1.) Formic acid, HFor, has a Ka value equal to about 1.8*10^-4. A student is ask
ID: 493698 • Letter: 1
Question
1.) Formic acid, HFor, has a Ka value equal to about 1.8*10^-4. A student is asked to prepare a buffer having a pH of 3.90 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20.0 mL of the HFor solution to make the buffer??4. Formic acid, HFor, has a Ka value equal to about 1.8 x 10 4. A student is asked to prepare a buffer having pH of 3.90 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20. mL of the HFor solution to make the buffer? (See discussion of buffers.) mL 5. How many mL of 0.10 M NaoH should the student add to 20 mL 0.10 M HFor if she wished to prepare a buffer with a pH of 3.90, the same as in Problem 4? mL
Explanation / Answer
4) Henderson equation is
pH = pKa + log([conjucate base]/[Acid])
3.90 = 3.74 + log([Conjucate base]/[Acid])
[Conjugate of base]/[Acid] = 1.45
Therefore, [conjucate base] = 1.45 × [ Acid]
[NaFor] = 1.45 × [ HFor ]
Volume of HFor = 20ml
Therefore, NaFor = 1.45 × 20 = 29ml
Therefore, 29 ml of NaFor should be taken for 20ml of HFor solution if both solutions are in same concentration.
5) 11.8 ml of 0.1M NaOH should be add
If 11.58 ml of NaOH added concentration of NaFor will be 0.0371M and concentration of HFor will be 0.0257M,and the resultant buffer solution will be of pH = 3.90
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