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 =
Moles BH⁺ = Moles HCl:
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
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
