1 = 1.26

(0.00882473 mol / x) = 1 + 1.26

(0.00882473 mol / x) = 2.26

x = (0.00882473 mol) / 2.26

x = 3.91 x 10^-3 mol

concentration HCl = (moles HCl) / (volume HCl in Liter)

concentration HCl = (3.91 x 10^-3 mol) / (0.02766 L)

concentration HCl = 0.1412 M