Question

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6 mL of the base is added. The concentration of acetic acid in the sample was ________ M.

Answer 1

The concentration of acetic acid in the sample is 0.186 M.

Step-by-step explanation:

The concentration of acetic acid in the sample can be determined using the concept of equivalence point in a titration. From the given information, we know that 26.6 mL of a 0.175 M NaOH solution is needed to reach the equivalence point when titrating a 25.0 mL sample of acetic acid solution.

The equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:

CH3COOH + NaOH → CH3COONa + H2O

Using the balanced equation, we can determine the moles of acetic acid in the sample and then calculate its concentration:

Moles of NaOH = concentration × volume (in liters)

Moles of acetic acid = moles of NaOH

Concentration of acetic acid = moles of acetic acid / volume of acetic acid (in liters)

Substituting the given values, we get:

Concentration of acetic acid = (0.175 M) × (0.0266 L) / (0.025 L) = 0.186 M

Answer 2

.186 M

Step-by-step explanation:

Remember, the equivalence point is reached when the number of moles of the two reactants is the same.

1. For this problem, obtain the number of moles of NaOH by multiplying the concentration and volume given.

2. Once obtain, the number of moles must be the same for the weak acid.

3. With step two in mind, simply solve for molarity by dividing the value obtained and 25 mL.

Calculation.

1. moles of NaOH = (.175)(.0266) = 4.66e-3

2. moles of Acetic acid = 4.66e-3

3. Concentration of acetic acid: (4.66e-3)/(.025) = .186 M