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3 years ago

The weak ionization constant (Ka)

Chemistry

1 answer:

Nostrana [21]3 years ago

4 0

Carbonic acid is a weak acid. Its weak ionization constant (Ka) is equal to:

Ka = [H⁺] [HCO₃⁻] / [H₂CO₃]

A weak acid is an acid that dissociates partially in the water.

The equation for the acid dissociation of carbonic acid is:

H₂CO₃ ⇄ H⁺ + HCO₃⁻

The weak ionization constant (Ka) is the equilibrium constant for this reaction, that is, the product of the concentration of the products divided by the product of the concentrations of the reactants, all of them raised to their stoichiometric coefficients.

Ka = [H⁺] [HCO₃⁻] / [H₂CO₃]

Carbonic acid is a weak acid. Its weak ionization constant (Ka) is equal to:

Ka = [H⁺] [HCO₃⁻] / [H₂CO₃]

Learn more about weak acids here: brainly.com/question/15192126

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Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

<u>Answer:</u> The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

<u>Explanation:</u>

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

<u>Initial:</u> 0.083

<u>At eqllm:</u> 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

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<h3>What is a reaction?</h3>

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