Question

What does the pKa of the conjugate acid of a base have to be to remove a hydrogen from water (pKa = 15.7) so that the ratio of hydroxide (OH-) to water is 100 to 1?

Answer 1

Answer:

The pKa of the conjugate acid is 17.7

Explanation:

If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,

Kₐ = [OH⁻]/[H₂O]

$K_a=\frac{100}{1} =100$

Now, we determine the equivalent pKa

pKa = -log[ka]

pKa = -log[100]

pKa = -2

Removal of hydrogen from water is reversible as shown below;

H₂O  ⇄    OH⁻ + H⁺

15.7             -2

This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];

pKa  of the conjugate acid = 15.7 - (-2) = 17.7

The pKa of the conjugate acid is 17.7