Answer:
Explanation:
Let the monoprotic acid be HX
HX ⇄ H⁺ + X⁻
pH = 2.53
Hydrogen ion concentration
[tex][ H^+]=10^{-2.53}[/tex]
[tex][ X^-]=10^{-2.53}[/tex]
Concentration of undissociated acid will remain almost the same as it is a weak acid
So
Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid
= [ H⁺] x [Cl⁻ ] / [ HX]
[tex]k_a=\frac{10^{-2.53}\times 10^{-2.53}}{.0166}[/tex]
[tex]k_a=\frac{.00295^2}{.0166}[/tex]
= 5.24 x 10⁻⁴ M .
