Question

a.) How many milliliters of 1.0 M benzoic acid need to be added to 550 mL of a solution already containing 6.00 grams of sodium benzoate, if you want the final pH to be 4.15? Ka for benzoic acid = 6.28x10^-5. b.) If you add 6.00 grams of sodium benzoate to 14.55 grams of benzoic acid, what is the pH of this solution? Is this a buffer, why or why not? c.) How many grams of potassium nitrite need to be added to 250 mL of a 0.1234 M HNO2 solution to produce a final buffer at pH = 3.65. Ka for HNO2 = 4.5x10^-4.

Answer 1

a) K_a for benzoic acid is K_a = 6.28*10^-5; pK_a = -log (K_a) = -log (6.28*10^-5) = 4.20.

Molar mass of sodium benzoate, C₆H₅COONa = (7*12.01 + 5*1.008 + 2*15.9994 + 1*22.9897) g/mol = 144.0985 g/mol.

Mole(s) of sodium benzoate corresponding to 6.00 g = (6.00 g)/(144.0985 g/mol) = 0.04164; molar concentration of sodium benzoate = (0.04164 mole)/[(550 mL)*(1 L/1000 mL)] =0.07571 mol/L ≈ 0.07571 M.

Use the Henderson-Hasslebach equation.

pH = pK_a + log [sodium benzoate]/[benzoic acid]

====> 4.15 = 4.20 + log (0.07571 M)/[benzoic acid]

====> -0.5 = log (0.07571)/[benzoic acid]

====> (0.07571 M)/[benzoic acid] = antilog (-0.5) = 0.31623

====> [benzoic acid] = (0.07571 M)/(0.31623) = 0.23941 M

The final concentration of benzoic acid in the buffer is 0.23491 M. Use the dilution equation to find the milliliters of 1.0 M benzoic acid required.

M₁*V₁ = M₂*V₂ where M₁ = 1.0 M; M₂ = 0.23941 M and V₂ = 550 mL; therefore,

(1.0 M)*V₁ = (0.23941 M)*(550 mL)

====> V₁ = (0.23941 M)*(550 mL)/(1.0 M) = 131.6755 mL ≈ 131.67 mL (ans).

b) Molar mass of benzoic acid, C₆H₅COOH = (7*12.01 + 6*1.008 + 2*15.9994) g/mol = 122.1168 g/mol.

Mole(s) of benzoic acid corresponding to 14.55 g = (14.55 g)/(122.1168 g/mol) = 0.11915 mole.

Molar concentration of benzoic acid = (0.11915 mole)/[(550 mL)*(1 L/1000 mL)] = 0.21664 M.

Use the Henderson-Hasslebach equation

pH = pK_a + log [sodium benzoate]/[benzoic acid] = 4.20 + log (0.07571 M)/(0.11915 M) = 4.20 + log (0.63542) = 4.20 + (-0.19694) = 4.00 (ans).