Post a New Question


posted by on .

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

  • chemistry - ,

    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.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question