What is the pH of a mixture of 0.042M NaH2PO4 and 0.054M Na2HPO4

Use the Henderson-Hasselbalch equation.

pH = pKa2 + log (base)/(acid)
base is HPO4^-; acid is H2PO4^-

To find the pH of a mixture of NaH2PO4 and Na2HPO4, you need to consider the acid-base properties of the phosphate ions. NaH2PO4 is a weak acid and Na2HPO4 is a weak base.

The phosphate ion, HPO4^2-, acts as a weak acid when it donates a proton to water, and becomes H2PO4^-. On the other hand, the phosphate ion can act as a weak base when it accepts a proton from water, and becomes HPO4^2-.

To find the pH of the mixture, you need to calculate the concentrations of the acid form (H2PO4^-) and the base form (HPO4^2-).

First, let's calculate the concentrations of H2PO4^- and HPO4^2- using the given concentrations of NaH2PO4 and Na2HPO4.

The equation for the dissociation of NaH2PO4 is:
NaH2PO4 ↔ Na+ + H2PO4^-

Since NaH2PO4 is a weak acid, it does not dissociate completely. Therefore, we can assume that the concentration of H2PO4^- is equal to the concentration of NaH2PO4.
So, [H2PO4^-] = 0.042 M

Next, let's calculate the concentration of HPO4^2-.
The equation for the dissociation of Na2HPO4 is:
Na2HPO4 ↔ 2Na+ + HPO4^2-

Again, since Na2HPO4 is a weak base, it does not dissociate completely. Therefore, we can assume that the concentration of HPO4^2- is equal to half the concentration of Na2HPO4.
So, [HPO4^2-] = 0.054 M / 2 = 0.027 M

Now, we need to use the Henderson-Hasselbalch equation to calculate the pH of the mixture:

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

The pKa value for H2PO4^-/HPO4^2- is 7.21.

Plugging in the values we calculated:
pH = 7.21 + log(0.027/0.042)
pH = 7.21 + log(0.6429)
pH ≈ 7.21 - 0.193
pH ≈ 7.02

Therefore, the pH of the mixture of 0.042 M NaH2PO4 and 0.054 M Na2HPO4 is approximately 7.02.

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