0.375 g of a monoprotic acid (MW = 245 g/mol)is dissolved in water to produce 25.0 mL of a solution with pH = 3.28.Determine the ionization constant of the acid. A)8.56 × 10^-3 B)4.5 × 10^-6 C)7.4 × 10^-5 D)4.5 × 10^-3 E)2.3 × 10^-2

Question

Quizplus · Accepted Answer

First, we need to find the concentration of the acid in the solution:

n(moles) = m/MW = 0.375 g / 245 g/mol = 0.00153 mol

c(molarity) = n/V = 0.00153 mol / 0.025 L = 0.0612 M

Next, we can use the pH and concentration of the acid to find the pKa:

pH = pKa + log([A-]/[HA])

3.28 = pKa + log([A-]/[HA])

We know that at equilibrium, the concentration of the acid that has ionized is equal to the concentration of the conjugate base formed:

[A-] = [H+]/Ka

[HA] = concentration of undissociated acid = c acid - [A-]

Substituting these into the Henderson-Hasselbalch equation gives:

3.28 = pKa + log([H+]/Ka / (c acid - [H+]/Ka))

Rearranging and solving for pKa:

pKa = 3.28 - log((c acid / [H+] - 1))

pKa = 3.28 - log((0.0612 / 10^(-3.28)) - 1)

pKa = 5.35

Finally, we can use the relationship between pKa and Ka to find the ionization constant:

pKa = -log(Ka)

Ka = 10^(-pKa)

Ka = 4.5 × 10^-6

Therefore, the answer is B.

0375 G of a Monoprotic Acid (MW = 245 G/mol)is