Given that K_a for the weak acid HA is 3.49 × 10^-8,calculate K for the reaction of HA with OH^-. HA(aq)+ OH−(aq)D A−(aq)+ H₂O( ) A) 3.49 B) 3.49 × 10⁶ C) 3.49 × 10^-22 D) 2.87 × 10²¹ E) 2.87 × 10^-7

Question

Quizplus · Accepted Answer

The reaction between HA and OH- can be represented as follows:
HA + OH- <-> A- + H2O

The equilibrium constant for this reaction, K, can be expressed in terms of the equilibrium constant for the dissociation of HA, Ka, as follows:
K = (A-/HA)(H2O/OH-) = [(A-/HA)/(HA/H2O)](H2O/OH-) = Ka(H2O/OH-)

Substituting the given value for Ka, we get:
K = (3.49 × 10^-8)(1/(10^-7)) = 3.49 × 10^-1 = 0.349

Option B (3.49 × 10^-16) is the closest value, as it only differs in the exponent. However, it is incorrect as the answer must be in the range of 0 to 1.

Given That Ka for the Weak Acid HA Is 3

Given That K_a for the Weak Acid HA Is 3