Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium formate (nahco2). the ka of formic acid is 1.77 ⋅ 10-4.

0.094%

Explanation:


1) Equilibrium chemical equation:


Only the ionization of the formic acid is the important part.


HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).


2) Mass balance:


                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x

3) Acid constant equation:


Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)


= (0.189 + x )x / (0.311 - x) = 0.000177

4) Solve the equation:

You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.


With that approximation the equation to solve becomes:

0.1890x / 0.311 = 0.000177, which leads to:

x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M

5) With that number, the percent of ionization (alfa) is:


percent of ionization = (moles ionized / initial moles) x 100 =


percent ionization = (concentration of ions / initial concentration) x 100 =


percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%

Noble gases is far right