M Chapter 9 Homework x c classify The Specles Giv x 92 ku to J Google Sea x 1 C
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M Chapter 9 Homework x c classify The Specles Giv x 92 ku to J Google Sea x 1 C D ezto mheducation.com/hmtpx Bookmarks Desmos Graphing ca L. Bontrager Replaceme o My pretslPrevi CHE 1121 Fa 2016 Home J Psy 1013 Int Chapter 19 Homework Question 200 14. 100 points 1 out of 2 attempts Be sure to answer all parts. Assistance Find the pH of the two equivalence points and the volume (ml) of 0.0514 M KOH needed to reach them in the View Hint titration of 17.3 mL of 0.130 M H2Co2. M My Work First equivalent point Second equivalent point mL KOH mL KOH pH pH asExplanation / Answer
14. H2CO3 titration with KOH
initial moles H2CO3 = 0.130 M x 17.3 ml = 2.249 mmol
volume KOH needed to reach first equivalence point = 2.249 mmol/0.0514 M = 43.75 ml
pH = 1/2(pKa1 + pKa2) = 1/2(6.35 + 10.33) = 8.34
volume KOH needed to reach second equivalence point = 2 x 2.249 mmol/0.0514 M = 87.51 ml
[CO3^2-] formed = 2.249 mmol/104.81 ml = 0.021 M
CO3^2- + H2O <==> HCO3- + OH-
let x amount has hydrolyzed
Kb1 = Kw/Ka2 = [HCO3-][OH-]/[CO3^2-]
1 x 10^-14/4.7 x 10^-11 = x^2/0.021
x = [OH-] = 2.14 x 10^-3 M
pOH = -log[OH-] = 2.67
pH = 14 - pOH = 11.33
20. pKa = pH - log(B/BH+)
= 8.99 - log(0.456/0.247)
= 8.72
when 0.002 mol HCl added
pH = pKa + log(B/BH+)
= 8.72 + log[(0.456 M x 0.25 L - 0.002 mol)/(0.247 M x 0.25 L + 0.002 mol)]
= 8.96
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