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A 0.035 M solution of a weak acid (HA) has a pH of 4.88, the Ka of the acid = [tex]4.98[/tex]×[tex]10^{-9}[/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.
We have given that pH = 4.88 and HA = 0.035 M
We know that pH = -log[H⁺]
pH = -log[H⁺]
[H⁺] = [tex]10^{-pH}[/tex]
= [tex]10^{-4.88}[/tex]
= 1.32 × 10⁻⁵
Lets now make the ICE table to get eq. for Ka
HA ⇆ H⁺ + A⁻
I 0.035 M 0 0
C -x +x +x
E 0.035 M-x +x +x
Now, we know that H⁺ = x = A = 1.32 × 10⁻⁵
Ka = [tex]\frac{[H_3O^+][A^+]}{HA}[/tex]
= [tex]\frac{x^2}{0.035-x}[/tex]
= [tex]\frac{(1.32*10^{-5})^2}{0.035}[/tex] (subitising the value of H₃O = x = A = 6.0 × 10⁻³)
= [tex]4.98[/tex]×[tex]10^{-9}[/tex]
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