Question

What mass of sodium benzoate should be added to 160.0 ml of a 0.17 m benzoic acid solution in order to obtain a buffer with a ph of 4.30?

Answer 1

Answer :

The mass of sodium benzoate that can be added to benzoic acid is 5.07 g

Explanation :

The problem can be solved with the help of Henderson-Hasselbalch equation.

Step 1 : Find concentration of sodium benzoate

The equation is given below.

$p^{H} = p^{Ka} + log \frac{[Base]}{[Acid]}$

We have been given,

pH = 4.30

[Acid] = 0.17 M

Ka of benzoic acid can be found out using standard reference table which is 6.5 x 10⁻⁵

pKa = - log (Ka)

pKa = - log (6.5 x 10⁻⁵)

pKa = 4.19

Let us plug in the above values in Henderson equation.

$4.30 = 4.19 + log \frac{[Base]}{(0.17M)}$

$0.11 = log \frac{[Base]}{0.17}$

$10^{0.11} = 10^{log \frac{[Base]}{0.17}}$

The functions $10^{log}$ cancel out each other.

$10^{0.11} = \frac{[Base]}{0.17}$

$1.29 = \frac{[Base]}{0.17}$

[Base] = 0.22 M

The base in this case is sodium benzoate.

Therefore concentration of sodium benzoate is 0.22 M.

Step 2 : Find moles of sodium benzoate

The volume of the solution is 160 mL.

Volume in L = $160 mL\times \frac{1L}{1000mL}= 0.16 L$

Moles of sodium benzoate can be found as,

$mole=concentration \times volume(L)$

$mole=0.22 mol/L \times 0.16 L$

$mole=0.0352 mol$

Moles of sodium benzoate are 0.0352

Step 3 : Find mass of sodium benzoate

Mass of sodium benzoate can be calculated as

Mass in grams = Moles x Molar mass

The formula of sodium benzoate is C₇H₅O₂Na.

Molar mass of sodium benzoate = 7 x 12.01 + 5 x 1.01 + 2 x 16 + 22.98

Molar mass of sodium benzoate = 144.1

Mass of sodium benzoate = $0.0352 mol \times 144.1 \frac{g}{mol}$

Mass of sodium benzoate = 5.07 g

5.07 grams of sodium benzoate should be added to the given volume of benzoic acid.