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18 Dec 2019

Question #46. A 0.1 M aqueous solution of formic acid was titrated at 25ÂºC with a 0.1 M

aqueous solution of NaOH. In this experiment, the equivalence point will be reached this pH

value (For formic acid, Ka = 1.80 x 10-4) :

Please show work

(1) pH < 7.0; (2) pH > 7.0; (3) pH = 7.0; (4) None of the above

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Keith LeannonLv2

31 Dec 2019

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Question #48. In a titration experiment, 50 mL of 0.1 M acetic acid (Ka = 1.75 x 10-5) was titrated with a 0.1 M NaOH (formal concentration) at 25 ÂºC. The pH of the solution at the equivalence point will be this: Please show work (1) 7.00; (2) 8.73; (3) 9.06; (4) None of the above

What is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.2567 M NaOH? Ka = 1.8 x 10^-4

How do you solve this without the M1V1 = M2V2 ?

Answer given: 8.44

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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