Respuesta :
Answer: The value of [tex]k_{a}[/tex] for the weak acid is [tex]3.07 \times 10^{-4}[/tex].
Explanation:
First, we will calculate the molarity of HA as follows.
[HA] = [tex]\frac{\text{no. of moles}}{\text{volume}}[/tex]
= [tex]\frac{0.03 mol}{2 L}[/tex]
= 0.015
At equilibrium,
[tex]HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}[/tex]
Initial: 0.015 0 0
Change: -x +x +x
Equilibm: 0.015 - x x x
It is given that, at equilibrium
[HA] = 0.015 - x = 0.013
- x = 0.013 - 0.015
= 0.002
Now, expression for [tex]k_{a}[/tex] of this reaction is as follows.
[tex]k_{a} = \frac{[A^{-}][H_{3}O^{+}]}{[HA]}[/tex]
= [tex]\frac{x^{2}}{[HA]}[/tex]
= [tex]\frac{(0.002)^{2}}{0.013}[/tex]
= [tex]3.07 \times 10^{-4}[/tex]
Thus, we can conclude that the value of [tex]k_{a}[/tex] for the weak acid is [tex]3.07 \times 10^{-4}[/tex].
The value of Ka for the weak acid should be 3.07*10^-4 when 0.030 moles of a weak acid.
Calculation of the value of Ka:
First determine the molarity of HA
= 0.03 mol / 2L
= 0.015
Now
HA = 0.015 -x = 0.013
-x = 0.013 - 0.015
= 0.002
Now the value should be
= (0.002)^2/0.013
= 3.07*10^-4
Hence, The value of Ka for the weak acid should be 3.07*10^-4 when 0.030 moles of a weak acid.
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