Problem broken down into 4 individual questions. Part 1: https://www.chegg.com/h
ID: 908953 • Letter: P
Question
Problem broken down into 4 individual questions.
Part 1: https://www.chegg.com/homework-help/questions-and-answers/problem-broken-4-individual-questions-part-1-4-supply-calculated-molarities-indicated-spec-q9055855
Part 2: https://www.chegg.com/homework-help/questions-and-answers/problem-broken-4-individual-questions-part-1-https-wwwcheggcom-homework-help-questions-ans-q9056129
Part 3 of 4:
Supply the calculated molarities of the indicated species in the solution mixtures.Assume the solutions are at 25 degrees Celsius.
Problem #C3: Calculate the Molarity for: H+(at equilibrium).
Given the following procedure (straight from lab book).
C3. pH of Deionized water: Obtain 19 mL of deionized water in each of two beakers. Read the pH of the deionized water. (NOTE: It will not be a pH of 7.00.)
Explanation / Answer
Part 1.
moles of sodium acetate trihydrate = g/molar mass = 3.3/136.08 = 0.024 mols
molarity of sodium acetate in solution = 0.024/(0.046+0.004) = 0.48 M
molarity of acetic acid in solution = 6 x 0.004/(0.046+0.004) = 0.48 M
pKa = 4.76
So, the concentration of [acid] = [base]
thus, pH = pKa = 4.76
Part 2.
To A : added 1 ml of 3 M HCl
moles HCl = molarity x volume = 3 x 0.0010.003 mols
new molarity of sodium acetate = (0.48 x 0.05 - 0.003)/0.051 = 0.412 M
new molarity of acetic acid = (0.48 x 0.05 + 0.003)/0.051 = 0.53 M
Feed in Hendersen-Hasselbalck equation,
pH = 4.76 + log(0.412/0.53) = 4.65
To B : added 1 ml of 3 M NaOH
moles NaOH = molarity x volume = 3 x 0.0010.003 mols
new molarity of sodium acetate = (0.48 x 0.05 + 0.003)/0.051 = 0.53 M
new molarity of acetic acid = (0.48 x 0.05 - 0.003)/0.051 = 0.412 M
Feed in Hendersen-Hasselbalck equation,
pH = 4.76 + log(0.53/0.412) = 4.87
#C3 This section is not clear, pH of water only is always 7.
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