I took General Chemistry a while ago, and haven't had to do many titration calculations since then, and I can't remember how to solve this problem.
In a titration experiment, 20.4mL of .883M HCOOH neutralize 19.3mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution?
Any help here is greatly appreciated, thanks!
2007-03-30 10:59:34 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
V1C1A = V2C2B
V = volume of solution
C = concentraion of solution
A = Number in front of H
B = Number in front of OH
So...
(20.4/1000) x 0.883 x 1 = (19.3/1000) x C2 x 2
So...
C2 = 0.018/0.0386
C2 = 0.466 M
2007-03-30 11:09:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
((20.4)(.883))/2 = (19.3)X
X = 0.467
mL times M of HCOOH divided by 2 (takes 2 moles of acid per mole of the dihydroxide) = mL Ba(OH)2 times Molarity
2007-03-30 18:11:45 · answer #2 · answered by dollhaus 7 · 0⤊ 0⤋