Advance Study Assignment: pH Measurements and the Properties of Butlers A soluti

Question

Advance Study Assignment: pH Measurements and the Properties of Butlers A solution of a weak acid was tested with the indicators used in this experiment. The colors observed were as follows: What is the approximate pH of the solution? The pH of a 0.10 M HCN solution is 5.2. a. What is [H^+] in that solution? b. What is [CN^-]? What is [HCM]? (Where do the H^+ and CN^- ions come from) c. What is the value of K_a for HCN? Formic acid, HFor, has a K_a value equal for about 1.8 times 10^-4. A student is asked to prepare a buffer having a pH of 3.40 from a 0.10 M HFor and a 0.10 M NaFor solution. How many milliliters of the NaFor solution should she add to 20 mL of the 0.10 M HFor lo make the buffer? When 5 drops of 0.10 M NaOH were added to 20 mL of the buffer in problem 3. the pH went from 3.40 to 3.43. Write a net ionic equation to explain why the pH didn't go up to about 10. as it would have if that amount of NaOH were added to distilled water or to 20 mL 0.00040 M HC1. which also would have a pH of 3.40.

Explanation / Answer

QUESTION 3

Foric acid HFor, Ka = 1.8*10^-4

HFor <-> H+ + For-

buffer at pH = 3.40

find pKa of acid, pKa = -log(Ka) = -log(1-8*10^-4) = 3.744

Stock solution

HFor = 0.1 M

NaFor = 0.1 M

find Volume of each for

V = 20 mL of HFor

we must relate via buffer equatons

pH = pKa + log(NaFor / HFor )

3.40 = 3.744 + log( mmol of NaFor / mmol of HFor )

mmol of Hfor = MV = 20*0.1 = 2 mmol of HFor

10^(3.4-3.744) = mmol of NaFor / mmol of HFor

0.45289= mmol of NaFor / 2

mol of NaFor = 0.45289*2

mmol of NaFo = 0.90578

M = mmol/mL

mL = mmol/M = 0.90578/0.1 = 9.057 mL of NaFor solution required

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