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Dr Bob

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What is the pH of the solution that results from mixing 5 ml of 0.2 M NaOH and 100 mL of 0.05 M lactic acid? (Pka of lactic acid= 3.86)

Use the Henderson-Hasselbalch equation. The mixture you have indicated will react to form a buffer, the base will be the sodium lactate and the acid will be the lactic acid not neutralized by the NaOH. Post your work if you get stuck.

I know that I'm supposed to use the equation:

pH= pKa+log(deprotnated, which case the lactic acid not neutralized)/(protonated, sodium lactate)

but I'm not sure where the numbers come into play. i got that initally there are 0.001 mol of NaOH and 0.005 mol of lactic acid. I tried doing limiting reagents and got that the products each get 0.001 mol? which would give the ratio of 1, which i don't think is right...

  • Dr Bob -

    So what is the reaction? I'll call lactic acid HL for simplicity.
    HL + NaOH ==> NaL + H2O
    HL = 0.005 mols. You are correct.
    NaOH = 0.001 mols. You are correct.

    Now you see that an acid + base = salt + water.
    So the smaller mols will react completely and leave an excess of what is left.
    0.001 mol NaOH will form 0.001 mol NaL, 0.001 mol H2O, and will leave 0 mol NaOH (all of it is reacted), but 0.005 - 0.001 mol = 0.004 mols HL.
    So you have an acid (0.004 mol HL) and a salt of the weak acid (0.001 NaL) so that is a buffered solution.
    (NaL) = (0.001 mol/0.105 L)
    (HL) = (0.004 mol/0.105 L)
    The H-H equation is
    pH = pKa + log (base/acid)
    base is NaL
    acid is HL.

  • Dr Bob -


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