Respuesta :
Answer : The value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
Explanation : Given,
Initial concentration = 0.090 M
pH = 1.80
First we have to calculate the hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]1.80=-\log [H^+][/tex]
[tex][H^+]=0.0158M[/tex]
Now we have to calculate the [tex]K_a[/tex] of the weak acid.
The dissociation reaction of weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Initial conc. 0.090 0 0
At eqm. (0.090-x) x x
x = 0.0158 M
The expression for dissociation constant is:
[tex]K_a=\frac{(x)\times (x)}{(0.090-x)}[/tex]
Now put all the given values in this expression, we get:
[tex]K_a=\frac{(0.0158)\times (0.0158)}{(0.090-0.0158)}[/tex]
[tex]K_a=3.36\times 10^{-3}[/tex]
Thus, the value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
The value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
How to calculate Ka of an acid?
To calculate the Ka of an acid, we have to calculate the hydrogen ion concentration of the acid using the following expression:
pH = -log {H+}
1.80 = -log {H+}
{H+} = 0.0158M
The dissociation equation is given as follows:
HA ⇌ H+ + A-
Ka = 0.0158²/(0.090 - 0.0158)
Ka = 2.49 × 10-⁴/7.42 × 10-²
Ka = 0.336 × 10-²
Ka = 3.36 × 10-³
Therefore, the value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
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