Question

How many grams of salt NaA (Mw = 145 g/mole) must be added into 573 mL of 0.591 M HA to prepare a buffer solution with pH =3.72? (weak acid HA: Ka = 4.71 × 10-5) Please keep your answer to two decimal places.

Answer 1

To prepare a buffer solution with a pH of 3.72, one needs to add approximately 20.28g of salt NaA to 573 mL of 0.591 M HA, according to the Henderson-Hasselbalch equation.

Step-by-step explanation:

To calculate the amount of salt needed to prepare a buffer solution, we'll use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base (the salt, NaA) and [HA] is the concentration of the acid (HA).

Given that the pH = 3.72 and the pKa (which is -log(Ka)) is -log(4.71*10^-5) = 4.33, we can work out the ratio [A-]/[HA] = 10^(pH-pKa) = 10^(3.72-4.33) = 0.413.

Because the final volume of the solution is 573 mL (or 0.573 L), the molarity of HA is 0.591 M, hence the moles of [HA] = 0.591 M * 0.573 L = 0.33873 mol. So, the moles of [A-] (base or the salt) should = [A-] * [HA] = 0.413 * 0.33873 mol = 0.13989 mol.

The molar mass of NaA is given as 145 g/mole. Therefore, the mass of NaA required = moles of NaA * molar mass = 0.13989 mol * 145 g/mol = 20.28 g.

Therefore, to prepare the buffer solution with a pH of 3.72, you will need to add approximately 20.28g of NaA to 573 mL of 0.591 M HA.

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