Respuesta :
A 0.15 M solution of butanoic acid, CH₃CH₂CH₂COOH, contains 1.51 ˣ 10⁻³MH₃O⁺ , the Ka of butanoic acid = [tex]1.52[/tex]× [tex]10^{-5}[/tex]
What is Ka?
The difference between strong and weak acids is specified by the acid dissociation constant (Ka). As Ka rises, the acid dissociates more. That is why, strong acids must dissociate more in water.
On the other hand, a weak acid has a lower propensity to ionise and release a hydrogen ion, resulting in a less acidic solution.
The equilibrium constant of dissociation reaction of an acid is known as the acid dissociation constant, or Ka.
The strength of acid in a solution is represented by this equilibrium constant. Ka is frequently expressed in mol/L units.
The equation of solution of butanoic acid is given, as
[C₃H₇COOH] + H₂O ⇆ H₃O + [C₃H₇COO]
Lets now make the ICE table to get eq. for Ka
[C₃H₇COOH] H₃O [C₃H₇COO]
I 0.15M 0 0
C -x +x +x
E 0.15 M-x +x +x
Now, we know that H₃O = x = [C₃H₇COO] = 1.51 ˣ 10⁻³M
Ka = [tex]\frac{[H_3O^+][C_3H_7COO]}{[C_3H_7COOH] }[/tex]
= [tex]\frac{x^2}{0.15-x}[/tex]
= [tex]\frac{(1.51*10^{-3})^2}{0.15}[/tex]
= [tex]1.52[/tex]× [tex]10^{-5}[/tex]
Learn more about butanoic acid
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