Question

ssume you dissolve 0.235 g of the weak acid benzoic acid, C₆H₅CO₂H, in enough water to make 7.00 ✕ 10² mL of solution and then titrate the solution with 0.133 M NaOH.

C₆H₅CO₂H(aq) + OH^-(aq) C₆H₅CO₂^-(aq) + H₂O(ℓ)

What are the concentrations of the following ions at the equivalence point?

Na⁺, H₃O⁺, OH^-
C₆H₅CO₂^-

_____M Na⁺
_____M H₃O⁺
_____M OH^-
_____M C₆H₅CO₂^-
What is the pH of the solution?

_____

Answer 1

0.00269 M Na⁺

1.53 x 10^-8 M H₃O⁺

6.54 x 10^-7 M OH^-

0.00269 M C₆H₅CO₂^-

What is the pH of the solution? : 7.82

Explanation

mass benzoic acid = 0.235 g

moles benzoic acid = (mass benzoic acid) / (molar mass benzoic acid)

moles benzoic acid = (0.235 g) / (122.1213 g/mol)

moles benzoic acid = 1.924 x 10^-3 mol

moles NaOH = moles benzoic acid

moles NaOH = 1.924 x 10^-3 mol

volume NaOH = (moles NaOH) / (concentration NaOH)

volume NaOH = (1.924 x 10^-3 mol) / (0.133 M)

volume NaOH = 0.0144685 L

volume NaOH = 14.5 mL

Total volume at equivalence point = (volume benzoic acid) + (volume NaOH)

Total volume at equivalence point = (700 mL) + (13.9 mL)

Total volume at equivalence point = 713.9 mL

Total volume at equivalence point = 0.7139 L

equivalence concentration benzoate = (moles benzoic acid) / (Total volume at equivalence point)

equivalence concentration benzoate = (1.924 x 10^-3 mol) / (0.7139 L)

equivalence concentration benzoate = 0.00269 M

benzoic acid K_a = 6.3 x 10^-5

K_b = (K_w) / (K_a)

K_b = (1.0 x 10^-14) / (6.3 x 10^-5)

K_b = 1.6 x 10^-10

ICE table	C6H5COO- (aq)	H2O (l)	$\rightleftharpoons$	C6H5COOH (aq)	OH- (aq)
Initial conc.	0.00269 M	-		0	0
Change	-x	-		+x	+x
Equilibrium conc.	0.00269 M - x	-		+x	+x

K_b = [C₆H₅COOH]_eq[OH^-]_eq / [C₆H₅COO^-]_eq

1.6 x 10^-10 = [(x) * (x)] / (0.00269 M - x)

Solving for x, x = 6.54 x 10^-7 M

[OH^-] = x = 6.54 x 10^-7 M

pOH = -log[OH^-]

pOH = -log(6.54 x 10^-7 M)

pOH = 6.18

pH = 14 - pOH

pH = 14 - 6.18

pH = 7.82