Question

The Weak Ionization constant (Ka) for HCOOH is equal to: Please help me lol

Answer 1

Answer:

$Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}$

Explanation:

Hello,

In this case, the weak ionization reaction for formic acid is:

$HCOOH(aq)\rightleftharpoons H^+(aq)+HCOO^-(aq)$

In such a way, we simply recall the law of mass action in order to represent the weak ionization constant, Ka, for such process, by taking into account that the concentration of products is divided over the concentration of reactants as shown below:

$Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}$

Best regards.

Answer 2

Answer:

$Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Explanation:

In this case we have to start with the ionization reaction of $CH_3COOH$, so:

$CH_3COOH~~CH_3COO^-~+~H^+$

With this in mind we can calculate the mathematical expression of Ka (The production of hydronium ions $H^+$), so:

$Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

I hope it helps!