***the measured pH of vinegar is 2.5*** Part B: Determination of Ka of Acetic Ac

Question

***the measured pH of vinegar is 2.5*** Part B: Determination of Ka of Acetic Acid 3. The concentration of distilled white vinegar is written as 5% (w/v) of acetic acid (CH3COOH). Convert the concentration units from % (w/v) into molarity. Express your answer in four significant figures. In other words, assume that the concentration is 5.000% (w/v). (Hint: See the Prelab Question 3.) 4. Calculate the concentration of hydronium ions [H3O+] of vinegar from the measured pH values. Express your answer in two significant figures.

Explanation / Answer

3. Concentration in %(w/v)={Mass of solute(g)/Volume of solution (mL)}x100

5.000%(w/v) of acetic acid means 5.000g acetic acid is present in 100.0 mL solution

So mass of solute=5.000 g

Volume of solution=100.0 mL=100.0 mL/1000.0 mL/L=0.1 L

Concentration in Molarity={Mass of solute/Molar mass of solute}/Volume (L) ........equation (1)

Molar mass of acetic acid=2xMolar mass of C+4xmolar mass of H+2xmolar mass of O

=2x12 g/mol + 4x1 g/mol + 2x16 g/mol

=24 g/mol +4 g/mol +32 g/mol = 60 g/mol

So using equation (1)

Concentration of given acetic acid solution in Molarity

=(5.000 g/60.000 g/mol)/0.1 L

=0.8333 mol/L=0.8333 M

4. Given measured pH of vinegar=2.5

pH=-log[H3O+]=2.5

So [H3O+]=10-2.5=0.0032

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