[LHM]

Suppose that 0.122 g of phosphorous acid, H₃PO₃, is dissolved in water and that the total volume of the solution is 50.0 mL. Estimate the pH of the solution if 10.00 mL of 0.175 M NaOH(aq) solution is added.

So first I figured out that there were 0.00149 moles of H₃PO₃, and there are 0.00175 moles of NaOH. After reacting, there was 0.00149 mol H₂PO₃^- and 0.00026 mol NaOH. Everything up to this point matched up with what the answer key had. After this, however, the answer key and I did different things and got different answers. I assumed the 0.00026 mol NaOH would continue to react with the H₂PO₃^- and there would be 0.00026 mol HPO₃^2- and 0.00123 mol H₂PO₃^- and then used Henderson-Hasselbach to get a pH of 5.92.

On the other hand the answer key calculated the concentration of H⁺ from the 0.00026 mol NaOH/0.060 L. Then set up the ICE table using
H₃PO₃ + H₃O⁺ H₂PO₃^- + H₂O and got that the answer was pH=11.6.

So why did they do this? Is there something wrong about what I did?

[Borek]

So why did they do this?

Hard to say. They were either before first morning coffee or after last night beer. Don't drink and derive.

Is there something wrong about what I did?

No.