A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. after the addition of the following amounts of KOH with sig figs.
a) 24.5 ml
b) 25.0 ml
c) 30.0 ml

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  1. I worked a problem similar to this earlier today. Here it is.

    The differences are this is an acid and pyridine is a base but that doesn't change the process at all. The equation of propanoic acid, if we call it HPr, is
    HPr + KOH ==> KPr + H2O
    KPr is a salt, potassium propionate
    a is 24.5 mL that is handled the same way as my other post.
    b is the equivalence point and handled the same way
    c. is the same as the 25, 28, 30 mL in the other post.
    For the pH of part b, the equation for the salt is
    KPr + HOH ==> HPr + OH^- + K^+

    Let me know if you run into problems or have questions.

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  2. Hi Dr.Bob, I can't see your other post for some reason, and I had a question about the same things and I still don't understand how to do it.

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