Question

(3) 36. What is the ratio (HCOOHCOOH) at pH 2.75? The pK, of formic acid is 3.75. 37. 1.42 L buffer solution consists of 0.181M butanoic acid and 0.310 M sodium butanoate. Calculate the pH of the solution following the addition of 0.069 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52x10-5 (6) 38. You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.140 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer? (6)

Answer 1

36.

According Henderson - Hasselbaltch's equation of acid buffers,

pH = pKa + Log[Salt] / [Acid]

2.75 = 3.75 + Log[HCOO-]/[HCOOH]

2.75 - 3.75 = Log[HCOO-]/[HCOOH]

- 1.00 = Log[HCOO-] / [HCOOH]

[HCOO-] / [HCOOH] = 1.00 x 10^-1

37.

Initial moles of acid = molarity x volume in L = 0.181 x 1.42 = 0.257 mol

Initial moles of salt = 0.310 x 1.42 = 0.440 mol

Addition of a base, NaOH increases the concentration of salt and decreases the concentration of acid.

SO,

Final moles of acid = 0.257 - 0.069 = 0.188 mol

Final moles of salt = 0.440 + 0.069 = 0.509 mol

Ka = 1.52 x 10^-5

pKa = - LogKa = - Log(1.52 x 10^-5) = 4.82

Formula,

pH = pKa + Log[salt]/[acid[

pH = 4.82 + Log(0.509 / 0.188)

pH = 4.82 + 0.432

pH = 5.25

38.

Let us consider the volume of acid = x mL

Then,

Volume of salt = 100 - x mL

pH = pKa + Log[salt]/[acid[

4.00 = 4.20 + Log(0.140 (100 - x) / (0.100 x)

Log0.140(100-x) / (0.100 x ) = - 0.20

0.140 ( 100 - x ) / 0.100 x = 10^-0.20

14.0 - 0.140 x = 0.631 * 0.100 x

14.0 = 0.0631 x + 0.140 x

14.0 = 0.203 x

x = 14.0 / 0.203

x = 69.0 mL

Therefore,

Volume of acid solution = 69.0 mL

Volume of salt solution = 100 - 69.0 = 31.0 mL