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Chapter 14: Problem 55

For an aqueous solution of acetic acid to be called "distilled white vinegar" it must contain \(5.0 \%\) acetic acid by mass. A solution with a density of \(1.05 \mathrm{~g} / \mathrm{mL}\) has a \(\mathrm{pH}\) of \(2.95 .\) Can the solution be called "distilled white vinegar"?

Short Answer

Expert verified
Answer: To determine if the solution can be considered "distilled white vinegar," calculate the mass percentage of acetic acid in the solution using the given density and pH. If the calculated mass percentage is equal to or greater than 5.0%, the solution is considered distilled white vinegar.

Step by step solution

01

Find the concentration of hydronium ions in the solution (H3O+)

To find the concentration of acetic acid, we first need to know the concentration of hydronium ions (H3O+) in the solution. This can be determined using the formula for pH: pH = -log10([H3O+]). Given: pH = 2.95 Let [H3O+] be the concentration of hydronium ions in the solution. We have the equation: 2.95 = -log10([H3O+]) Solving for [H3O+], we get: [H3O+] = 10^(-2.95)
02

Calculate the concentration of acetic acid using the ionization constant Ka

Acetic acid (HC2H3O2) ionizes in water to produce hydronium ions (H3O+) and acetate ions (C2H3O2-): HC2H3O2 (aq) + H2O (l) <-> H3O+ (aq) + C2H3O2- (aq) The ionization constant (Ka) for acetic acid can be obtained from a standard table of Ka values: Ka = 1.8 x 10^(-5) Using Ka and the concentration of H3O+ ions, we can determine the concentration of acetic acid ([HC2H3O2]): Let [HC2H3O2] be the concentration of acetic acid in the solution Let [C2H3O2-] be the concentration of acetate ions in the solution Since the concentrations of H3O+ and C2H3O2- are equal, [C2H3O2-] = [H3O+] Hence, Ka = ([H3O+] * [C2H3O2-]) / [HC2H3O2]
03

Solve for the concentration of acetic acid

Replacing the known values into the Ka equation: 1.8 x 10^(-5) = ( [H3O+] * [H3O+] ) / [HC2H3O2] [H3O+] = 10^(-2.95) HC2H3O2] = x 1.8 x 10^(-5) = (10^(-2.95) * 10^(-2.95)) / x Solve for x: x = (10^(-2.95) * 10^(-2.95)) / (1.8 x 10^(-5))
04

Determine the mass percentage of acetic acid in the solution

Now that we have the concentration of acetic acid (x mol/L), We want to find out if this solution meets the 5.0% by mass requirement for distilled white vinegar. First, find the mass of acetic acid in 1L of solution: Mass of acetic acid = concentration of acetic acid * molar mass of acetic acid Mass of acetic acid = x * 60.05 g/mol Then, find the mass of 1L of solution using the given density (1.05 g/mL): mass of 1L solution = density * volume mass of 1L solution = 1.05 g/mL * 1000 mL = 1050 g Finally, calculate the mass percentage of acetic acid as follows: % acetic acid = (mass of acetic acid / mass of 1L solution) * 100 % acetic acid = (x * 60.05 / 1050) * 100
05

Compare the calculated mass percentage with the requirement

If the calculated mass percentage of acetic acid in the solution is equal to or greater than 5.0%, then the solution can be considered distilled white vinegar. Compare the calculated value to the required value: % acetic acid ≥ 5.0% Based on the result, we can conclude whether the solution is distilled white vinegar or not.

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Most popular questions from this chapter

Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{H}^{+}\) ions. (a) sodium fluoride (b) barium hydroxide (c) potassium dihydrogen phosphate \(\left(\mathrm{KH}_{2} \mathrm{PO}_{4}\right)\)

A \(20.00\) -mL sample of \(0.220 \mathrm{M}\) triethylamine, (CH \(\left._{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\), is titrated with \(0.544 \mathrm{M} \mathrm{HCl}\). $$\left(\mathrm{Kb}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}=5.2 \times 10^{-4}\right)$$ (a) Write a balanced net ionic equation for the titration. (b) How many \(\mathrm{mL}\) of \(\mathrm{HCl}\) are required to reach the equivalence point? (c) Calculate \(\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right],\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{Cl}^{-}\right]\) at the equivalence point. (Assume that volumes are additive.) (d) What is the \(\mathrm{pH}\) at the equivalence point?

Explain why it is not possible to prepare a buffer with a pH of \(6.50\) by mixing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\).

A buffer is made up of \(355 \mathrm{~mL}\) each of \(0.200 \mathrm{M} \mathrm{NaHCO}_{3}\) and \(0.134 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Assuming that volumes are additive, calculate (a) the \(\mathrm{pH}\) of the buffer. (b) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0300 \mathrm{~mol}\) of \(\mathrm{HCl}\) to \(0.710 \mathrm{~L}\) of buffer. (c) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0300 \mathrm{~mol}\) of \(\mathrm{KOH}\) to \(0.710 \mathrm{~L}\) of buffer.

At \(25^{\circ} \mathrm{C}\) and \(1.00\) atm pressure, one liter of ammonia is bubbled into \(725 \mathrm{~mL}\) of water. Assume that all the ammonia dissolves and the volume of the solution is the volume of the water. A \(50.0-\mathrm{mL}\) portion of the prepared solution is titrated with \(0.2193 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.
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