ph titration help. please please. :) Assume you dissolve 0.383 g of the weak aci
ID: 964387 • Letter: P
Question
ph titration help. please please.
:)
Assume you dissolve 0.383 g of the weak acid benzoic acid, C_6H_5CO_2H, in enough water to make 1.00 times 10^2 ml of solution and then titrate the solution with 0.179 M NaOH. (K_a for benzoic acid = 6.3 times 10^-5.) C_6H_5CO_2H(aq) + OH (aq) C_6H_5CO_2 - (aq) + H_2O(l) a. What was the pH of the original benzoic acid solution? pH = b. What are the concentrations of all of the following ions at the equivalence point: Na +, H_3O +, OH-, and C_6H_5CO_2-? [NA +] = [H_3O+] = [OH-] = [C_6H_5CO_2-] = c. What is the pH of the solution at the equivalence point? pH =Explanation / Answer
moles of benzoic acid = 0.383 / 122 = 3.14 x 10^-3
molarity of benzoic aicd = moles / volume = 3.14 x 10^-3 / 0.1 = 0.0314 M
pH = 1/2 [pKa -logC]
pH = 1/2 [4.20 - log 0.0314]
pH = 2.85 -------------------------> original pH
C6H5COOH + OH- --------------------> C6H5COO- + H2O
0.0314 0.179 0 0 ----------------> initial
0 0.1476 0.0314 - ---------------> equilence point
[Na+] = 0.179 M
[OH-] = 0.1476 M
[H3O+] = Kw / [OH-]
[H3O+] = 6.78 x 10^-14 M
[C6H5CO2-] = 0.0314 M
pH = -log [H3O+]
pH = 13.17
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