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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: LHM on April 07, 2011, 09:44:55 PM

Title: pH of the solution
Post by: LHM on April 07, 2011, 09:44:55 PM

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⁺ :rarrow: 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?

Title: Re: pH of the solution
Post by: Borek on April 08, 2011, 03:11:34 AM

Quote from: LHM on April 07, 2011, 09:44:55 PM

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.

Quote

Is there something wrong about what I did?

No.