Question

The pH of a 0.150 molar solution of a weak acid is 4.10. What is the pKa of the acid?

Answer 1

Answer:

D. 7.4

Explanation:

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Answer 2

Answer: 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