Calculate the pH of a 2.0 *10^-⁵ M solution of HOC₆H₅. (K_a = 1.6 * 10^-¹⁰)

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The correct answer is A because the pH of a 2.0 F1F1F1S1F1F1F10 10^-⁵ M solution of HOC₆H₅ can be calculated using the Henderson-Hasselbalch equation:pH = pK_a + log([A^-]/[HA])where pK_a is the negative logarithm of the acid dissociation constant, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the pK_a of HOC₆H₅ is 1.6 F1F1F1S1F1F1F10 10^-¹⁰, and the concentration of HOC₆H₅ is 2.0 F1F1F1S1F1F1F10 10^-⁵ M.Substituting these values into the Henderson-Hasselbalch equation, we get:pH = 1.6 F1F1F1S1F1F1F10 10^-¹⁰ + log([A^-]/[HA])pH = 1.6 F1F1F1S1F1F1F10 10^-¹⁰ + log([2.0 F1F1F1S1F1F1F10 10^-⁵ M]/[2.0 F1F1F1S1F1F1F10 10^-⁵ M])pH = 1.6 F1F1F1S1F1F1F10 10^-¹⁰ + log(1)pH = 1.6 F1F1F1S1F1F1F10 10^-¹⁰ + 0pH = 1.6 F1F1F1S1F1F1F10 10^-¹⁰Therefore, the pH of a 2.0 F1F1F1S1F1F1F10 10^-⁵ M solution of HOC₆H₅ is 1.6 F1F1F1S1F1F1F10 10^-¹⁰.

Calculate the PH of a 2