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

Answer:

.186 M

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

Answer 2

The concentration of acetic acid titrated with a 0.175 M NaOH solution. in the sample was 0.186M

HOW TO CALCULATE CONCENTRATION:

• The concentration of acid in a titration experiment can be calculated using the formula as follows:

C1V1 = C2V2

Where;

1. C1 = concentration of acid (M)
2. V1 = volume of acid (ml)
3. C2 = concentration of base (M)
4. V2 = volume of the base (mL)

• According to this question;

1. C1 = ?
2. V1 = 25.0ml
3. V2 = 26.6ml
4. C2 = 0.175M

C1 × 25 = 0.175 × 26.6

25C1 = 4.655

C1 = 4.655 ÷ 25

C1 = 0.186M

• Therefore, the concentration of acetic acid titrated with a 0.175 M NaOH solution. in the sample was 0.186M.

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