Question

Consider the titration of 20.00 mL of 0.754 M sodium benzoate with a solution of 0.525 M nitric acid. a. Calculate the equivalence volume in mL. b. Calculate the pH at the equivalence point. c. Calculate the pH of the solution after addition of 16.20 mL nitric acid. d. Calculate the pH of the solution after addition of 39.82 mL nitric acid.

Answer 1

a. 28,7 mL

b. pH = 2,16

c. pH = 4,09

d. pH = 1,01

Step-by-step explanation:

a. The titration of sodium benzoate with nitric acid is:

C₆H₆COONa + HNO₃ → C₆H₆COOH + NaNO₃

The moles of sodium benzoate (NaBz) are:

0,0200L ×
`(0,754mol)/(1L)` = 0,01508mol of NaBz

As molar ratio in the titration is 1:1, moles of nitric acid you need to add at the equivalence point are 0,01508mol, in mL:

0,01508mol×
`(1L)/(0,525mol)` = 0,0287L = 28,7 mL

b. In equivalence point, all 0,754M of NaBz were converted in benzoic acid.

As acidic equilibrium of benzoic acid (Bz) is:

C₆H₆COOH ⇄ C₆H₆COO⁻ + H⁺; pka = 4,20; ka = 10^-4,20; ka = [C₆H₆COO⁻][H⁺]/[C₆H₆COOH]

In equilibrium:

[C₆H₆COOH] = 0,754-x

[C₆H₆COO⁻] = x

[H⁺] = x;

Replacing in ka:

`10^(-4,20) = (x^2)/(0,754-x)`

4,757x10⁻⁵ - 6,31x10⁻⁵x = x²

x² + 6,31x10⁻⁵x - 4,757x10⁻⁵ = 0

Solving for x:

x = -0.00692872 No physical sense.

x = 0.00686562 Real answer.

As pH = -log [H⁺]; pH = 2,16

c. The moles of nitric acid in 16,20mL are:

0,01620L×0,525M = 8,505x10⁻³ moles, that are the same moles of benzoic acid. Thus, moles of NaBz are:

0,01508mol - 8,505x10⁻³ moles = 6,575x10⁻³ moles

Using Henderson-Hasselbalch formula:

pH = pka + log [C₆H₆COONa] / [C₆H₆COOH]

pH = 4,20 + log 6,575x10⁻³/8,505x10⁻³

pH = 4,09

d. After addition of 28,7mL of nitric acid, the additional volume will produce free H⁺. Additional volume is: 39,82mL - 28,7mL = 11,12mL. In moles of HNO₃ ≡ moles of H⁺:

0,01112L×
`(0,525mol)/(1L)` = 5,838x10⁻³moles of H⁺. The volume is 20,0mL + 39,82mL = 59,82mL ≡ 0,05982L. Thus, [H⁺] is:

5,838x10⁻³moles / 0,05982L = 0,09759 M. pH = -log [0,09759] = 1,01

I hope it helps!