Question

1. A 0.130 M solution of a weak base is titrated with a 0.130 M HCl solution. After the addition of 8.50 mL of the HCl solution to 25.00 mL of the weak base solution, the pH of the solution is 9.36 . Determine the pKb of the weak base.2. Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to (a) 33.0 mL of 0.230 M NaOH(aq).

Answer 1

Answer:

1. 4.93 = pKb of the weak base

2. pH = 12.61

Explanation:

1. When a weak base, B, is being titrated with HCl, the reaction occurs as follows:

B + HCl → BH⁺ + Cl⁻

That means the moles added of HCl are the moles of BH⁺ produced and moles of B are initial moles of B - Moles of HCl

Thus:

Moles B:

Initial moles:

0.0250L * (0.130mol / L) = 3.25x10⁻³ moles B

Moles HCl:

8.5x10⁻³L * (0.130mol / L) = 1.105x10⁻³ moles HCl

3.25x10⁻³ - 1.105x10⁻³ =

2.145x10⁻³ moles B

Moles BH⁺ = Moles HCl:

1.105x10⁻³ moles BH⁺

pH of the buffer made from B/BH⁺ is determined using H-H equation for weak bases:

pOH = pKb + log [BH⁺] / [B]

Where pOH is 14-pH = 14-9.36 = 4.64

pKb is pKb of the weak base, our unknown.

[BH⁺] could be taken as moles of BH⁺ =  1.105x10⁻³ moles

And [B] as moles of B = 2.145x10⁻³ moles B

Replacing:

pOH = pKb + log [BH⁺] / [B]

4.64 = pKb + log [1.105x10⁻³ moles] / [2.145x10⁻³ moles]

4.64 = pKb -0.288

4.93 = pKb of the weak base

2. When HCl and NaOH are in solution the reaction that occurs is:

HCl + NaOH → H₂O + NaCl

To find pH we need to determine, first, which reactant is in excess:

Moles HCl:

0.023L * (0.230mol / L) = 5.29x10⁻³ moles

Moles NaOH:

0.033L * (0.0230mol / L) = 7.59x10⁻³ moles

That means NaOH is in excess and after the reaction will remain:

7.59x10⁻³ moles - 5.29x10⁻³moles = 2.3x10⁻³ moles NaOH = Moles of OH⁻

In 23+33mL = 56mL = 0.056L:

2.3x10⁻³ moles OH⁻ / 0.056L = 0.0411M [OH-]

As pOH = -log [OH-]

pOH = 1.39

pH = 14 - pOH

pH = 12.61