Calculate the molarity.?

Question

A solution of formic acid is found to have a pH of 2.0. What is the molarity of the solution of formic acid (Ka = 1.78 x 10-4)

H(CHO2) <-> H+ + CHO2-

Answer 1

1.78*10^-4 = (H+)^2/0.01
= 1.78*10^-6 and then take the square root of that
= 1.33*10^-3
= 0.00133 mol dm^-3

Answer 2

Let's assume that the concentration is C. Then you have

.. . . . . .HCOOH <=> H+ +HCOO-
Initial.. .. .. .. .C
dissociate. .. x
produce.. . . . .. .. .. .. .. x. . . ..x
equilibrium. C-x.. .. . . . .x.. .. .. .x

and Ka=[H+][HCOO-]/[HCOOH]
thus Ka=x^2/(C-x)

But pH=2.0 thus x=[H+]=10^-pH=10^-2

and therefore 1.78 * 10^-4= (10^-2)^2 /(C-0.01)
and thus C=0.57 M

Answer 3

56M