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Question

H3A is a weak triprotic acid (Ka1=10−5, Ka2=10−9, Ka3=10−13).
What is the value of pX of 0.1 M H3A(aq.) solution, where pX=−logX and X=[A3−][HA2−]?

Answer 1

First we will calculate [H₃O⁺] from Ka₁

At equilibrium :

\(\underset{0.1-Y}{H_3A(aq.) }+ H_2O(I) \rightleftharpoons \underset{Y}{H_3O^ + (aq.)} + \underset Y{H_2A^-(aq.)}\)

\(Ka_1 = \frac{[H_2A^-][H_3O^+]}{[H_3A]}\)

\(10^{-5} = \frac{Y \times Y}{0.1 - Y}\)

since Ka₁ is small , 0.1−Y ≈ 0.1

\(10^{-5} = \frac{Y\times Y}{0.1}\)

10⁻⁶ = Y × Y

⇒ Y² = 10⁻⁶

⇒ Y = 10⁻³ = [H₃O⁺]

Now from Ka₃, we will calculate X

where,

\(X= \frac{[A^{3-}]}{[HA^{2-}]}\)

At equilibrium,

\(HA^{2-}(aq) + H_2O (I) \rightleftharpoons H_3O^+(aq) + A^{3-}(aq)\)

\(Ka_3 = \frac{[A^{3-}][H_3O^+]}{[HA^{2-}]}\)

⇒ Ka₃ = X × [H₃O⁺]

Putting value of [H₃O⁺] and Ka₃,

⇒ 10⁻¹³ = X × 10⁻³

⇒ X = 10⁻¹⁰

⇒ pX = −logX = −log(10⁻¹⁰) = 10