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chemistry

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We titrated Sprite with 0.1051 M standard NaOH. I've ploted three different titrations and am supposed to use the titration curve to identify the acid. I assume the acid is Citric since it's Sprite, but the lab manual also says "to recall that two K values must be far enough apart ( at least 10^ -3) to see two equivalence points. What am I supposed to use to figure out the K values. I have pH's vs. mL of NaOH and started with 25mL of Sprite. Help PLEASE!

  • chemistry - ,

    Why not plot the graph and see what you get? The pK value is the pH at the half-way point to the equivalence point. If you can identify the equivalence point, then take half that volume and read the pH for the pKa. If you have two equivalence points just do that twice. Could the acid be carbonic and not citric acid? Carbonic acid has two ks that are separated by at least 10^3.

  • chemistry - ,

    Okay, thanks. The choices offered are phosphoric acid, citric acid, a oxalic.
    Let me make sure I'm understaning you, for example: I have an equilvenc point at 4.80mL and the pH at that point read 6.93. So I should go to 3.5 and the pH at that point is 5.69. So K= 5.69? or am I still missing something? I did plot the graphs, but honestly, they look a LOT alike, so I'm thinking that's not a big help. Besides I really would like to know the math behind this. Thank you for all you help.

  • chemistry - ,

    Close but not quite there. If the equivalence point is at 4.80 mL, then go to 2.40 mL (the half-way point) and the pH at that point equals the pKa (not Ka). The theory is simple enough.
    A weak acid HA ==> H^+ + A^- and
    Ka = (H^+)(A^-)/(HA). Solve for
    (H^+)=Ka*(HA)/(A^-). At the half way point, the acid, HA, has been exactly half neutralized forming an equal amount of A^-; therefore, (HA) = (A^-) so they cancel and (H^+) = Ka. If we take the negative log of both sides, we get pH = pKa. I can't help much more, and this may not be much help, since I can't see the data or graph.

  • chemistry - ,

    That does help greatly. Thank you so much!

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