These take a while to work, but there are two main parts to each. Halfway to tbe equivalence point, pH = pKa, that is right.
I will work the first one as an example: What is the pH after adding 15 mL of 0.20 M NaOH to 35.0 mL of 0.20 M HC2H3O2? Once you get the bugs worked out of question 1, question 2 will make more sense (I hope!).
Part I. Reaction of the acid with the NaOH.
I like to write the reaction with the moles of substances under the formulas so it's easier to tell what's going on. Since NaOH is a strong base, this reaction will go 100% to products. We also need to calculate the moles of HC2H3O2 and NaOH that are initially present.
moles of HC2H3O2 = volume in liters X molarity
= 0.0350 L X 0.20 mol/L = 0.0070 mol
moles of NaOH = 0.0150 L X 0.20 mol/L = 0.0030 mol
HC2H3O2 (aq) + NaOH (aq) ––�?gt; NaC2H3O2 (aq) + H2O (l)
Initially: 0.0070 mol 0.0030 mol 0 (solvent)
(excess) (limiting)
After Rxn: 0.0040 mol 0 0.0030 mol (solvent)
Since NaOH is the limiting reactant, none of it will be left after the reaction.
The amount of the exceess reactant, HC2H3O2, that remains after the reaction is the initial amount minus the amount that reacts. The moles of HC2H3O2 that react is equal to the moles of NaOH since the reaction is one-to-one.
The moles of NaC2H3O2 that form is equal to the moles of NaOH that react, since NaOH is the limiting reactant and the reaction is one-to-one for these substances as well.
OK, ready for....
Part II. Calculation of the pH.
In this case, since we have a weak acid (HC2H3O2) and its conjugate base (C2H3O2�?/SUP> from NaC2H3O2) in the same solution (which by definition is a buffer solution), we can simply plug into the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log { [C2H3O2�?/SUP>] / [HC2H3O2] }
Ordinarily, we would convert moles of substances back to molarity units by dividing the moles by the total volume in liters, which is 0.0500 L. However, in the Henderson-Hasselbalch equation, you would be dividing the numerator and the denominator by that same total volume. This results in the volumes canceling. So, we can simply plug the moles of C2H3O2�?/SUP> and HC2H3O2 into the formula. We have to be careful about doing that, because normally we want amounts in units of molarity, not moles, when doing pH calculations.
pH = –log(1.8 X 10�?) + log (0.0030 mol / 0.0040 mol)
= 4.7447 + �?.1249
= 4.6198 or about 4.62 (I habitually give pH values to two places after the decimal.)
Remember, the Henderson-Hasselbalch equation can only be used if both a weak acid and its conjugate base are present. If only one of these is present, you have to calculate the pH differently (pH of a solution of a weak acid or weak base only).
In this problem, it is hard to tell if the pH "makes sense." The pH will depend on the amounts of weak acid and weak base present of course, but also on their relative strengths, which is given by the Ka of the weak acid and Kb of the weak base. Here, the weak acid (acetic acid) is still in excess over the weak base (acetate ion), so we might expect the pH to be less than 7. Ka for acetic acid, 1.8 X 10�?, is much larger than Kb for the acetate ion, 5.6 X 10�?0. Therefore, the weak acid "dominates" the solution as well.
As I said, this is a more involved problem involving a number of concepts, but once you get the method down, they can be worked much faster!
Try working question 2. The reaction for Part I is,
NH3 (aq) + HCl (aq) ––�?gt; NH4Cl (aq)
For the Henderson-Hasselbalch calculation in Part II, the weak acid is ammonium ion, NH4+, and the weak base is ammonia, NH3. Good luck!
Steve
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