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Chapter 4: Problem 94

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base were required for the titration?

Short Answer

Expert verified
The concentration of acetic acid in the vinegar is 0.115 M.

Step by step solution

01

Calculate moles of Sodium Hydroxide

First, calculate the number of moles of the base, sodium hydroxide. We know from the question that the molarity (M) of NaOH is 1.00M, and the volume used in the titration is 5.75 mL (or 0.00575 L, because 1L = 1000 ml). The formula to calculate moles is: Moles = Volume(L) x Molarity(M). So, by substituting the values, we get the moles of NaOH = 1M x 0.00575 L = 0.00575 moles.
02

Find moles of Acetic Acid

Next, look at the stoichiometry of the reaction. In the balanced equation given, we see that one mole of NaOH reacts with one mole of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)). Hence, at the end point of the titration, the moles of NaOH that reacted should be equal to the moles of acetic acid in the vinegar solution. Therefore, we have 0.00575 moles of acetic acid present in our vinegar solution.
03

Determine concentration of Acetic Acid

Finally, calculate the molarity of the acetic acid. Remember that Molarity(M) = Moles/volume(L). From the question, we know that volume of the vinegar solution is 50.0 mL or 0.050 L. And from step 2,the moles of acetic acid found was 0.00575 moles. So substituting in the formula, we get Molarity of acetic acid: M = 0.00575 moles / 0.050 L = 0.115 M.

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Most popular questions from this chapter

List the general properties of acids and bases.

Describe hydration. What properties of water enable its molecules to interact with ions in solution?

A 22.02 -mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64 -mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH} .\) Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Classify these reactions according to the types discussed in the chapter: (a) \(\mathrm{Cl}_{2}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO}_{3}\) (c) \(\mathrm{NH}_{3}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{4}^{+}\) (d) \(2 \mathrm{CCl}_{4}+\mathrm{CrO}_{4}^{2-} \longrightarrow\) \(2 \mathrm{COCl}_{2}+\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{Cl}^{-}\) (e) \(\mathrm{Ca}+\mathrm{F}_{2} \longrightarrow \mathrm{CaF}_{2}\) (f) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (g) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{BaSO}_{4}\) (h) \(\mathrm{CuO}+\mathrm{H}_{2} \longrightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (i) \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (j) \(2 \mathrm{FeCl}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}\)

Give a chemical explanation for each of these: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. Explain. (b) In the activity series aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. Why? (c) Sodium and potassium lie above copper in the activity series. Explain why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series?
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