# Hint even though this is a titration of a weak base

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Hint: Even though this is a titration of a weak base with a strong acid, pH still equals pK a at halfway to the equivalence point. But in this titration, what is the acid in the buffer system? When you obtain the pKa, you will be determining it for this acid. ½ equivalence point = 2.5 drops pH = 9.2 = pKa Ka = 10^-9.2 = 6.31 * 10^-10 Kb = Kw/Ka Kb = 1*10^-14/6.31*10^-10 = 1.58 * 10^-5 pKb = -log(Kb) = 4.8 Since the pH still equals pKa at ½ equivalence point and the equivalence point is at 5 drops, I was able to use this to determine the pKa of the titration. With a known value of pKa, you can find Ka which would be the value for the acid. Hence, I would have to find Kb (for the base) through a known quantity of Kw. With Kb you can determine pKb in the same method as pKa and Ka.
7 Question 18: Why is the equivalence point acidic for the titration between ammonia and HCl? Hint: Think about what species are present at the equivalence point.
Question 19: Suppose you want to make a buffer solution of pH 8.2 for a clinical physiological experiment. Using Figure 14.9, describe how you would do this. Include the K b of the base used as well as the amount of added HCl.
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