Calculate the ph at the equivalence point when 10.0 ml of 0.10 m hc9h7o2 (ka = 3.6 × 10–5) is titrated against 0.20 m sodium hydroxide 8.63 5.37 2.72 4.44

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

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.