Answer:
[H⁺] = 0 M
[CH₃COOH] = 0 M
[OH⁻] = 4.55x10⁻²M
[Na⁺] = 9.55x10⁻²M
[CH₃COO⁻] = 5.00x10⁻²M
Explanation:
CH₃COOH is a weak acid, and when in aqueous solution, it is in equilibrium with its conjugate base CH₃COO⁻:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
When NaOH is at solution, it dissolves as the ions Na⁺ and OH⁻. The number of moles of each compound is:
nCH₃COOH = 0.5 L * 0.100 mol/L = 0.05 mol
nNaOH = 0.5 L * 0.191 mol/L = 0.0955 mol
Thus, all the CH₃COOH reacts, forming NaCH₃COO, and H₂O. The stoichiometry is 1:1, and the remaining NaOH will be:
nNaOH = 0.0955 - 0.05 = 0.0455 mol
nNa⁺ = 0.0455 mol
nOH⁻ = 0.0455 mol
And nNaCH₃COO = 0.05 mol, nNa⁺ = 0.05 mol, and nCH₃COO⁻ = 0.05 mol. Because all the H⁺ forms water, there'll be no H⁺ left to form CH₃COOH, thus, the final volume of the solution is 1 L, and the concentration is the number of moles divided by the volume:
[H⁺] = 0 M
[CH₃COOH] = 0 M
[OH⁻] = 0.0455/1 = 4.55x10⁻²M
[Na⁺] = (0.0455 + 0.05)/1 = 9.55x10⁻²M
[CH₃COO⁻] = 0.05/1 = 5.00x10⁻²M
