A solution is made by mixing exactly 500 mL of 0.191 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 * 10-5 [H+] * 10 M Enter your answer in scientific notation. [OH-] M [CH3COOH] * 10 M Enter your answer in scientific notation. [Na+] M [CH3COO-] M

[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