. 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.

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answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.