Given:
Volume of HC9H7O2 = 10.0 ml = 0.010 L
Molarity of HC9H7O2 = 0.10 M
Ka (HC9H7O2) = 3.6 * 10-5
Molarity of NaOH = 0.20
To determine:
pH at the equivalence point
Step-by-step explanation:
The titration equation is:
C9H7OOH + NaOH ↔ C9H7OONa + H2O
As per the reaction stoichiometry-
1 mole of the weak acid reacts with 1 mole of base to form 1 mole of the conjugate base
# Moles of C9H7COOH = M * V = 0.10 M * 0.010 L = 0.001 moles
Now, based on the stoichiometry:
# moles of C9H7COOH = # moles of NaOH = # moles of C9H7COONa = 0.001 moles
volume of NaOH required to reach the eq. pt is:
V(NaOH) = Moles NaOH/Molarity NaOH
= 0.001 moles/0.20 M = 0.005 L = 5ml
Total volume = 10.0 ml + 5.0 ml = 15 ml
The concentration of the conjugate base the equivalence point is:
[C9H7COONa] = 0.001/0.015 = 0.0667 M
To find pH set up the ICE table as follows:
C9H7COO- + H2O ↔ C9H7COOH + OH-
I 0.0667 0 0
C -x +x +x
E 0.0667-x +x +x
Kb = [C9H7COOH][OH-]/[C9H7COO-]
Kw/Ka = x²/(0.0667-x)
10⁻¹⁴/3,6*10⁻⁵ = x²/(0.0667-x)
x = [OH-] = 4.30*10⁻⁶ M
p[OH-] = -log[OH-] = -log[4.30*10⁻⁶] = 5.37
pH = 14-p[OH] = 14-5.37 = 8.63
Ans (a)
pH = 8.63