3.(30pts) A buffer is prepared by mixing 100 mL of 0.800 M formic acid (HCOOH),

Question

3.(30pts) A buffer is prepared by mixing 100 mL of 0.800 M formic acid (HCOOH), a monoprotic acid, with 200 mL of 0.100 M sodium formate (NaHCOO). Ks of formic acid 1.7 x 104 at 25°C. (i) What is the pH of this buffer? (ii) If 0.75 mL of 10.0 M HCl is added to this buffer, what is the pH of the resulting solution? 4.(30pts)A scientist wants to make an efficient buffer solution having a pH 3.82. Which one of the following weak acids along with the sodium salt of its conjugate base would probably be best to use: m clorobenzioic acid, Ka 1.04 x 104 , p chlorocinnimic acid, K. 3.89 x 10, 2,5 dihydroxybenzioic acid, Ka- 1.08 x 103, acetoacetic acid, K,- 2.62 x 104. Justify your selection

Explanation / Answer

Ans. #3. Part I:

Total volume after mixing the two solution = 100 mL + 200 mL = 300 mL

# Now, using             C1V1 (original solution) = C2V2 (Mixed solution),

            [HCOOH] in mixed solution = (0.800 M x 100.0 mL) / 300.0 mL = 0.2667M

            [NaHCOO] in mixed solution = (0.100 M x 200.0 mL) / 300.0 mL = 0.0667M

Since 1 mol NaHCOO yields 1 mol HCOO-, the concertation of both these chemical species must be the same.

            So, [HCOO-] = 0.0667M

# Using HH equation:           pH = pKa + log ([A-] / [AH])

Where,            AH = weak acid, A- = Conjugate acid

# pKa = -log Ka = -log 1.7 x 10-4 = 3.77

Putting the value sin above expression-

            pH = 3.77 + log (0.0667 / 0.2667) = 3.77 + log 0.250 = 3.77 + (-0.60) = 3.17

# Part II: # Concertation after mixing HCl to the Buffer:

Total volume = 300.0 mL + 0.75 mL = 300.75 mL

Now,

            [HCOOH] = (0.800 M x 100.0 mL) / 300.75 mL = 0.2660 M

            [HCOO-] = (0.100 M x 200.0 mL) / 300.75 mL = 0.0665 M

            [H+] from HCl = (10.0 M x 0.75 mL) / 300.75 mL = 0.02494 M

# The Reaction in the total mixed solution:

Being strong acid, HCl undergoes complete dissociation in aqueous solution. 1 mol H+ from HCl reacts with 1 mol HCOO- to form 1 mol HCOOH.

So,

            [A-] neutralized = [H+] from HCl = 0.02494 M

            [AH] formed = [A-] neutralized = 0.02494 M

            Total [AH] = Initial [AH] + [AH] formed = 0.2660 M + 0.02494 M = 0.29094M

            Remaining [A-] at equilibrium = Initial [A-] – [A-] neutralized

                                                            = 0.0665 M – 0.02494 M = 0.04156 M

#Using HH equation –

            pH = 3.77 + log (0.04156 / 0.29094) = 3.77 + (-0.85) = 2.92

#4. pKa of m-chlorobenzoic acid = -log (1.04 x 10-4) = 3.98

pKa of p-chlorocinnimic acid = -log (3.89 x 10-5) = 4.41

pKa of 2,5-dihydrozybenzoic acid = -log (1.08 x 10-3) = 2.96

pKa of acetoacetic acid = -log (2.62 x 10-4) = 3.58

# A buffer exhibits maximum buffer capacity when the pKa of the weak acid is equal to that of the required pH.

Since the pKa of m-chlorobenzoic acid is closest to the required pH 3.82, it is the most suitable option to make the buffer.

So, the best option is- m-chlorobenzoic acid and its conjugate base.

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