Question

Assume you dissolve 0.175 g of the weak acid benzoic acid, CH, CO,H, in enough water to make 1.00 x 102 mL of solution and then titrate the solution with 0.182 M NaOH. K, for benzoic acid = 6.3 x 10-5.) CH3CO, H(aq) + OH(aq) = CHCO2 (aq) + H2O(C) What was the pH of the original benzoic acid solution? pH = What are the concentrations of all of the following ions at the equivalence point: Na, HO OH, and CH,CO2? (Nat] = (130+) - DNIM [OH^] = M (C6H3CO2-) = c. What is the pH of the solution at the equivalence point? pH =
Please help. The answer to this question has been wrong both times I have submitted it on here. I really need to understand this.

Answer 1

Answer A

The molecular weight of Benzoic acid is 122 so 0.175 gm is (0.001434 mole)

Solution concentration is  Exactly 0.01434426 M

The dissociation constant is given 6.3 X 10^-5

K_a = [H⁺][PhCOO^-]/[PhCOOH]

By solving the equation [H+] is approximately  0.000934

So pH = - log [H+] = 3.0298

Answer B

The concentration of NaOH is 0.182 M

So for the equivalence point to reach we need same molar amount of NaOH as of Benzoic Acid.

Benzoic acid present is 0.0014344 Mole

So NaOH required in ml is(1000 X 0.0014344)/0.182 = 7.88 ml

So the l volume of the solution is 100 ml water + 7.88 ml = 107.88 ml of solution at equivalance point

So molarity M = (0.0014344 X 1000) / 107.88 = 0.01329 M

Answer to question B is 0.01329 M for all ions

Answer C:

Dissociation constant is 6.3 X 10^-5

So PKa = 4.2 (-log[Ka])  

So PKb = 14-PKa = 14-4.2= 9.8

So[OH] =(Kb X Cb)^1/2 = 1.451X 10 ^-6

Then P[OH] = 5.8

pH = 14-5.8 = 8.2

Answer is 8.2