Be able to calculate the pH of a solution prepared by mixing 50.0mL of 0.200M NaH2PO4 (pKa2=7.20) with 50.0mL of 0.120M NaOH. I did work out a pH of 7.38 but I am not sure if i need to take into account any dilution of the mixing of solutions and do you need to know pKa1 to use the Henderson-Hasselbalch equation?
For this problem you do not need pKa1.(Note that pK1 has H3PO4 in it and that isn't one of the ingredients.)
I obtained the same answer and I did it with mols ONLY (which doesn't involve dilution and I like that way) but I also did it with molarity and took into account dilution. Same answer either way.So I think the answer is that you do not need to worry about dilution UNLESS you plug in concns in M (and not mols).
To calculate the pH of the solution prepared by mixing the two solutions, you can use the Henderson-Hasselbalch equation. However, before applying the equation, it is important to consider the dilution effects and whether pKa1 is necessary.
First, let's address the dilution effect. When the two solutions are mixed, the final volume will be the sum of the volumes used for each solution, which is 50.0 mL + 50.0 mL = 100.0 mL.
Since the two solutions are mixed in equal volumes, their concentrations will be diluted by a factor of 2. Therefore, the final concentrations of NaH2PO4 and NaOH will be:
- NaH2PO4 concentration = (0.200 M) / 2 = 0.100 M
- NaOH concentration = (0.120 M) / 2 = 0.060 M
Now, let's address the use of pKa1 in the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is typically used to calculate the pH of a solution containing a weak acid and its conjugate base. In this case, NaH2PO4 is a weak acid and its conjugate base is HPO4^2-.
pKa1 is the acid dissociation constant for the first ionization step of phosphoric acid (H3PO4 -> H+ + H2PO4^-). Since we are working with NaH2PO4, which is a salt of the first ionization step, pKa1 is not directly needed in the calculation. Instead, we will use pKa2, which is the acid dissociation constant for the second ionization step (H2PO4^- -> H+ + HPO4^2-).
Now, let's apply the Henderson-Hasselbalch equation using pKa2:
pH = pKa2 + log ([HPO4^2-] / [NaH2PO4])
Substituting the values we obtained earlier:
pH = 7.20 + log (0.060 M / 0.100 M)
Calculating the ratio:
pH = 7.20 + log (0.6)
pH ≈ 7.20 + (-0.22)
pH ≈ 7.38
Therefore, your calculation of pH = 7.38 is correct, considering the dilution effects and the use of pKa2 instead of pKa1 in the Henderson-Hasselbalch equation.