Question

Prepare 3 liters of 0.20M acetate buffer pH 5.00 starting from solid sodium acetate trihydrate (MW 136) and 1.0M solution of acetic acid.

Ans. Add 51.4 g of solid + 222mL of the stock and bring to total volume of 3L with water.

Answer 1

Here we have to make the buffer which should have pH = 5 and volume = 3.0 L . Solution should be of 0.20 M

Chemicals:

Sodium acetate trihydrate.

1.0 M solution of acetic acid.

Solution:

First ratio of conjugate base (CH3COO- ) to weak acid (CH3COOH) is found by using Henderson Hasselbalch equation.

We know the concentration of acetate and that is used to get the concentration of sodium acetate trihydrate.

Since the volume of the solution is 3.0 L we can get their mores by using mo0.11403larity and volume. Mass of sodium acetate trihydrate is found by using molar mass and moles. Volume of acetic acid is found by using its moles and molarity.

Ratio calculation :

pH = pka + log ([CH3COO-]/[CH3COOH])

Lets use the value of pka of acetic acid from reference table.

5.0 = 4.756 +log ([CH3COO-]/[CH3COOH])

Log (([CH3COO-]/[CH3COOH]) ) = 0.244

([CH3COO-]/ 0.20 ) = 0.57016427

[CH3COO-]= 0.114 M

Mol acetate ions = 0.114

We know mole ratio between CH3COO- to CH3COONa. 3H2O is 1 : 1

Therefore, Number of moles of sodium acetate trihydrate = 0.114 x 3.0 L = 0.3421 mol

Now calculation of mass of sodium acetate trihydrate

= 0.3421 mol x 136 g /mol

= 46.52 g