A solution of 2.9 M weak acid is 0.52% ionized.What is the K_a value of this acid? A)7.9 $\times$ 10^-⁵ B)1.5 C)1.5 $\times$ 10^-² D)5.2 $\times$ 10^-³ E)none of these

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Quizplus · Accepted Answer

The degree of ionization is 0.52%, which means 0.0052 in decimal form. For a weak acid HA, the ionization can be represented as HA ⇌ H⁺ + A⁻. The concentration of H⁺ ions is 2.9 M * 0.0052 = 0.01508 M. The equilibrium concentration of HA is approximately 2.9 M. The Kₐ is calculated as [H⁺][A⁻]/[HA] = (0.01508)(0.01508)/(2.9) ≈ 7.9 × 10⁻⁵.

A Solution of 2 ×\times× 10-5 B)1 ×\times× 10-2 D)5 ×\times× 10-3 E)none of These

A Solution of 2 $\times$ 10^-⁵
B)1 $\times$ 10^-²
D)5 $\times$ 10^-³
E)none of These