Respuesta :
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(-) ]
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(-) ]
The Ka₂ expression for H₂CO₃ in an aqueous solution
[tex]Ka_2\:=\:\frac{[H_3O^+][CO_3^{2-}]}{HCO_3^{-}}[/tex]
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
[tex]\large{\boxed{\bold{Ka\:=\:\frac{[H^+][A^-]}{[HA]} }}}[/tex]
while the concentration of H⁺ and HA there is a relationship:
[tex]\large{\boxed{\bold{[H^+]\:=\:\sqrt{Ka. [HA]}}}}[/tex]
So the value of Ka₂
[tex]Ka_2\:=\:\frac{[H_3O^+][CO_3^{2-}]}{HCO_3^{-}}[/tex]
Learn more
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Keywords: weak acid, protected, Ka, carbonic acid
