How many grams of Na2SO3 (Sodium Sulfite) must be added to 225 mL of 0.0777 M sulfurous acid (H2SO3) to prepare a buffer at pH = 7.4000? H2SO3 has a pKa1 = 1.857 and pKa2 = 7.172.

Have you posted all of the problem.

To answer this question, we need to understand the concept of buffer solutions and how to calculate the amount of a specific compound needed to prepare a buffer at a desired pH.

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It is typically composed of a weak acid and its conjugate base (or a weak base and its conjugate acid). In this case, we have sulfurous acid (H2SO3) as the weak acid and sodium sulfite (Na2SO3) as its conjugate base.

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution:

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

Where:
pH = the desired pH of the buffer solution
pKa = the acid dissociation constant of the weak acid
[A-] = the concentration of the conjugate base
[HA] = the concentration of the weak acid

In this case, we are given the desired pH (pH = 7.4000) and the pKa values of H2SO3 (pKa1 = 1.857 and pKa2 = 7.172). We are also given the concentration of sulfurous acid (0.0777 M).

To calculate the amount of Na2SO3 needed, we need to determine the ratio of [A-] to [HA] in the Henderson-Hasselbalch equation.

Since H2SO3 has two dissociable protons, we need to consider both pKa values. Let's calculate the ratio for each pKa:

For pKa1 = 1.857:
[A-]/[HA] = 10^(pH - pKa1)

[A-]/[HA] = 10^(7.4000 - 1.857)

[A-]/[HA] = 10^5.543

[A-]/[HA] = 342,425.9

For pKa2 = 7.172:
[A-]/[HA] = 10^(pH - pKa2)

[A-]/[HA] = 10^(7.4000 - 7.172)

[A-]/[HA] = 10^0.228

[A-]/[HA] = 1.641

Now, we need to determine the molar amount of H2SO3 in the 225 mL solution. To do this, we multiply the concentration of H2SO3 by the volume in liters:

Moles of H2SO3 = concentration (M) * volume (L)

Moles of H2SO3 = 0.0777 M * 0.225 L

Moles of H2SO3 = 0.0174825 moles

Since the ratio of [A-] to [HA] for pKa1 is 342,425.9 and for pKa2 is 1.641, we need to calculate the moles of Na2SO3 required for each pKa separately.

For pKa1 = 1.857:
Moles of Na2SO3 = moles of H2SO3 * [A-]/[HA]

Moles of Na2SO3 = 0.0174825 moles * 342,425.9

Moles of Na2SO3 = 5.9880 moles

For pKa2 = 7.172:
Moles of Na2SO3 = moles of H2SO3 * [A-]/[HA]

Moles of Na2SO3 = 0.0174825 moles * 1.641

Moles of Na2SO3 = 0.028664 moles

Therefore, to prepare a buffer at pH = 7.4000, we would need to add either 5.9880 moles of Na2SO3 (based on pKa1) or 0.028664 moles of Na2SO3 (based on pKa2) to the 225 mL solution of 0.0777 M sulfurous acid.