Question

Calculate the mass of solid sodium acetate and how much of 1.0 M of acetic acid required to mix with 100.0 mL to prepare a pH 4 buffer and total common ion concentration= 0.200 M (Show all work). 1. How many grams of solid sodium acetate is needed? 2. How much mL of 1.0 M acetic acid is needed?

Answer 1

From the Henderson-hasselbalch equation.

pH=pka +log[Salt ]/[Acid]

we know that pKa of acetic acid = 4.76 and required pH = 4

If substituted, 4 = 4.76 + log[Sodium acetate]/[Acetic acid]

log[Sodium acetate]/[Acetic acid] = -0.76

[Sodium acetate]/[Acetic acid] = 10^-0.76

[CH3COONa] = 10^-0.76 x [CH3COOH], this relation we can rewrite as

[CH3COO-] concentration in CH3COONa =  10^-0.76 times that of in CH3COOH

Let us assume [X] = Concentration of [CH3COO-] ion in CH3COONa

                        [Y] = Concentration of [CH3COO-] ion in CH3COOH

So, we can say [X] = 10^-0.76 [Y]

We know that the Total Common ion concentration = 0.2 M = [X]+[Y]

10^-0.76 [Y] + [Y] = 0.2 , if solved we will get [Y] = 0.1704 M

So [X] = 0.2 - 0.1704 = 0.0296 M

1) 1. How many grams of solid sodium acetate is needed?

[X] = Concentration of [CH3COO-] ion in CH3COONa = 0.0296 moles/Litre

0.0296 mole/L = (weight in g / 82.03g/mole) / 0.1 L

Weight in g = 0.2428 g

2. How much mL of 1.0 M acetic acid is needed?

    [Y] = Concentration of [CH3COO-] ion in CH3COOH = 0.1704 M

     Let us use M1V1 = M2V2 Formula

    Then V1 = 0.1704 x 100 / 1.0 = 17.04 ml