Answer to Question #349587 in General Chemistry for Angelo Abboud

Question

25.0 mL of 0.100 M HC2H3O2 is titrated by a 0.100 M NaOH. A) Write the net ionic equation. B) calculate the pH of the solution after adding 10 mL of NaOH

Answer 1

Solution:

(A):

Molecular equation:

HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(l)

Total ionic equation:

H⁺(aq) + C₂H₃O₂⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + C₂H₃O₂⁻(aq) + H₂O(l)

Net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)

(B):

Moles of HC₂H₃O₂ = Molarity of HC₂H₃O₂ × Volume of HC₂H₃O₂

Moles of HC₂H₃O₂ = (0.100 M) × (25.0 mL) × (1 L / 1000 mL) = 0.0025 mol

Moles of HC₂H₃O₂ = 0.0025 mol

Moles of NaOH = Molarity of NaOH × Volume of NaOH

Moles of NaOH = (0.100 M) × (10 mL) × (1 L / 1000 mL) = 0.001 mol

Moles of NaOH = 0.001 mol

Balanced chemical equation:

HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(l)

ICE table for the reaction:

According to the ICE table:

n(HC₂H₃O₂) = 0.0015 mol

n(NaC₂H₃O₂) = 0.0010 mol

Total volume of solution = Volume of HC₂H₃O₂ + Volume of NaOH

Total volume of solution = (25.0 mL + 10.0 mL) × (1 L / 1000 mL) = 0.035 L

Molarity of solute = Moles of solute / Liters of solution

Therefore,

[HC₂H₃O₂] = (0.0015 mol) / (0.035 L) = 0.04286 M

[NaC₂H₃O₂] = (0.0010 mol) / (0.035 L) = 0.02857 M

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution from the concentration of the buffer components:

where pK_a is the acid dissociation constant and [base] and [acid] are the base and acid concentrations, respectively, in the chemical equation.

In this specific reaction, the base is NaC₂H₃O₂ and the acid is HC₂H₃O₂

K_a for acetic acid (HC₂H₃O₂) is 1.8×10⁻⁵

pK_a = −log(K_a) = −log(1.8×10⁻⁵) = 4.745

Therefore,

pH = 4.745 + log(0.02857 M / 0.04286 M) = 4.569 = 4.57

pH = 4.57

Answer:

A) Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)

B) pH = 4.57