Help with 4 & 5 4. Formic acid, HFor, has a K, value equal to about 1.8 x 10 . A

Question

Help with 4 & 5 4. Formic acid, HFor, has a K, value equal to about 1.8 x 10 . 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. 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 41

Explanation / Answer

Formic acid

HCOOH : ka = 1.8 E -4 , buffer pH = 3.90

We use Henderson equation to calculate mol ratio of sodium formate to formic acid

pH = pka + log ([sodium formate ] / [ formic acid ] )

pka = -log ka

3.90 = -log ( 1.8E-4) + log ( mol sodium formate / mol formic acid)

log ( mol sodium formate / mol formic acid) = 0.1552

now by taking antilog of both side

( mol sodium formate / mol formic acid) = Antilog (0.1552)

( mol sodium formate / mol formic acid) = 1.43

According to the problem condition we assume 1 M each.

Moles of formic acid = volume in L * molarity

= 0.020 L * 1.0 M

= 0.020 mol HFor

Lets find moles of NaFor by using mol ratio

mol sodium formate / 0.020 mol = 1.43

mol sodium formate = 0.0286 mol

We know molarity is 1.0 M we use molarity and number of moles of sodium formate

Volume in L = mol / Molarity = 0.0286 mol / 1.0 M

= 0.0286 L

Volume in mL = 28.6 mL

Question 2

Now we have mol ratio

mol sodium formate / mol formic acid) = 1.43

If we add the x mole of NaOH then x mol of sodium formate will be formed and x mole of formic acid will be decreased

(x mol sodium formate / formic acid –x ) = 1.43

We mol of formic acid = 0.10 M * 0.020 L = 0.002 mol

Lets plug this value in above equation

x / (0.002 –x ) = 1.43

x = 1.43 * (0.002-x)

x= 0.00286 – 1.43 x

x + 1.43 x= 0.00286

2.43 x = 0.00286

X = 0.001777

Now mol of NaOH = 0.001777 mol

Volume of NaOH = mol / molarity = 0.001777 mol / 0.10 M = 0.01177 L

= 11.8 mL

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