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Chapter 4: Problem 84

The distinctive odor of vinegar is due to aceticacid, $\mathrm{CH}_{3} \mathrm{COOH}$, which reacts with sodium hydroxide according to: \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q) $$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of $0.115 \mathrm{M} \mathrm{NaOH}$ to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00-\) qt sample of this vinegar?

Short Answer

Expert verified
The mass of acetic acid in a 1.00-quart sample of this vinegar is 80.67 grams.

Step by step solution

01

Find the moles of NaOH at the equivalence point

We are given that the volume of NaOH needed to reach the equivalence point is 42.5 mL, and the concentration of NaOH solution is 0.115 M. To find the moles of NaOH, we can use the formula: moles of NaOH = volume (in liters) × concentration First, we need to convert the volume of NaOH from mL to L: 42.5 mL = 42.5 / 1000 L = 0.0425 L Now we can find the moles of NaOH: moles of NaOH = 0.0425 L × 0.115 mol/L = 0.0048875 mol This is the moles of NaOH at the equivalence point.
02

Use stoichiometry to find the moles of acetic acid

From the balanced chemical equation: CH3COOH(aq) + NaOH(aq) → H2O(l) + NaCH3COO(aq) At the equivalence point, the moles of acetic acid are equal to the moles of NaOH. Therefore, moles of acetic acid = 0.0048875 mol
03

Calculate the concentration of acetic acid in the vinegar sample

We are given that the volume of vinegar sample used in the titration is 3.45 mL. First, we need to convert the volume of vinegar from mL to L: 3.45 mL = 3.45 / 1000 L = 0.00345 L Now, we can calculate the concentration of acetic acid in the vinegar sample using the formula: concentration of acetic acid = moles of acetic acid / volume (in liters) concentration of acetic acid = 0.0048875 mol / 0.00345 L = 1.4167 M
04

Convert the volume of vinegar from milliliters to quarts

We need to find the amount of acetic acid in a 1.00-quart sample of this vinegar. First, we need to convert the volume of vinegar from quarts to liters: 1.00 qt = 1.00 × 0.946353 L/qt = 0.946353 L
05

Calculate the mass of acetic acid in the 1.00-quart sample

To find the mass of acetic acid in the 1.00-quart sample, we can use the formula: mass of acetic acid = volume (in liters) × concentration × molar mass The molar mass of acetic acid is 60.05 g/mol. Therefore, mass of acetic acid = 0.946353 L × 1.4167 mol/L × 60.05 g/mol = 80.67 g So, there are 80.67 grams of acetic acid in a 1.00-quart sample of this vinegar.

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Most popular questions from this chapter

A \(1.248-g\) sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

A solution of \(105.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) is mixed with a solution of \(150.0 \mathrm{~mL}\) of \(0.060 \mathrm{MAlCl}_{3} .\) (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the aqueous \(\mathrm{HCl}\) in the stomach and each of the following substances used in various antacids: (a) $\mathrm{Al}(\mathrm{OH})_{3}(s),(\mathbf{b}) \mathrm{Mg}(\mathrm{OH})_{2}(s),(\mathbf{c}) \mathrm{MgCO}_{3}(s),$ (d) \(\mathrm{NaAl}\left(\mathrm{CO}_{3}\right)(\mathrm{OH})_{2}(s)\) (e) \(\mathrm{CaCO}_{3}(s)\)

Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(s) ;\) solutions of metals like this that are solids are called alloys. There is a range of compositions over which the solution is considered a bronze. Bronzes are stronger and harder than either copper or tin alone. (a) A \(100.0-\mathrm{g}\) sample of a certain bronze is \(90.0 \%\) copper by mass and \(10.0 \%\) tin. Which metal can be called the solvent, and which the solute? (b) Based on part (a), calculate the concentration of the solute metal in the alloy in units of molarity, assuming a density of $7.9 \mathrm{~g} / \mathrm{cm}^{3} .$ (c) Suggest a reaction that you could do to remove all the tin from this bronze to leave a pure copper sample. Justify your reasoning.

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\), and \(\mathrm{Fe}^{2+}\), which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+} .\) Keep in mind that charge balance must be maintained. (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of Nat are needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?
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