Respuesta :
Answer :
(a) The ratio of acetate to acetic acid is, [tex]4.47\times 10^2[/tex]
(b) The ratio of ethylamine to ethylammonium ion is, [tex]3.89\times 10^{-4}[/tex]
Explanation :
Part (a) :
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[CH_3COO^-]}{[CH_3COOH]}[/tex]
Given:
[tex]pK_a[/tex] for acetic acid = 4.75
pH = 7.4
Now put all the given values in this expression, we get:
[tex]7.4=4.75+\log \frac{[CH_3COO^-]}{[CH_3COOH]}[/tex]
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]}=446.68=4.47\times 10^2[/tex]
Thus, the ratio of acetate to acetic acid is, [tex]4.47\times 10^2[/tex]
Part (b) :
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[\text{ethylamine}]}{[\text{ethylammonium ion}]}[/tex]
Given:
[tex]pK_a[/tex] for ethylamine = 10.81
pH = 7.4
Now put all the given values in this expression, we get:
[tex]7.4=10.81+\log \frac{[\text{ethylamine}]}{[\text{ethylammonium ion}]}[/tex]
[tex]\frac{[\text{ethylamine}]}{[\text{ethylammonium ion}]}=3.89\times 10^{-4}[/tex]
Thus, the ratio of ethylamine to ethylammonium ion is, [tex]3.89\times 10^{-4}[/tex]
