36. Repeat the procedure from problem #35 but for the titration
of 25.0 mL of 0.100 M NH3 (Kb = 1.8 ×
10-5)
with 0.100 M HCl.
ANS: at e.p. pH = 5.28
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37. A sample of a certain monoprotic weak acid was dissolved
in water and titrated with 0.125 M NaOH, requiring 16.00 mL to
reach the equivalence point. During the titration, the pH after
adding 2.00 mL of NaOH was 6.912. Calculate Ka for
the weak acid. ANS: 1.74 x 10-8
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38. What volume of 0.0100 M NaOH must be added to 1.00 L of
0.0500 M HOCl to achieve a pH of 8.00?
(The pKa for HOCl is 7.46)
ANS: 3.8 L
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This is another titration curve calculation, but for the case of a base being titrated with an acid.
Hints:
Recall that this is the titration curve for a Weak Base, so you're best directed to use the pH line at all stages of titration except for the very beginning and the equivalence point. Beyond the equivalence point, of course, you're only adding strong acid.
Clueless:
This is a "backwards" calculation, wherein you are using titration data to come up with a value of pKa (and thus Ka). This is a plausible way of actually coming up with a value for pKa for a newly synthesized acid, and the Organic Chemists actually do this.
Hints:
Again you need to set up a template, this time the H-H Equation. You also need to know how much acid is actually present (which comes from the full titration data). Finally, since you're told the pH at one stage, you need to determine how many mmoles of acid and conjugate are actually present at that pH value. This leaves you with only 1 unknown in the H-H Eqn, the pKa.
Clueless:
A buffer preparation problem which is very pragmatic. Here you are starting with the weak acid and using a strong base to generate the necessary amount of conjugate. Labs actually do something very similar to this.
Hints:
Again you need to know how many mmoles of weak acid you have on hand at first. You also need to know the ratio of conjugate to weak acid at the desired pH. When you learn this you can set up an expression in the H-H Eqn wherein there is only one value of X, the amount of acid which was reacted with the strong base and thus the amount of conjugate formed. This will permit a calculation for the number of mmoles of acid which react, which will tell you how many mmoles of strong base have to be added. From this point you are back to another Chem 111 solution calculation problem.