I will do the first one and you can do the others.
pH = pKa + log[(base)/(acid)]
I found Ka of 7.2E-4 for HF but you need to use the number in your text. pKa from this value is 3.14.
pH = 3.14 + log(NaF/HF)
pH = 3.14 + log(0.4545/0.4105) = 3.18
[Note: I have divided the molarity by 2 to find moles in 0.5L of each since the buffer capacity is defined as the moles of base or acid that can be added to a LITER of buffered solution without changing the pH more than +/- 1. The answer to your question is to determine how much base (or acid) that can be added to each without changing the pH more than +/- 1; i.e., between range 4.18 to 2.18)
......HF +..... NaOH ==> NaF +... H2O
start 0.4105... 0........0.4545...0
change -x......+x........+x.......+x
end...0.4105-x..0........0.4545+x....N/A
4.18 = 3.14 + log(0.4545+x/0.4105-x)
1.04 = log etc.
10.96 = (0.4545+x)/(0.4105-x)
solve for x and I obtained
0.338 for moles NaOH I can add.
If you chose to go the other way (adding acid) the answer will be slightly different. I found 0.369 moles acid which is why some profs don't like to use this definitions because +/- 1 pH unit isn't the same for base and acid. At any rate, compare the numbers.
Of the following solutions, which has the greatest buffering capacity?
A) 0.821 M HF and 0.909 M NaF
B) They are all buffer solutions and would all have the same capacity.
C) 0.100 M HF and 0.217 M NaF
D) 0.121 M HF and 0.667 M NaF
E) 0.821 M HF and 0.217 M NaF
how do i know when something has greater buffer capacity?
p.s. happy thanksgiving to everyone! sucks to be studying today of all days!
1 answer
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