Answer :
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
The concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter .
Ionization of lactic acid can be represented as:
CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻ + H⁺
Given:
ionization constant = 1.36 x 10⁻⁴
mass= 20.0 g
Now, Molecular weight of lactic acid = 90 g
[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]
It is dissolved in 1.00L so molar concentration of lactic acid formed will be
C = 0.22M
Consider "n" to be the fraction of moles ionized
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionization constant Ka
[tex]K_a =\frac{nC*nC}{C-nC}[/tex]
[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )
On substituting the values we will get:
[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]
To find the concentration of hydronium ion in the solution,
[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]
So, concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter.
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