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Chapter 16: Problem 27

A 0.1401-g sample of a purified carbonate was dissolved in \(50.00 \mathrm{~mL}\) of \(0.1140 \mathrm{M} \mathrm{HCl}\) and boiled to eliminate \(\mathrm{CO}_{2}\). Back-titration of the excess \(\mathrm{HCl}\) required \(24.21 \mathrm{~mL}\) of \(0.09802 \mathrm{M} \mathrm{NaOH}\). Identify the carbonate.

Short Answer

Expert verified
The carbonate is sodium carbonate (\(\text{Na}_2\text{CO}_3\)).

Step by step solution

01

Calculate Initial Moles of HCl

First, determine how many moles of hydrochloric acid were initially present. Use the formula: \[ \text{moles of } \text{HCl} = \text{volume (L)} \times \text{molarity} \]Given, the volume is \(50.00 \ \text{mL} = 0.05000 \ \text{L}\) and the molarity is \(0.1140\ \text{M}\).\[\text{moles of HCl} = 0.05000 \ \text{L} \times 0.1140\ \text{mol/L} = 0.0057 \ \text{mol} \]
02

Calculate Moles of NaOH Used for Back-titration

Next, find out how many moles of NaOH were used to titrate the excess HCl. Use the formula:\[ \text{moles of } \text{NaOH} = \text{volume (L)} \times \text{molarity} \]Given, the volume of NaOH is \(24.21 \ \text{mL} = 0.02421 \ \text{L}\) and the molarity is \(0.09802\ \text{M}\).\[\text{moles of NaOH} = 0.02421 \ \text{L} \times 0.09802\ \text{mol/L} = 0.00237274 \ \text{mol} \]
03

Calculate Moles of Unreacted HCl

For the excess hydrochloric acid, which reacts with the NaOH, the moles of HCl will be equivalent to the moles of NaOH because they react in a 1:1 ratio.Thus, \[\text{moles of unreacted } \text{HCl} = 0.00237274 \ \text{mol} \]
04

Determine Moles of HCl Reacted with Carbonate

Subtract the moles of unreacted HCl from the initial moles of HCl to find the moles that reacted with the carbonate.\[\text{moles of } \text{HCl (reacted)} = 0.0057 \ \text{mol} - 0.00237274 \ \text{mol} = 0.00332726 \ \text{mol} \]
05

Relate Moles of HCl to Moles of Carbonate

Carbonates \((\text{CO}_{3}^{2-})\) may react with hydrochloric acid according to:\[\text{CO}_{3}^{2-} + 2\text{HCl} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O} + 2\text{Cl}^{-}\]From the stoichiometry of the reaction, 1 mole of carbonate reacts with 2 moles of HCl. Therefore, solve for moles of carbonate:\[\text{moles of carbonate} = \frac{0.00332726 \ \text{mol}}{2} = 0.00166363 \ \text{mol} \]
06

Determine Molar Mass of Carbonate and Identify

Use the molar mass of the carbonate to identify it. Given that \(\text{mass} = 0.1401\ \text{g}\), solve for molar mass:\[\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.1401 \ \text{g}}{0.00166363 \ \text{mol}} = 84.2\ \text{g/mol} \]The molar mass corresponds to sodium carbonate \(\text{Na}_2\text{CO}_3\), which has a molar mass of approximately 84.0 g/mol.

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Key Concepts

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

Molarity Calculations
Molarity is a way of expressing concentration, specifically the number of moles of a solute in a liter of solution. To calculate molarity, use the formula:
  • \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in Liters}} \]
Molarity helps us determine how many moles of a substance, such as hydrochloric acid (HCl), are present in a given volume.
In the exercise, a solution with a molarity of 0.1140 M indicates there are 0.1140 moles of HCl per liter. By multiplying this molarity by the volume of the solution (in liters), we can find the moles of HCl.
These calculations form the basis for many titration problems, allowing us to subsequently determine the amounts of reactants involved in a chemical reaction.
Acid-Base Reaction
Acid-base reactions are chemical reactions involving the transfer of protons (H+) between reactants. Such reactions are central to titration studies, where an acid reacts with a base.
In this exercise, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH).
The essential reaction in the back-titration is:
  • \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
The reaction follows a 1:1 molar ratio, which means one mole of HCl reacts with one mole of NaOH.
By measuring how much NaOH is required to neutralize the remaining HCl, we can understand how much HCl did not initially react with the carbonate, allowing us to deduce the amount that did to identify the carbonate present.
Stoichiometry
Stoichiometry enables the calculation of reactants and products in a chemical reaction, based on balanced chemical equations. For this carbonate reaction:
  • \[ \text{CO}_3^{2-} + 2 \text{HCl} \rightarrow \text{CO}_2 + \text{H}_2\text{O} + 2 \text{Cl}^- \]
The stoichiometric coefficients (the numbers in front of molecules) indicate that 2 moles of HCl react with 1 mole of carbonate.
This is crucial for evaluating the reacting amounts. By finding out how many moles of HCl reacted, and using the stoichiometry, the moles of carbonate are calculated.
Understanding these relationships is fundamental for correctly solving titration problems, where the quantity of an unknown reactant is determined through reaction with a known quantity of another substance.
Molar Mass Determination
Molar mass is the weight of one mole of a substance, expressed in grams. Identifying substances by their molar mass is common in chemical titrations.
Given the moles of a reactant (like the carbonate in the exercise) and the sample's mass, we determine its molar mass using:
  • \[ \text{Molar mass} = \frac{\text{mass of substance}}{\text{moles of substance}} \]
In the exercise, the calculated molar mass is 84.2 g/mol, matching sodium carbonate (Na\(_2\)CO\(_3\)) with an approximate molar mass of 84.0 g/mol.
Recognizing a compound based on its molar mass from experimental data is an integral part of identifying compositions in a titration. Understanding how molar mass relates to the components and how to accurately compute it is key to mastering chemistry practicals.

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Most popular questions from this chapter

A dilute solution of an unknown weak acid required a 28.62-mL titration with \(0.1084 \mathrm{M} \mathrm{NaOH}\) to reach a phenolphthalein end point. The titrated solution was evaporated to dryness. Calculate the equivalent mass of the acid if the sodium salt was found to weigh \(0.2110 \mathrm{~g}\).

What types of organic nitrogen-containing compounds tend to yield low results with the Kjeldahl method unless special precautions are taken?

The digestion of a 0.1417-g sample of a phosphoru: containing compound in a mixture of \(\mathrm{HNO}_{3}\) an \(\mathrm{H}_{2} \mathrm{SO}_{4}\) resulted in the formation of \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}\), an \(\mathrm{H}_{3} \mathrm{PO}_{4}\). Addition of ammonium molybdate yielded solid having the composition \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoC}\) (1876.3 \(\mathrm{g} / \mathrm{mol}\) ). This precipitate was filtered, washed and dissolved in \(50.00 \mathrm{~mL}\) of \(0.2000 \mathrm{M} \mathrm{NaOH}\) : $$ \begin{gathered} \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}(s)+26 \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} \\ +12 \mathrm{MoO}_{4}^{2-}+14 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{NH}_{3}(g) \end{gathered} $$ After the solution was boiled to remove the \(\mathrm{NH}_{3}\), the excess \(\mathrm{NaOH}\) was titrated with \(14.17 \mathrm{~mL}\) of \(0.1741 \mathrm{M} \mathrm{HCl}\) to a phenolphthalein end point. Calculate the percentage of phosphorus in the sample.

Air was bubbled at a rate of \(30.0 \mathrm{~L} / \mathrm{min}\) through a trap containing \(75 \mathrm{~mL}\) of \(1 \% \mathrm{H}_{2} \mathrm{O}_{2}\left(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{SO}_{2} \rightarrow\right.\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ). After \(10.0 \mathrm{~min}\), the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) was titrated with \(11.70 \mathrm{~mL}\) of \(0.00197 \mathrm{M} \mathrm{NaOH}\). Calculate concentration of \(\mathrm{SO}_{2}\) in parts per million (that is, \(\mathrm{mL} \mathrm{SO}_{2} / 10^{6}\) \(\mathrm{mL}\) air) if the density of \(\mathrm{SO}_{2}\) is \(0.00285 \mathrm{~g} / \mathrm{mL}\).

A 1.219-g sample containing \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}, \mathrm{NH}_{4} \mathrm{NO}_{3}\), and nonreactive substances was diluted to \(200 \mathrm{~mL}\) in a volumetric flask. A \(50.00\)-mL aliquot was made basic with strong alkali, and the liberated \(\mathrm{NH}_{3}\) was distilled into \(30.00 \mathrm{~mL}\) of \(0.08421 \mathrm{M} \mathrm{HCl}\). The excess \(\mathrm{HCl}\) required \(10.17 \mathrm{~mL}\) of \(0.08802 \mathrm{M} \mathrm{NaOH}\) for neutralization. A \(25.00-\mathrm{mL}\) aliquot of the sample was made alkaline after the addition of Devarda's alloy, and the \(\mathrm{NO}_{3}{ }^{-}\)was reduced to \(\mathrm{NH}_{3}\). The \(\mathrm{NH}_{3}\) from both \(\mathrm{NH}_{4}{ }^{+}\)and \(\mathrm{NO}_{3}{ }^{-}\)was then distilled into \(30.00 \mathrm{~mL}\) of the standard acid and back-titrated with \(14.16 \mathrm{~mL}\) of the base. Calculate the percentage of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in the sample.
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