Question

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to make a buffer with a pH of 5.00. Assume no volume change with addition of the solid. ( Ka of acetic acid = 1.8 x 10 -5 )

Answer 1

Answer: The mass of sodium acetate that must be added is 30.23 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$  

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$  

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams