Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Titration acid Problem, really really need help !?
A 50.00 mL solution of .2500M niacin, a monoporitc weak acid with Ka of 1.5X 10^-5 was tritrated with .3125M NaOH. What is the pH of the solution after adding 40.00mL of .3125M NaOH?
2 Answers
- skipperLv 71 decade agoFavorite Answer
50.00 mL x 0.2500 M = 12.5 millimoles of niacin
40.00 mL x 0.3125 M = 12.5 millimoles of NaOH
This is the equivalence point; you have 12.5 millimoles of the salt of a weak acid in 90.00 mL of solution; concentration of the salt = 0.139 M (12.5 mmol / 90.00 mL = 0.139 M)
Let HA = niacin and NaA = the sodium salt of niacin
[A-] + H2O = HA + [OH]
Kb = [HA][OH-]/[A-]
Kb = Kw/Ka = 1x10^-14 / 1.5x10^-5 = 6.7x10^-10
let X = [OH-]
6.7x10^-10 = X^2/(0.139 - X)
X = 9.62x10^-6
pOH = -log(9.62x10^-6) = 5.02
pH = 14.00 - 5.02 = 8.98
