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March 28, 2017

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when a 25.0 mL sample of an unknown acid was titrated with a 0.100 M NaOH solution. Determine Ka for the unknown acid. Volume NaOH = 12.5 mL; pH=3.80

please help me explain in details!!!

  • chem--please help me!! - ,

    This problem needs a lot of work. In its present form, there are too many details not known.
    1. Technically, this can't be solved since we don't know if the acid is monoprotic, diprotic, or triprotic.
    Assuming it is monoprotic,
    HA + NaOH ==> NaA + H2O

    2. The problem SHOULD say that 25.0 mL is the equivalence point OR
    3. That the 12.5 mL represents titration to the half-way point. All of that information is missing.
    However, if we supply that missing information, then
    Ka = (H^+)(A^-)/(HA)

    At the exact half-way point in the titration, (A^-) = (HA), so Ka = (H^+).
    Convert pH to (H^+) and you will have Ka.

  • chem--please help me!! - ,

    25.00 mL sample of this sulfuric acid solution. You then titrate it with a 0.1322 M potassium hydroxide solution. A volume of 41.22 mL of potassium hydroxide was required to complete the titration. What is the concentration of this unknown sulfuric acid solution?

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