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Chapter 5: Problem 58

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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Most popular questions from this chapter

Briefly describe (a) half-equation method of balancing redox equations; (b) disproportionation reaction; (c) titration; (d) standardization of a solution.

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

A \(\mathrm{KMnO}_{4}(\) aq) solution is to be standardized by titration against \(\mathrm{As}_{2} \mathrm{O}_{3}(\mathrm{s}) .\) A \(0.1078 \mathrm{g}\) sample of \(\mathrm{As}_{2} \mathrm{O}_{3}\) requires \(22.15 \mathrm{mL}\) of the \(\mathrm{KMnO}_{4}(\) aq) for its titration. What is the molarity of the \(\mathrm{KMnO}_{4}(\) aq)? \(5 \mathrm{As}_{2} \mathrm{O}_{3}+4 \mathrm{MnO}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O}+12 \mathrm{H}^{+} \longrightarrow\) \(10 \mathrm{H}_{3} \mathrm{AsO}_{4}+4 \mathrm{Mn}^{2+}\)

Predict in each case whether a reaction is likely to occur. If so, write a net ionic equation. (a) $\mathrm{HI}(\mathrm{aq})+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \longrightarrow$ (b) $\operatorname{CuSO}_{4}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow$ (c) $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow$

Explain the important distinctions between (a) a strong electrolyte and strong acid; (b) an oxidizing agent and reducing agent; (c) precipitation reactions and neutralization reactions; (d) half-reaction and overall reaction.
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