/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 The concentration of \(\mathrm{C... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 20

The concentration of \(\mathrm{CO}_{2}\) in air is determined by an indirect acid-base titration. A sample of air is bubbled through a solution that contains an excess of \(\mathrm{Ba}(\mathrm{OH})_{2},\) precipitating \(\mathrm{BaCO}_{3} .\) The excess \(\mathrm{Ba}(\mathrm{OH})_{2}\) is back titrated with HCl. In a typical analysis a 3.5-L sample of air is bubbled through \(50.00 \mathrm{~mL}\) of \(0.0200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2} .\) Back titrating with \(0.0316 \mathrm{M} \mathrm{HCl}\) requires \(38.58 \mathrm{~mL}\) to reach the end point. Determine the \(\mathrm{ppm} \mathrm{CO}_{2}\) in the sample of air given that the density of \(\mathrm{CO}_{2}\) at the temperature of the sample is \(1.98 \mathrm{~g} / \mathrm{L}\).

Short Answer

Expert verified
The CO_2 concentration in the air sample is approximately 4.91 ppm.

Step by step solution

01

Calculate moles of Ba(OH)_2 used

Start by calculating the total moles of \( \mathrm{Ba(OH)}_2 \) initially present. Use the volume and molarity of the \( \mathrm{Ba(OH)}_2 \) solution.\[\text{Moles of } \mathrm{Ba(OH)}_2 = 50.00 \text{ mL} \times \frac{0.0200 \text{ mol}}{1000 \text{ mL}}\approx 0.00100 \text{ mol}\]
02

Calculate moles of HCl used in titration

Convert the volume of \( \mathrm{HCl} \) used in titration from mL to L, then use the molarity of \( \mathrm{HCl} \) to find the moles.\[\text{Moles of } \mathrm{HCl} = 38.58 \text{ mL} \times \frac{0.0316 \text{ mol}}{1000 \text{ mL}} \approx 0.00122 \text{ mol}\]
03

Determine moles of excess Ba(OH)_2

Since \( \mathrm{HCl} \) neutralizes only the excess \( \mathrm{Ba(OH)}_2 \), the moles of excess \( \mathrm{Ba(OH)}_2 \) can be found. Each mole of \( \mathrm{HCl} \) neutralizes one mole of \( \mathrm{OH}^- \), and since \( \mathrm{Ba(OH)}_2 \) provides 2 \( \mathrm{OH}^- \) ions, divide the moles of \( \mathrm{HCl} \) by 2.\[\text{Excess moles of } \mathrm{Ba(OH)}_2 = \frac{0.00122}{2} \approx 0.00061 \text{ mol}\]
04

Calculate moles of CO_2 absorbed

Subtract excess \( \mathrm{Ba(OH)}_2 \) from total initial moles to find moles of \( \mathrm{Ba(OH)}_2 \) that reacted with \( \mathrm{CO}_2 \). These are the same as the moles of \( \mathrm{CO}_2 \) absorbed, since one mole of \( \mathrm{Ba(OH)}_2 \) reacts with one mole of \( \mathrm{CO}_2 \).\[\text{Moles of } \mathrm{CO}_2 \approx 0.00100 - 0.00061 = 0.00039 \text{ mol}\]
05

Convert moles of CO_2 to grams

Use the molar mass of \( \mathrm{CO}_2 \) (44.01 g/mol) to convert moles to grams.\[\text{Mass of } \mathrm{CO}_2 = 0.00039 \text{ mol} \times 44.01 \text{ g/mol} \approx 0.0172 \text{ g}\]
06

Calculate ppm of CO_2 in air

Since ppm is mg of \( \mathrm{CO}_2 \) per liter of air, first convert mass of \( \mathrm{CO}_2 \) to milligrams, then divide by the volume of the air sample (3.5 L).\[\text{ppm } \mathrm{CO}_2 = \frac{0.0172 \text{ g} \times 1000}{3.5 \text{ L}} \approx 4.91 \mathrm{ppm}\]

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Key Concepts

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

Carbon Dioxide Concentration
Carbon dioxide concentration in the air can be measured through a clever use of acid-base titration. This method involves bubbling air into a solution that can react with carbon dioxide. In this exercise, the air sample is passed through a solution of barium hydroxide, which is known for its ability to precipitate carbon dioxide in the form of barium carbonate. This precipitate is an indication that carbon dioxide has been absorbed.
afterwards, to ensure all carbon dioxide has reacted, the remaining solution is back titrated with hydrochloric acid. The amount of hydrochloric acid used in the titration helps to determine the remaining excess barium hydroxide, and thus calculate the initial concentration of carbon dioxide.
By knowing how much air has been bubbled through the solution and the amount of carbon dioxide that was absorbed, one can understand the concentration of carbon dioxide in the air. It is fascinating how chemical reactions can help us understand what is seemingly invisible in our atmosphere.
Precipitation Reaction
A precipitation reaction occurs when two solutions are mixed and an insoluble solid forms. This solid, known as a precipitate, separates from the liquid solution. In the context of this exercise, when carbon dioxide gas is passed through the barium hydroxide solution, barium carbonate solid is formed. This reaction proceeds as:
\[\text{CO}_2(g) + \text{Ba(OH)}_2(aq) \rightarrow \text{BaCO}_3(s) + \text{H}_2\text{O}(l)\]
The barium carbonate is the precipitate. Its formation implies that carbon dioxide was absorbed from the air into the solution. Understanding precipitation reactions is crucial in chemistry as they help remove certain impurities or indicate the presence of specific elements or compounds. In this exercise, it tells us that carbon dioxide is present and has interacted with barium hydroxide to form an insoluble compound.
Molarity Calculation
Molarity is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Calculating molarity involves knowing both the number of moles of a substance and the total volume of the solution. In our exercise, we begin by calculating the moles of barium hydroxide and hydrochloric acid used in the titration process.
Knowing the molarity of the hydrochloric acid allows us to determine how much of it was needed to neutralize the excess barium hydroxide. From these computations, we infer the amount of carbon dioxide initially present in the solution.
This process highlights why molarity is an important concept. It gives a clear view of the chemical dynamics in the solution. Once we have determined the molarity, we can further calculate the concentration of carbon dioxide in parts per million (ppm), useful for expressing very dilute concentrations typical of gaseous components in the atmosphere.

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Most popular questions from this chapter

A 0.1036 -g sample that contains only \(\mathrm{BaCl}_{2}\) and \(\mathrm{NaCl}\) is dissolved in \(50 \mathrm{~mL}\) of distilled water. Titrating with \(0.07916 \mathrm{M} \mathrm{AgNO}_{3}\) requires \(19.46 \mathrm{~mL}\) to reach the Fajans end point. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{BaCl}_{2}\) in the sample.

The concentration of \(\mathrm{CO}\) in air is determined by passing a known volume of air through a tube that contains \(\mathrm{I}_{2} \mathrm{O}_{5}\), forming \(\mathrm{CO}_{2}\) and \(\mathrm{I}_{2}\). The \(\mathrm{I}_{2}\) is removed from the tube by distilling it into a solution that contains an excess of \(\mathrm{KI}\), producing \(\mathrm{I}_{3}^{-}\). The \(\mathrm{I}_{3}^{-}\) is titrated with a standard solution of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). In a typical analysis a 4.79 - \(\mathrm{L}\) sample of air is sampled as described here, requiring \(7.17 \mathrm{~mL}\) of \(0.00329 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to reach the end point. If the air has a density of \(1.23 \times 10^{-3} \mathrm{~g} / \mathrm{mL},\) determine the parts per million CO in the air.

Prada and colleagues described an indirect method for determining sulfate in natural samples, such as seawater and industrial effluents. \({ }^{12}\) The method consists of three steps: precipitating the sulfate as \(\mathrm{PbSO}_{4}\); dissolving the \(\mathrm{PbSO}_{4}\) in an ammonical solution of excess EDTA to form the soluble \(\mathrm{Pb} \mathrm{Y}^{2-}\) complex; and titrating the excess EDTA with a standard solution of \(\mathrm{Mg}^{2+}\). The following reactions and equilibrium constants are known $$ \begin{array}{ll} \mathrm{PbSO}_{4}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) & K_{\mathrm{sp}}=1.6 \times 10^{-8} \\ \mathrm{~Pb}^{2+}(a q)+\mathrm{Y}^{4-}(a q) \rightleftharpoons \mathrm{Pb} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=1.1 \times 10^{18} \\ \mathrm{Mg}^{2+}(a q)+\mathrm{Y}^{4-}(a q) \rightleftharpoons \mathrm{Mg} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=4.9 \times 10^{8} \\ \mathrm{Zn}^{2+}(a q)+\mathrm{Y}^{4-}(a q)=\mathrm{Zn} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=3.2 \times 10^{16} \end{array} $$ (a) Verify that a precipitate of \(\mathrm{PbSO}_{4}\) will dissolve in a solution of \(\mathrm{Y}^{4-}\). (b) Sporek proposed a similar method using \(\mathrm{Zn}^{2+}\) as a titrant and found that the accuracy frequently was poor. \(^{13}\) One explanation is that \(\mathrm{Zn}^{2+}\) might react with the \(\mathrm{PbY}^{2-}\) complex, forming \(\mathrm{ZnY}^{2-}\). Show that this might be a problem when using \(\mathrm{Zn}^{2+}\) as a titrant, but that it is not a problem when using \(\mathrm{Mg}^{2+}\) as a titrant. Would such a displacement of \(\mathrm{Pb}^{2+}\) by \(\mathrm{Zn}^{2+}\) lead to the reporting of too much or too little sulfate? (c) In a typical analysis, a 25.00 -mL sample of an industrial effluent is carried through the procedure using \(50.00 \mathrm{~mL}\) of \(0.05000 \mathrm{M}\) EDTA. Titrating the excess EDTA requires \(12.42 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) \(\mathrm{Mg}^{2+}\). Report the molar concentration of \(\mathrm{SO}_{4}^{2-}\) in the sample of effluent.

Before the introduction of EDTA most complexation titrations used \(\mathrm{Ag}^{+}\) or \(\mathrm{CN}^{-}\) as the titrant. The analysis for \(\mathrm{Cd}^{2+},\) for example, was accomplished indirectly by adding an excess of \(\mathrm{KCN}\) to form \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\), and back titrating the excess \(\mathrm{CN}^{-}\) with \(\mathrm{Ag}^{+},\) forming \(\mathrm{Ag}(\mathrm{CN})_{2}^{-} .\) In one such analysis a \(0.3000-\mathrm{g}\) sample of an ore is dissolved and treated with \(20.00 \mathrm{~mL}\) of \(0.5000 \mathrm{M} \mathrm{KCN}\). The excess \(\mathrm{CN}^{-}\) requires \(13.98 \mathrm{~mL}\) of \(0.1518 \mathrm{M} \mathrm{AgNO}_{3}\) to reach the end point. Determine the \(\% \mathrm{w} / \mathrm{w}\) Cd in the ore.

The EDTA titration of mixtures of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) can be followed thermometrically because the formation of the \(\mathrm{Ca}^{2+}-\mathrm{EDTA}\) complex is exothermic and the formation of the \(\mathrm{Mg}^{2+}\) -EDTA complex is endothermic. Sketch the thermometric titration curve for a mixture of \(5.00 \times 10^{-3} \mathrm{MCa}^{2+}\) and \(5.00 \times 10^{-3} \mathrm{M} \mathrm{Mg}^{2+}\) using \(0.0100 \mathrm{MEDTA}\) as the titrant. The heats of formation for \(\mathrm{CaY}^{2-}\) and \(\mathrm{MgY}^{2-}\) are, respectively, \(-23.9 \mathrm{~kJ} / \mathrm{mole}\) and \(23.0 \mathrm{~kJ} / \mathrm{mole}\).
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