Calculate the pH of the buffer solution after the addition of NaOH. 5.07 0.30 4.

Question

Calculate the pH of the buffer solution after the addition of NaOH. 5.07 0.30 4.74 8.17 7.40 Use the information below to answer questions 28-30. An Erlenmeyer (E-flask) containing 35.0 ml of 0,20 M CH_3COOH was titrated with 0.15 M NaOH. K_3(CH_3COOH) = 1.8 Times 10 Identify the major species that were present in the E-flask during the titration. Identify the major the species that were present in the E-flask at the equivalence point. Predict the pH of the solution at the equivalence point. pH 7 pH cannot be predicted

Explanation / Answer

28) CH3COOH (aq) + NaOH(aq) CH3COO (aq) + H2O(l) + Na+ (aq). at the time of initialy titration some reaction overed and CH3COOH remaining in solution

so mejor species will be  CH3COO (aq) , H2O(l) , Na+, CH3COOH

option will be B

29) at the time of equivalence point reaction will over so species will be

so option will be C

30) at equivalence point concentration will equal both acid and base

calculate the pH at the equivalence point. we need 46.66 ml NaOH

You have reacted 0.007mol CH3COOH with 0.007mol NaOH to produce 0.007 mol CH3COONa dissolved in 81.6mL = 0.0816L solution:
Molarity of CH3COONa solution = 0.007 /0.0816 L = 0.0857M solution.
Calculate pH as follows:
The aim is to find the [OH-] of the solution You will know that a salt of a strong base and weak acid is basic.
The CH3COONa dissociates in water :
CH3COONa CH3COO- + Na+ The CH3COO- reacts with water:
CH3COO- + H2O CH3COOH + OH-

Ka for CH3COOH = 1.8*10-5
Kw = Ka * Kb - we want Kb
Kb = Kw* Ka
Kb = 10-14 / (1.8*10-5)
Kb = 5.55*10-10

Kb = [CH3COO-] *[[OH] / [CH3COONa]
We know that [CH3COO-] = [OH-] so product is [OH-]²
[CH3COONa] = 0.0857 M

Kb = [OH]² / 0.0857 M
(5.55*10-10 ) * 0.0857 = [OH[²
[OH]² = 4.76*10-11
[OH-] = 6.9*10-6
To calculate pH you require [H+]
[OH-] *[H+] = 10-14
[H+] = 10-14 / [OH-]
[H+] = 10-14 / (6.9*10-6)
[H+] = 1.45*10-9
pH = -log ( 1.45*10-9)
pH = 8.83

option C

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