I have a big test in a couple of days. I really need steps on how to do this. This is a comprehensive question style that if I can master will be fine for the test. This is the second time I have had to post this question because it was both incorrect and incomplete the first time. Please only answer if you have time and understand it. The first few items are easy, the last few are more difficult.
Consider a titration of 25.00 mL solution of formic acid [Ka(HCOOH) = 1.8 x 10-4] with 33.55 mL of 0.1211 M solution of sodium hydroxide. Please show work.
a) The concentration of the formic acid solution before the titration.
b) The pH of the formic acid solution before the titration.
c) The pH of the solultion at half-equivalence point.
d) The pH of the solution at the equivalence point.
e) The pH of the solution when 1.00 mL of the NaOH have been added after the equivalence point was reached.
f) The volume (in mL) of 0.1211 M solution of barium hydroxide that would have to be added to reach the equivalence point if it were used instead of the 0.1211 M NaOH solution?
I have a big test in a couple of days. I really need steps on how to do this. This is a comprehensive question style that if I can master will be fine for the test. This is the second time I have had to post this question because it was both incorrect and incomplete the first time. Please only answer if you have time and understand it. The first few items are easy, the last few are more difficult.
Consider a titration of 25.00 mL solution of formic acid [Ka(HCOOH) = 1.8 x 10-4] with 33.55 mL of 0.1211 M solution of sodium hydroxide. Please show work.
a) The concentration of the formic acid solution before the titration.
b) The pH of the formic acid solution before the titration.
c) The pH of the solultion at half-equivalence point.
d) The pH of the solution at the equivalence point.
e) The pH of the solution when 1.00 mL of the NaOH have been added after the equivalence point was reached.
f) The volume (in mL) of 0.1211 M solution of barium hydroxide that would have to be added to reach the equivalence point if it were used instead of the 0.1211 M NaOH solution?
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