The ph of a 0.150 molar solution of a weak acid is 4.10. what is the pka of the acid?

4.21×10⁻⁸

Explanation:

1) Assume a general equation for the ionization of the weak acid:

Let HA be the weak acid, then the ionization equation is:

HA ⇄ H⁺ + A⁻

2) Then, the expression for the ionization constant is:

Ka = [H⁺][A⁻] / [HA]

There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)

3) So, you need to determine [H⁺] which you do from the pH.

By definition, pH = - log [H⁺]

And from the data given pH = 4.1

⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵

4) Now you have all the values to calculate the expression for Ka:

ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸

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