How many ml of 0.2M NaH2PO4 should b added to 40 mL of 0.1M Na3PO4 to make the buffer pH 7.2.

Question

How many ml of 0.2M NaH2PO4 should b added to 40 mL of 0.1M Na3PO4 to make the buffer pH 7.2.

Ka1:1.1 x 10^-2
Ka2:7.5 x 10^-8
Ka3:4.8 x 10^-13

Answer 1

To solve this problem, we'll use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of its acidic and conjugate base components.

The Henderson-Hasselbalch equation is given by:

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

Where:
- pH is the desired buffer pH (7.2 in this case).
- pKa is the negative logarithm (base 10) of the acid dissociation constant for the acid component. In this case, we have three dissociation constants (Ka1, Ka2, and Ka3), and we'll use the one most relevant to the pH range we desire (pKa1 for the first dissociation step).
- [A-] is the concentration of the conjugate base component (Na2HPO4).
- [HA] is the concentration of the acid component (NaH2PO4).

First, we need to determine the concentrations of Na2HPO4 and NaH2PO4 in the buffer solution. Let's assume that x mL of 0.2M NaH2PO4 is added to 40 mL of 0.1M Na3PO4.

The initial concentration of Na3PO4 (conjugate base) is 0.1M * 0.04L = 0.004 mol.

The concentration of Na2HPO4 (acid) after adding the desired volume (x mL) is (0.2M * x/1000) mol.

Now, let's calculate the final concentrations ([A-] and [HA]) by considering the dilution factor. The total volume of the buffer solution will be 40 mL + x mL.

[A-] = moles of Na2HPO4 / total volume of buffer solution (L)
= (0.2M * x/1000) mol / (40 mL + x mL)

[HA] = moles of Na3PO4 / total volume of buffer solution (L)
= 0.004 mol / (40 mL + x mL)

Now, let's substitute the values into the Henderson-Hasselbalch equation:

7.2 = pKa1 + log ([A-] / [HA])

Substituting the expressions for [A-] and [HA]:

7.2 = pKa1 + log ((0.2M * x/1000) mol / (40 mL + x mL) / (0.004 mol / (40 mL + x mL)))

Before proceeding, we need to convert the pKa1 value to a Ka value using the equation Ka = 10^(-pKa1).

Ka1 = 10^(-pKa1)

Now, we have everything we need to solve for x.

Rearranging the equation, we get:

Ka1 = (0.2M * x/1000) mol / (0.004 mol)

Simplifying:

Ka1 * 0.004 mol = 0.2M * x/1000

0.004 * 10^(-pKa1) = 0.2M * x/1000

x = (0.004 * 10^(-pKa1) * 1000) / 0.2M

Now, substituting the given pKa1 value (1.1 x 10^-2):

x = (0.004 * 10^(-1.1 x 10^-2) * 1000) / 0.2M

x ≈ 14.65 mL

So, approximately 14.65 mL of 0.2M NaH2PO4 should be added to 40 mL of 0.1M Na3PO4 to make the buffer solution with a pH of 7.2.