Asked by rodrigue
What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.74 at equilibrium? ka =1.7x10^-4
Answers
Answered by
DrBob222
HCOOH ==> H^+ + HCOO^-
Ka = (H^+)(HCOO^-)/(HCOOH) = 1.7 x 10^-4
Convert pH = 3.74 to (H^+). Substitute for (H^+) and (HCOO^-) (they will be the same). For (HCOOH), substitute X-(H^+). Solve for X.
Ka = (H^+)(HCOO^-)/(HCOOH) = 1.7 x 10^-4
Convert pH = 3.74 to (H^+). Substitute for (H^+) and (HCOO^-) (they will be the same). For (HCOOH), substitute X-(H^+). Solve for X.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.