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    Buffers and Buffer Type Solutions

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    1. The Common Ion Effect. What is the pH of a solution that contains 0.250-M benzoic acid and 0.600-M sodium benzoate?

    2. Buffers. A buffer solution is made of 0.100 M HOCl and 0.250 M NaOCl. What is the pH of the resulting solution when a 15.0-mL portion of 0.200 M HCl is added to 100.0 mL of the buffer?

    3. Preparing a Buffer. The pH of blood is 7.40. Calculate the [HCO3-] / [H2CO3] ratio of a bicarbonate-carbonic acid buffer system that will effectively maintain a sample of blood at a pH of 7.40. Hint: the Henderson-Hasselbalch equation is useful for this type of problem.

    4. Weak Acid - Strong Base Titration. A 50.0-mL portion of 0.200 M HCOOH is titrated with 35.0 mL of 0.200 M KOH. What is the pH of the resulting solution?

    5. Weak Acid - Strong Base Titration. Now consider a titration that is similar to problem 4, except that a 50.0-mL portion of 0.200 M HCOOH has been titrated with 0.200 M KOH to the equivalence point. What is the pH at the equivalence point?

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    Solution Preview

    These questions are all about buffers and buffer type situations. Here we will use a similar equation to a weak acid, but it will be much simpler. This is the buffer equation, and you should try your best to use it without doing many chemical equations, at least on your first attempt at a question:

    Buffer based on an Acid
    Ka = [H+] * A-/HA

    (notice here that the A- and the HA terms are not in concentration brackets, you can use concentrations, or moles here)

    For a Buffer based on a Base
    Kb = [OH-] * HB+/B (very similar to the acid equation)

    The only tough thing to do is identify which numbers go where, the hint here is that the species you have the K value for, goes on the bottom of the ratio.

    1. The Common Ion Effect. What is the pH of a solution that contains 0.250-M benzoic acid and 0.600-M sodium benzoate?

    Applying this to your first question

    [HA] = 0.250
    [A-] = 0.600
    Ka = 6.3 x 10^-5 (this value should be available in any organic chemistry textbook or website)

    Using the equation above we can see that we only have one variable left which is the [H+], which will tell us the pH, which is our answer.

    6.3 x 10^-5 = [H+] * 0.600/0.250 rearrange and solve for [H+]

    [H+] = (6.3 x 10-5)*0.250/0.600
    = 0.00151
    With this value we can easily calculate the pH by using the pH = -log[H+] = ...

    Solution Summary

    The solution carefully explains the steps for each of the sub-problems, using relevant theory and equations.

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