Question

100. ml of a 0.025 M solution of benzoic acid(ka=6.3x10^5) is tritated with 0.100M NaOH to the equivalence point.

a.What volume of NaOH will be needed to completely neutralize the acid?

b. What is the pH when 10.0 ml of 0.100 M NaOH is added to the acid in the flask?

c. What is the pH when 12.5 ml of 0.100 M NaOH is added to the flask?

d. What is the pH at the equivalence point?

e. What is the pH when 35.0 ml of 0.100 NaOH has been added to the flask?

Answer 1

(a) According to normality equation, N₁V₁ = N₂V₂

As valence factor for both NaOH and PhCOOH = 1

M₁V₁ = M₂V₂

Where, M₁ and V₁ = molarity and volume of PhCOOH respectively and M₂ and V₂ = molarity and volume of NaOH

0.025 X 100 = 0.1 X V₂

V₂ = 25 mL

So, 25 mL of 0.1 M NaOH is required to neutralize 100 mL of 0.025 M PhCOOH

(b) M₁ = 0.025, V₁ = 100 mL, M₂ = 0.1 and V₂ = 10 mL

first determine the number of moles of PhCOOH and PhCOO^- that are present after the neutralization reaction. We then calculate pH using Ka together with [PhCOOH] and [PhCOO^-].

we know, moles = molarity X volume

number of moles of acid = 0.025 X 0.1 = 2.5 X 10^-3

number of moles of alkali = 0.01 X 0.1 = 1 X 10^-3

PhCOOH + OH^- PhCOO^- + H₂O

Initial  2.5 X 10^-3

after addition of 1 X 10^-3 moles of alkali, number of moles of acid left = (2.5 - 1) X 10^-3

Final 1.5 X 10^-3 1 X 10^-3

The volume of solution after addition of alkali = 110 mL

The resulting molarities of PhCOOH And PhCOO^- after the reaction are therefore

[PhCOOH] = 1.5 X 10^-3 / 0.110 = 0.136 M

[PhCOO^-] = 1 X 10^-3 / 0.110 = 0.009 M

We know, K_a = [H⁺] [PhCOO^-] / [PhCOOH] = 6.3 X 10^-5

[H⁺] = 9.519 X 10^-4

Now, pH = -log [H⁺] = 3.02

(c) Similarly, for 12.5 mL addition of NaOH

number of moles of alkali = 0.0125 X 0.1 = 1.25 X 10^-3

after addition of 1.25 X 10^-3 moles of alkali, number of moles of acid left = 1.25 X 10^-3

The volume of solution after addition of alkali = 112.5 mL

[PhCOOH] = 1.25 X 10^-3 / 0.110 = 0.011 M

[PhCOO^-] = 1.25 X 10^-3 / 0.110 = 0.011 M

[H⁺] = 6.3 X 10^-5

pH = 4.2

(d) The initial number of moles of benzoic acid will equal the number of moles of benzoate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of benzoate ion. Because the benzoate ion is a weak base, we can calculate the pH using K_b and [PhCOO^-].

number of moles of acid = 2.5 X 10^-3 = number of moles of PhCOO^-

Since 25 mL of NaOH will be required to reach the equivalent point the volume of total solution = 125 mL

[PhCOO^-] = 2.5 X 10^-3 / 0.125 = 0.02 M

We know, K_b = K_w / K_a = (1 X 10^-14) / (6.3 X 10^-5) = 1.587 X 10^-10

Also, K_b = [PhCOOH] [OH^-] / [PhCOO^-] = (x) (x) / (0.02-x) = 1.587 X 10^-10

Making the approximation that 0.0200 – x is equivalent to 0.0200, and then solving for x, we have x = [OH^-] = 1.781 X 10^-6 M

pOH = -log [OH^-] = 5.749

pH = 14 - 5.749 = 8.25

(e) part can be solved similarly