/*! 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 52 The concentration of \(\mathrm{C... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 52

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} M .\) The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

Short Answer

Expert verified
To reduce the level of Ca虏鈦� to 20% of its original level in the given water supply, \(1.69 \times 10^{7} \mathrm{g}\) of Ca(OH)鈧� and \(4.5 \times 10^{6} \mathrm{g}\) of Na鈧侰O鈧� must be added.

Step by step solution

01

Analyze the target concentration for Ca虏鈦�

We are told to reduce the Ca虏鈦� concentration to 20% of its original level. The original concentration is \(5.7 \times 10^{-3} M\). So, the target concentration of Ca虏鈦� will be: \[C_{Ca^{2+}} =0.20 \times 5.7 \times 10^{-3} \mathrm{M} = 1.14 \times 10^{-3} \mathrm{M}\]
02

Find the amount of Ca虏鈦� that needs to be removed

To find how much Ca虏鈦� concentration needs to be removed, subtract the target concentration from the initial concentration: \[\Delta C_{Ca^{2+}} = 5.7 \times 10^{-3} \mathrm{M} - 1.14 \times 10^{-3} \mathrm{M} = 4.56 \times 10^{-3} \mathrm{M}\]
03

Analyze the amounts of each substance in moles

Calculate the moles of calcium ions (Ca虏鈦�), bicarbonate ions (HCO鈧冣伝), calcium hydroxide (Ca(OH)鈧�), and sodium carbonate (Na鈧侰O鈧�) needed for this change: \[\text{moles of } Ca^{2+} = \Delta C_{Ca^{2+}} \times V = 4.56 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 2.28 \times 10^{5} \mathrm{mol}\] \[\text{moles of } HCO_3^{-} = C_{HCO_3^-} \times V = 1.7 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 8.5 \times 10^{4} \mathrm{mol}\] Since, one mole of Ca(OH)鈧� reacts with one mole of Ca虏鈦�, we have: \[\text{moles of } Ca(OH)_2 = \text{moles of } Ca^{2+} = 2.28 \times 10^{5} \mathrm{mol}\] As we know, one mole of Na鈧侰O鈧� reacts with two moles of HCO鈧冣伝, so we get: \[\text{moles of } Na_2CO_3 = \frac{1}{2} \times \text{moles of } HCO_3^{-} = \frac{1}{2} \times 8.5 \times 10^{4} \mathrm{mol} = 4.25 \times 10^{4} \mathrm{mol}\]
04

Calculate the masses of Ca(OH)鈧� and Na鈧侰O鈧�

With the moles of each substance determined, calculate the masses of Ca(OH)鈧� and Na鈧侰O鈧� needed: \[\text{mass of } Ca(OH)_2 = \text{moles of } Ca(OH)_2 \times M_{Ca(OH)_2} = 2.28 \times 10^{5} \mathrm{mol} \times 74.09 \frac{\mathrm{g}}{\mathrm{mol}} = 1.69 \times 10^{7} \mathrm{g}\] \[\text{mass of } Na_2CO_3 = \text{moles of } Na_2CO_3 \times M_{Na_2CO_3} = 4.25 \times 10^{4} \mathrm{mol} \times 105.99 \frac{\mathrm{g}}{\mathrm{mol}} = 4.5 \times 10^{6} \mathrm{g}\] Thus, we need to add \(1.69 \times 10^{7} \mathrm{g}\) of Ca(OH)鈧� and \(4.5 \times 10^{6} \mathrm{g}\) of Na鈧侰O鈧� to the water to achieve the desired reduction in Ca虏鈦� concentration.

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Key Concepts

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

Calculating Molar Concentration
Understanding molar concentration is fundamental in chemistry and pivotal for tasks such as reducing water hardness. Molar concentration, often represented by the symbol 'M', indicates the moles of a substance per liter of solution. It's a way of expressing how much of a solute is present in a given volume of solvent.

For example, when given a concentration of \(5.7 \times 10^{-3} M\) for calcium ions (\(\mathrm{Ca}^{2+}\)), we can interpret this to mean we have 0.0057 moles of \(\mathrm{Ca}^{2+}\) for every liter of water. When reducing water hardness, understanding this concept helps us calculate how much of a reacting agent we need to introduce to achieve the desired concentration level of ions in the water.

To calculate the target concentration, we simply multiply the initial molar concentration by the percentage of the reduction desired鈥攊n this case, 20%, leading to the target concentration formula: \[ C_{Ca^{2+}} = 0.20 \times \text{Initial Concentration} \]. This step is crucial for determining the amount of reactive agent needed for the treatment process.
Stoichiometry
Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of the mole. It allows us to predict how much reactants are required to produce a certain amount of product, or conversely, how much product can be formed from a certain amount of reactants.

In water hardness treatment, for instance, stoichiometry helps us understand that a precise ratio of calcium hydroxide (\(\mathrm{Ca(OH)_2}\)) is needed to react with the present calcium ions to reduce hardness. Similarly, sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3}\)) is used to react with bicarbonate ions. Since one mole of \(\text{Ca(OH)_2}\) reacts with one mole of \(\text{Ca}^{2+}\), and one mole of \(\text{Na}_2\mathrm{CO}_3\) reacts with two moles of \(\text{HCO}_3^-\), stoichiometry allows us to calculate the exact amounts needed for these reactions to happen efficiently without excess reactant.
Water Treatment Chemistry
Water treatment chemistry involves various processes and reactions to remove impurities and make water suitable for a specific purpose, like consumption, industrial use, or releasing back into the environment. Hard water, which contains high levels of minerals such as calcium and magnesium, can cause problems like scaling in pipes and reduced effectiveness of soap.

To soften hard water, chemical processes such as the addition of calcium hydroxide or sodium carbonate are used, which react with the calcium ions and bicarbonate ions to form precipitates that can be removed from the water. The exercise described utilizes these principles to calculate the precise amounts of chemicals needed to achieve a desired level of water hardness reduction.

Water treatment professionals rely on these calculations and the stoichiometry of the reactions involved to ensure they add the correct amounts of chemicals. This maintains the efficiency and safety of the water treatment process, ensuring that the treated water meets regulatory standards and is fit for its intended use.

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Most popular questions from this chapter

An impurity in water has an extinction coefficient of \(3.45 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(280 \mathrm{nm}\), its absorption maximum Closer Look, p. 564). Below 50 ppb, the impurity is not a problem for human health. Given that most spectrometers cannot detect absorbances less than 0.0001 with good reliability, is measuring the absorbance of water at \(280 \mathrm{nm}\) a good way to detect concentrations of the impurity above the 50 -ppb threshold?

A reaction for converting ketones to lactones, called the Baeyer-Villiger reaction, is used in the manufacture of plastics and pharmaceuticals. 3 Chloroperbenzoic acid is shock-sensitive, however, and prone to explode. Also, 3 -chlorobenzoic acid is a waste product. An alternative process being developed uses hydrogen peroxide and a catalyst consisting of tin deposited within a solid support. The catalyst is readily recovered from the reaction mixture. (a) What would you expect to be the other product of oxidation of the ketone to lactone by hydrogen peroxide? (b) What principles of green chemistry are addressed by use of the proposed process?

The degradation of \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of \(k=1.6 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(4{ }^{\circ} \mathrm{C}\). If the tropo- spheric concentrations of \(\mathrm{OH}\) and \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) are \(8.1 \times 10^{5}\) and \(6.3 \times 10^{8}\) molecules/cm \(^{3}\), respectively, what is the rate of reaction at this temperature in \(M / s ?\)

(a) With respect to absorption of radiant energy, what distinguishes a greenhouse gas from a nongreenhouse gas? (b) \(\mathrm{CH}_{4}\) is a greenhouse gas, but \(\mathrm{N}_{2}\) is not. How might the molecular structure of \(\mathrm{CH}_{4}\) explain why it is a greenhouse gas?

A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, (b) greenhouse gas, \((\mathrm{c})\) photochemical smog, \((\mathbf{d})\) ozone depletion. Give a brief explanation of each term and identify one or two of the chemicals associated with each.
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