Posted by rodrigue on .
What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.74 at equilibrium? ka =1.7x10^-4
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.