When 2.5 $\times$ 10^-²mol of nicotinic acid (a monoprotic acid)is dissolved in 350 mL of water,the pH is 3.05.Calculate the K_a of nicotinic acid. A)1.3 $\times$10^-² B)1.1 $\times$ 10^-⁵ C)7.1 $\times$ 10^-² D)3.3 $\times$ 10^-⁵ E)none of these

Question

Quizplus · Accepted Answer

First, we can use the pH to calculate the concentration of H+ ions:
pH = -log[H+] 
3.05 = -log[H+] 
[H+] = 7.1 x 10^-4 M

Next, we can use the concentration of the acid and the concentration of H+ to calculate the Ka: 
Ka = [H+]^2 / [A-] 
We know that the concentration of the acid is equal to the concentration of the conjugate base: 
[A-] = [HA] = 2.5 x 10^-3 mol / 0.350 L = 7.14 x 10^-3 M 
Plugging in the values: 
Ka = (7.1 x 10^-4)^2 / (7.14 x 10^-3) = 1.1 x 10^-5 
Therefore, the answer is B.

When 25 ×\times× 10-2 mol of Nicotinic Acid (A Monoprotic Acid)is Dissolved in Acid)is

When 25 $\times$ 10^-²mol of Nicotinic Acid (A Monoprotic Acid)is Dissolved in Acid)is