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n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a phenolphthalein end point. Calculate the molarity of the acetic acid solution, and the percent (by weight) of acetic acid in the solution (assuming its density to be 1.00 g/mL).

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Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

The molarity of acetic acid is 0.11M and the percent by weight is 0.666%.

How we calculate molarity?

Molarity of any solution is used to define their concentration and it will be calculated as:

M = n/V, where

n = moles

V = volume

Molarity of acetic acid will be calculated as:

M₁V₁ = M₂V₂, where

M₁ = molarity of acetic acid = ?

V₁ = volume of acetic acid = 37.54mL = 0.037L

M₂ = molarity of NaOH = 0.1062M

V₂ = volume of NaOH = 39.26mL = 0.039L

On putting all these values on the above equation we can calculate the molarity as:

M₁ = (0.1062)(39.26) / (37.54) = 0.11M

Now we calculate the moles of acetic acid by using the molarity formula as:

n = 0.11M × 0.037L = 0.00407 moles

Molar mass of acetic acid = 60g/mole

Mass of 0.00407 moles of acetic acid = 4.1x10⁻³ moles×(60g / mol) = 0.2501 g

Density of solution = 1.00 g/mL

So, 37.54mL in 1g/mL = 37.54g/mL

Percent by weight will be calculated as:

%w/w = 0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

Hence, molarity and %(w/w) of acetic acid is 0.11M and 0.666% respectively.

To know more about percent weight, visit the below link:

https://brainly.com/question/5493941

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