107 760
Assignments Done
100%
Successfully Done
In May 2024

# Answer to Question #349029 in Organic Chemistry for Sibusiso

Question #349029

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

1
2022-06-08T17:02:03-0400

Solution:

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

where pKa 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

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!