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A \(7.55 \mathrm{g}\) sample of \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is added to \(125 \mathrm{mL}\) of a vinegar that is \(0.762 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} .\) Will the resulting solution still be acidic? Explain.

Short Answer

Expert verified
No, the resulting solution will not be acidic. All of the acetic acid in the vinegar (\(\mathrm{CH}_{3} \mathrm{COOH}\)) will react with sodium carbonate (\(\mathrm{Na}_{2} \mathrm{CO}_{3}\)), thus neutralizing the solution.

Step by step solution

01

Convert mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to moles

Since the molar mass of sodium carbonate (\(\mathrm{Na}_{2} \mathrm{CO}_{3}\)) is approximately \(106 \mathrm{g/mol}\), use this conversion factor to convert the mass of the sodium carbonate to moles: \(\frac {7.55 g} {106 g/mol} = 0.0712 mol\).
02

Determine the initial moles of \(\mathrm{CH}_{3} \mathrm{COOH}\)

You have \(0.762 M\) \(\mathrm{CH}_{3} \mathrm{COOH}\) in \(125 mL\) of solution. To find the initial moles, multiply the molarity by the volume in liters: \((0.762 M) * (0.125 L) = 0.0953 mol\).
03

Determine the moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) after reaction

When \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) reacts with \(\mathrm{CH}_{3} \mathrm{COOH}\), two moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) are needed for each mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Therefore, the moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) that react equals to \(2 * 0.0712 mol = 0.1424 mol\). After the reaction, the remaining moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) is found by subtracting the initial moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) by the moles that reacted: \(0.0953 mol - 0.1424 mol = -0.0471 mol\).
04

Determine if the solution remains acidic

The amount of \(\mathrm{CH}_{3} \mathrm{COOH}\) after the reaction is negative, which implies that there is no \(\mathrm{CH}_{3} \mathrm{COOH}\) remaining and that all of it has reacted. Therefore, the solution will no longer be acidic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Carbonate
Sodium carbonate is a common chemical compound often found in household cleaning products. It is known by its chemical formula \( \mathrm{Na}_2\mathrm{CO}_3 \) and sometimes referred to as soda ash or washing soda.
Sodium carbonate is an alkaline substance, meaning it has a basic nature. It is commonly used to neutralize acids in various chemical reactions.
When sodium carbonate is added to an acidic solution, such as vinegar which contains acetic acid, a reaction occurs, producing carbon dioxide gas, water, and a salt.
  • Sodium carbonate in reactions: acts as a base to neutralize acids.
  • It releases carbon dioxide gas upon reacting with acids.
  • Used in various industries for its cleaning and neutralizing properties.
Acetic Acid
Acetic acid is the key component of vinegar, giving it a distinctive sour taste and smell. Chemically, it is represented as \( \mathrm{CH}_3\mathrm{COOH} \). This compound is a weak organic acid, which means it does not completely dissociate into ions in water.
Acetic acid is widely used in cooking and cleaning, but it also plays an important role in various chemical reactions involving bases like sodium carbonate.
  • Weak acid: only partially ionizes in solution.
  • Reacts with bases to form salts and water, a typical acid-base neutralization reaction.
  • Used as a preservative, solvent, and in the synthesis of chemical compounds.
This characteristic makes it valuable in industries from food to plastics, and as seen in reactions like with sodium carbonate, it plays critical roles in educational chemistry settings.
Molarity Calculation
Molarity is a measurement of concentration and is defined as moles of solute per liter of solution. It indicates how much of a substance is present in a given volume of liquid.
The calculation of molarity is crucial in preparing solutions and reacting them in the correct stoichiometric proportions.
You calculate molarity using the formula:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
  • This allows chemists to ensure they are using the right amounts for reactions.
  • Eases the process of scaling reactions up or down depending on the desired output.
  • Ensures that the reactants are in the necessary proportions to react completely without excess.
In the provided exercise, knowing the molarity of acetic acid helped determine the amount of substance present before the reaction commenced.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It allows us to predict how much of each substance is needed or produced in a chemical reaction.
This concept is rooted in the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
In an acid-base reaction, stoichiometry helps determine how much of each reactant will interact. When sodium carbonate reacts with acetic acid, stoichiometry tells us exactly how many moles of acetic acid are required to completely react with the given amount of sodium carbonate.
  • Uses balanced chemical equations to relate moles of reactants to moles of products.
  • Ensures no excess reactants are wasted and that products are formed efficiently.
  • Critical in predicting reaction outcomes and for industrial applications where yield and cost are significant.
This concept was pivotal in calculating the reaction's outcome, determining that all the acetic acid would be consumed, leaving a non-acidic solution.

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Most popular questions from this chapter

Blood alcohol content (BAC) is often reported in weight-volume percent (w/v\%). For example, a BAC of \(0.10 \%\) corresponds to \(0.10 \mathrm{g} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) per 100 mL of blood. Estimates of BAC can be obtained from breath samples by using a number of commercially available instruments, including the Breathalyzer for which a patent was issued to R. F. Borkenstein in 1958\. The chemistry behind the Breathalyzer is described by the oxidation- reduction reaction below, which occurs in acidic solution: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{g})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad(\text { not balanced })\)A Breathalyzer instrument contains two ampules, each of which contains \(0.75 \mathrm{mg} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) dissolved in \(3 \mathrm{mL}\) of \(9 \mathrm{mol} / \mathrm{L} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) .\) One of the ampules is used as reference. When a person exhales into the tube of the Breathalyzer, the breath is directed into one of the ampules, and ethyl alcohol in the breath converts \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) into \(\mathrm{Cr}^{3+} .\) The instrument compares the colors of the solutions in the two ampules to determine the breath alcohol content (BrAC), and then converts this into an estimate of BAC. The conversion of BrAC into BAC rests on the assumption that 2100 mL of air exhaled from the lungs contains the same amount of alcohol as \(1 \mathrm{mL}\) of blood. With the theory and assumptions described in this problem, calculate the molarity of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in the ampules before and after a breath test in which a person with a BAC of \(0.05 \%\) exhales 0.500 Lof his breath into a Breathalyzer instrument.

Assuming the volumes are additive, what is the \(\left[\mathrm{Cl}^{-}\right]\) in a solution obtained by mixing \(225 \mathrm{mL}\) of \(0.625 \mathrm{M}\) \(\mathrm{KCl}\) and \(615 \mathrm{mL}\) of \(0.385 \mathrm{M} \mathrm{MgCl}_{2} ?\)

Warfarin, \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4},\) is the active ingredient used in some anticoagulant medications. The amount of warfarin in a particular sample was determined as follows. A 13.96 g sample was first treated with an alkaline I_ solution to convert \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) to \(\mathrm{CHI}_{3}\). This treatment gives one mole of \(\mathrm{CHI}_{3}\) for every mole of \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) that was initially present in the sample. The iodine in \(\mathrm{CHI}_{3}\) is then precipitated as \(\mathrm{AgI}(\mathrm{s})\) by treatment with excess \(\mathrm{AgNO}_{3}(\mathrm{aq}):\) $$\begin{aligned} \mathrm{CHI}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow & \longrightarrow 3 \mathrm{AgI}(\mathrm{s})+3 \mathrm{HNO}_{3}(\mathrm{aq}) &+\mathrm{CO}(\mathrm{g}) \end{aligned}$$ If \(0.1386 \mathrm{g}\) solid \(\mathrm{AgI}\) were obtained, then what is the percentage by mass of warfarin in the sample analyzed?

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

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