. A 25.00 mL sample of 0.100 M acetic acid is titrated with 0.100 M NaOH. Consider the titration curve, and predict the main chemical species present after adding 50.00 mL NaOH titrant. Example choices More acetic acid than acetate ion; Equal amounts of acetic acid and acetate ion; More acetate ion than acetic acid; Mainly acetate ion; Mainly the excess titrant

More acetate ion than acetic acid.

Explanation:

Step 1: Data given

Volume of a 0.100 M acetic acid = 25.00 mL = 0.025 L

Volume of a 0.100 M NaOH = 50.00 mL = 0.050 L

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Moles CH3COOH = 0.100 M * 0.025 L

Moles CH3COOH = 0.0025 moles

Moles NaOH = 0.100 M * 0.050 L

Moles CH3COOH = 0.0050

Step 4: Calculate limiting reactant

CH3COOH is the limiting reactant. It will completely be consumed (0.0025 moles). NaOH is in excess. There will react 0.0025 moles . There will remain 0.0050 - 0.0025 = 0.0025 moles

Step 5: Calculate moles CH3COONa

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 1 mol H2O

For 0.0025 moles CH3COOH we'll have 0.0025 moles CH3COONa  (= CH3COO-)

The acetic acid is completely consumed, so we will have more acetate ion than acetic acid. But the amount of acetate ion is the same as amount of excess titrant.