Question

Complete the ka2 expression for h2co3 in an aqueous solution.

Answer 1

The Ka₂ expression for H₂CO₃ in an aqueous solution

$Ka_2\:=\:\frac{[H_3O^+][CO_3^{2-}]}{HCO_3^{-}}$

Further explanation

According to Arrhenius, acids are substances which, when dissolved in water, release H  ions.

An HₓY acid in water will ionize:

HₓY (aq) --------> xH⁺ (aq) + Yˣ- (aq)

Example:

HCl -------> H⁺ + Cl⁻

The amount of H⁺ ions produced by 1 acid molecule is called valence acid, whereas acidic residual ions are ions formed after the release of H⁺ ions.

Usually, the name acid begins with the word acid followed by the name of the remaining acidic ion

The ion concentration of a weak acid is determined by the value of the acid ionization constant (Ka).

The greater the value of Ka, the greater the dissociated acid produces its H⁺ion and the greater its acidity

Based on the number of protons (H ⁺) released, acids can be divided into:

• 1. Monoprotic acid is an acid which releases one H + ion such as HCl, HCN, HNO₃

• 2. Polyprotic acid which is an acid that releases more than one H + ion.

This acid can be divided into two:

• a. Diprotic acid: releases 2 H + ions, such as H₂SO₄, H₂CO₃

• b. Triprotic acid: releases 3 H + ions such as H₃PO₄

Carbonic acid includes diprotic acid and weak acid which will be ionized in two stages

• 1. H₂CO₃ + H₂O ---> H₃O⁺ + HCO₃⁻ Ka₁

• 2. HCO₃⁻ + H₂O ---> CO₃²⁻ + H₃O⁺ Ka₂

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

Ka's value

$\large{\boxed{\bold{Ka\:=\:\frac{[H^+][A^-]}{[HA]} }}}$

while the concentration of H⁺ and HA there is a relationship:

$\large{\boxed{\bold{[H^+]\:=\:\sqrt{Ka. [HA]}}}}$

So the value of Ka₂

$Ka_2\:=\:\frac{[H_3O^+][CO_3^{2-}]}{HCO_3^{-}}$

Learn more

What is the reaction to HCN

brainly.com/question/9012584

base would not effectively deprotonate acetylene

brainly.com/question/5613072

strong enough acids

brainly.com/question/5273689

Keywords: weak acid, protected, Ka, carbonic acid

Answer 2

The complete question is: complete the ka2 expression for H2CO3 in an aqueous solution:

Ka2 = 4.69*10^-11 = ____________________


Ka2 is the constant for the second deprotonation.

H2CO3 has two H, then it can undergo two deprotonation.

First deprotonation is: H2CO3(aq) + H2O (l) ⇄ HCO3 (-) + H(+)

=> Ka1 = [ HCO3(-) ]*[ H(+) ] / [H2CO3(aq)]

Second deprotonation si HCO3 (-) + H2O(l) ⇄ CO3(2-) + H(+)

=> Ka2 = [ CO3(2-) ] * [H(+) ] / [ HCO3(-) ]

And the answer is

Ka2 = 4.69*10^-11 =  [ CO3(2-) ] * [H(+) ] / [ HCO3(-) ]