Chemistry, 18.09.2019 07:20 nmm32

A0.50 m solution of formic acid, hcooh, has a ph of 2.02. calculate the percent ionization of hcooh

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Answer from: EllaSue

Answer : 1.91 %

Explanation : The steps to solve this problem are explained below;

1. HCOOH ⇄ $HCOO^{-} + H^{+}$

Here Ka =( $[HCOO^{-}]_{eq} X [H^{+}]_{eq}$ )/ $[HCOOH]_{eq}$

As the equilibrium concentration of $H^{+}$ will be the pH of the solution.

∴ $[H^{+}]_{eq}$ = $10^{(-2.02)} = 9.55 x <img src=$ M

2. The initial concentration of HCOOH. When it loses x moles from it as the acid undergoes dissociation to form $HCOO^{-}$ and $H^{+}$ .

3. The moles present will be as

[HCOOH] (M) $[H^{+}]$ (M) $[HCOO^{-}]$ (M)

Initial 0.50 0.00 0.00

After Change -x +x +x

Equilibrium ( 0.50 -x) x x

∴ Ka = (x) x (x) / (0.50 - x)

4. Assuming that all of the $H^{+}$ comes from the acid, and none from water.

As $[H^{+}]_{eq}$ = 9.55 x $10^{-3}$ which is much higher than the 1.0 x $10^{-7 }$ M [tex[H^{+}" alt=" 10^{-3}" /> M

2. The initial concentration of HCOOH. When it loses x moles from it as the acid undergoes dissociation to form $HCOO^{-}$ and $H^{+}$ .

3. The moles present will be as

[HCOOH] (M) $[H^{+}]$ (M) $[HCOO^{-}]$ (M)

Initial 0.50 0.00 0.00

After Change -x +x +x

Equilibrium ( 0.50 -x) x x

∴ Ka = (x) x (x) / (0.50 - x)

4. Assuming that all of the $H^{+}$ comes from the acid, and none from water.

As $[H^{+}]_{eq}$ = 9.55 x $10^{-3}$ which is much higher than the 1.0 x $10^{-7 }$ M [tex[H^{+}" /> from water.

Also, the concentration of HCOOH will change very little, from 0.50 to 0.50 - 9.55 x $10^{-3}$ .

The change in concentration can be ignored if it is less than 5% of the original concentration.

∴ 0.50 M x 5% = 0.025, so the change in [HCOOH] in this problem can be ignored.

Now, Ka = (x)(x)/0.50 = (9.55 x $10^{-3})^{2}$ /0.50= 1.82 x $10^{-4}$

Now, calculating the percent ionization for this problem.

which will represent the relative number of acid molecules which dissociate. It is calculated as :

$[H^{+}]_{eq}$ x 100 / $[HCOOH]_{i}$

∴ percent ionization = {(9.55 x $10^{-3})$ / (0.50)}x 100 = 1.91 %

This value of 1.91 % indicates that very little of this acid dissociates (ionizes) under these conditions.

For strong acids and bases, the percent ionization is 100%.

Answer from: Quest

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Answer from: Quest

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A0.50 m solution of formic acid, hcooh, has a ph of 2.02. calculate the percent ionization of hcooh...

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