Answer to Question #349029 in Organic Chemistry for Sibusiso

Question

A 400 mL buffer solution was prepared by reacting sodium acetate, CH₃COONa, with 1.00 M acetic acid, CH₃COOH, with a pH of 4.56 Calculate the mass of CH₃COONa required to prepare the solution. Ka CH₃COOH= 1.75 x 10^-5  

Answer 1

Solution:

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution from the concentration of the buffer components:

where pK_a is the acid dissociation constant and [base] and [acid] are the base and acid concentrations, respectively.

In this case, the base is sodium acetate (CH₃COONa) and the acid is acetic acid (CH₃COOH).

Therefore,

4.56 = −log(1.75×10⁻⁵) + log([CH₃COONa] / 1.00)

log([CH₃COONa]) = −0.19696

[CH₃COONa] = 10^−0.19696 = 0.635

Molarity of CH₃COONa = 0.635 M

Moles of solute = Molarity × Liters of solution

Therefore,

Moles of CH₃COONa = (0.635 M) × (400 mL) × (1 L / 1000 mL) = 0.254 mol

The molar mass of sodium acetate (CH₃COONa) is 82.0 g/mol

Therefore,

(0.254 mol CH₃COONa) × (82.0 g CH₃COONa / 1 mol CH₃COONa) = 20.8 g CH₃COONa

Answer: 20.8 grams of CH₃COONa required to prepare the solution