Question

Exactly 25.00 ml of a 0.0685 M pentanoic acid solution,(C₄H₈O₂), is placed in a flask and titrated with standardized 0.050 M NaOH. (the pKa of this acid is 4.84).         

      A. Determine the endpoint of the titration

      B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution:

                  a.) Initially, 0.00 ml NaOH.                 b.) 5.00 ml                          c.) 15.00 ml

                  d.) 25.00 ml                                      e.) 34.00 ml                             f.) At the endpoint of the titration        

                      g.) 34.35 ml                                      h.) 45.00 ml                        i.) 50.00 ml

Answer 1

Pentanoic acid (say PA) is a weak monobasic acid and it reacts with NaOH as follows-

CH₃CH₂CH₂CH₂COOH + NaOH → CH₃CH₂CH₂CH₂COONa + H₂O

Given, pKa = 4.84

Or, -log K_a = 4.84

Or, K_a = 10^(-4.84)

∴ K_a = 1.4 × 10^-5

(A) Determination of end point:

Total conc. (mmol) of PA before addition of NaOH = 25 × 0.0685 = 1.7125 mmol

At the end point mmol of pentanoic acid must be equal to the mmol of NaOH

So,          x × 0.05 mmol = 1.7125 mmol

Therefore, x = 34.25 mL

Thus, on addition of 34.25 mL of 0.05 M NaOH solution end point will come.

(B) The dissociation of pentanoic acid occurs as follows:

CH₃CH₂CH₂CH₂COOH → CH₃CH₂CH₂CH₂COO^- + H⁺

(a) Addition of 0.00 mL of NaOH

When there will no trace of NaOH and pH will be determined only from dissociation of weak acid and concentration of acid is given C_a = 0.0685 M

We know that conc. of [H⁺] = √(K_a × C_a) = √{(1.4×10^-5)×(0.0685)} = 0.00097

pH = -log [H⁺] = -log (0.00097) = 3.01

∴ pH = 3.01

(b) Addition of 5.00 mL of NaOH

Total conc. of NaOH = 5.0 × 0.05 M = 0.25 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is much higher than NaOH the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pH = pK_a + log {[salt]/[acid]}

[salt] = 0.25 / (25+5) = 0.25 / 30 M

[acid] = (1.7825-0.25) / (25+5) = 1.4625 / 30 M

pH = 4.84 + log [0.25/30] / [1.4625/30] = 4.84 – 0.76 = 4.08

∴ pH = 4.08

(c) Addition of 15.00 mL of NaOH

Total conc. of NaOH = 15.0 × 0.05 M = 0.75 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is much higher than NaOH the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pH = pK_a + log {[salt]/[acid]}

[salt] = 0.75 / (25+15) = 0.75 / 40 M

[acid] = (1.7825-0.75) / (25+15) = 1.0325 / 40 M

pH = 4.84 + log [0.75/30] / [1.0325/30] = 4.84 – 0.13 = 4.71

∴ pH = 4.71

(d) Addition of 25.00 mL of NaOH

Total conc. of NaOH = 25.0 × 0.05 M = 1.25 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is much higher than NaOH the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pH = pK_a + log {[salt]/[acid]}

[salt] = 1.25 / (25+25) = 0.75 / 50 M

[acid] = (1.7825-1.25) / (25+25) = 0.5325 / 50 M

pH = 4.84 + log [1.25/50] / [0.5325/50] = 4.84 +0.37 = 5.21

∴ pH = 5.21

(e) Addition of 34.00 mL of NaOH

Total conc. of NaOH = 34.0 × 0.05 M = 1.70 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is much higher than NaOH the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pH = pK_a + log {[salt]/[acid]}

[salt] = 1.70 / (25+34) = 1.70 / 59 M

[acid] = (1.7825-1.70) / (25+34) = 0.0825 / 59 M

pH = 4.84 + log [1.70/59] / [0.0825/59] = 4.84 + 1.31 = 6.15

∴ pH = 6.15

(f) At the end point i.e. addition of 34.25 mL of NaOH

Total conc. of NaOH = 34.25 × 0.05 M = 1.7125 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is equal to the conc. of NaOH.

pH of this mixture can be calculated from salt hydrolysis of

CH₃CH₂CH₂CH₂COONa + H₂O → CH₃CH₂CH₂CH₂COOH (weak acid) + NaOH (strong base)

pH = 7 + ½ pK_a + ½ log [C]

[salt] = 1.7125 / (25+34.25) = 1.7125 / 59.25 =0.028 M

pH = 7+ 2.42 + ½ log [0.028] = 7+ 2.42 –(0.5×1.55) = 8.64

∴ pH = 8.64

(g) Addition of 34.35 mL of NaOH

Total conc. of NaOH = 34.35 × 0.05 M = 1.7175 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is less than that of conc. of NaOH.

pOH of this mixture can be calculated from the conc. of OH- ions. Then pH can be calculated from pH + pOH =14

conc. of base after complete neutralization

[OH^-] = (1.7175-1.7125) / (25+34.35) = 0.005 / 59.35 M = 0.000084 M

pOH = -log [OH^-] = -log (0.000084) = 4.07

pH = 14 - 4.07 = 9.93

(h) Addition of 45 mL of NaOH

Total conc. of NaOH = 45 × 0.05 M = 2.25 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is less than that of conc. of NaOH.

pOH of this mixture can be calculated from the conc. of OH- ions. Then pH can be calculated from pH + pOH =14

conc. of base after complete neutralization

[OH^-] = (2.25-1.7125) / (25+45) = 0.5375 / 70 M = 0.0076 M

pOH = -log [OH^-] = -log (0.0076) = 2.11

pH = 14 – 2.11 = 11.89

(i) Addition of 50 mL of NaOH

Total conc. of NaOH = 50 × 0.05 M = 2.5 M

Total conc. PA before addition of NaOH = 25 × 0.0685 M = 1.7125 M

Since conc. of PA is less than that of conc. of NaOH.

pOH of this mixture can be calculated from the conc. of OH- ions. Then pH can be calculated from pH + pOH =14

conc. of base after complete neutralization

[OH^-] = (2.5-1.7125) / (25+50) = 0.7875 / 75 M = 0.0105 M

pOH = -log [OH^-] = -log (0.0105) = 1.97

pH = 14 – 1.97 = 12.03