Question

You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 4.90 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answer 1

1)

9.0075g

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where

A^- = conjucate base, CH3COO^-

HA = weak acid , CH3COOH

Total acetate concentration of the buffer = [CH3COO^-] + [CH3COOH] = 0.300M

Volume of buffer = 500ml

molarity is defined as the number of moles of solute per liter of solution

total number of moles of acetate = (0.300mol/1000ml) ×500ml = 0.150mol

moles of CH3COOH required = 0.150mol

mass = number of moles × molar mass

mass of CH3COOH required = 0.150mol × 60.05g/mol = 9.0075g