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Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}_{3}\), (b) \(1.3 \times 10^{-2} \mathrm{M} \mathrm{MgSO}_{4},(\mathrm{c}) 0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (d) a mixture of \(45.0 \mathrm{~mL}\) of \(0.272 \mathrm{M} \mathrm{NaCl}\) and \(65.0 \mathrm{~mL}\) of \(0.0247 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\). Assume that the volumes are additive.

Short Answer

Expert verified
The concentrations of ions and molecules in the solutions are: (a) 0.250 M NaNO鈧 Na鈦: 0.25 M NO鈧冣伝: 0.25 M (b) 1.3 x 10^{-2} M MgSO鈧 Mg虏鈦: 1.3 x 10^{-2} M SO鈧劼测伝: 1.3 x 10^{-2} M (c) 0.0150 M C鈧咹鈧佲倐O鈧 C鈧咹鈧佲倐O鈧: 0.0150 M (d) A mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH鈧)鈧侰O鈧 Na鈦: 0.1113 M Cl鈦: 0.1113 M NH鈧勨伜: 0.02912 M CO鈧兟测伝: 0.01456 M

Step by step solution

01

Identify ions in the solution

In NaNO鈧, we have Na鈦 ions and NO鈧冣伝 ions.
02

Calculate the concentration of ions

Since it is a 1:1 ratio, the concentration of Na鈦 and NO鈧冣伝 ions would also be 0.25 M. Na鈦: 0.25 M NO鈧冣伝: 0.25 M #b) Finding the concentration of ions in 1.3 x 10^(-2) M MgSO鈧 solution. #
03

Identify ions in the solution

In MgSO鈧, we have Mg虏鈦 ions and SO鈧劼测伝 ions.
04

Calculate the concentration of ions

Since it is a 1:1 ratio, the concentration of Mg虏鈦 and SO鈧劼测伝 ions would also be 1.3 x 10^(-2) M. Mg虏鈦: 1.3 x 10^(-2) M SO鈧劼测伝: 1.3 x 10^(-2) M #c) Finding the concentration of molecules in 0.0150 M C鈧咹鈧佲倐O鈧 solution. #
05

Identify molecules in the solution

In C鈧咹鈧佲倐O鈧, there are no ions, it's a covalent molecular compound (glucose).
06

Calculate the concentration of molecules

Glucose does not dissociate into ions, so it remains as a molecule at a concentration of 0.0150 M. C鈧咹鈧佲倐O鈧: 0.0150 M #d) Finding the concentration of ions in the mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH鈧)鈧侰O鈧 solution. #
07

Identify ions in the solutions

In NaCl, we have Na鈦 ions and Cl鈦 ions. In (NH鈧)鈧侰O鈧, we have 2 NH鈧勨伜 ions and 1 CO鈧兟测伝 ion.
08

Calculate the moles of ions

For NaCl: moles of Na鈦 and Cl鈦 ions = 0.272 M 脳 0.045 L = 0.01224 mol For (NH鈧)鈧侰O鈧: moles of NH鈧勨伜 ions = 2 脳 (0.0247 M 脳 0.065 L) = 0.003203 mol moles of CO鈧兟测伝 ions = 0.0247 M 脳 0.065 L = 0.001602 mol
09

Calculate the total volume of the mixture

Total volume = 45.0 mL + 65.0 mL = 110.0 mL
10

Calculate the concentration of ions in the mixture

Concentration of Na鈦 ions = 0.01224 mol / 0.110 L = 0.1113 M Concentration of Cl鈦 ions = 0.01224 mol / 0.110 L = 0.1113 M Concentration of NH鈧勨伜 ions = 0.003203 mol / 0.110 L = 0.02912 M Concentration of CO鈧兟测伝 ions = 0.001602 mol / 0.110 L = 0.01456 M The concentrations of ions and molecules in the solutions are: (a) 0.250 M NaNO鈧 Na鈦: 0.25 M NO鈧冣伝: 0.25 M (b) 1.3 x 10^(-2) M MgSO鈧 Mg虏鈦: 1.3 x 10^(-2) M SO鈧劼测伝: 1.3 x 10^(-2) M (c) 0.0150 M C鈧咹鈧佲倐O鈧 C鈧咹鈧佲倐O鈧: 0.0150 M (d) A mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH鈧)鈧侰O鈧 Na鈦: 0.1113 M Cl鈦: 0.1113 M NH鈧勨伜: 0.02912 M CO鈧兟测伝: 0.01456 M

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Key Concepts

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

Molar Concentration
Understanding molar concentration is crucial when studying chemistry. It indicates how much of a substance is present in a given volume of solution. It鈥檚 often expressed in moles per liter (M), which can be seen in the exercise with various substances like NaNO3 and MgSO4. To calculate the molar concentration, you divide the number of moles of solute by the volume of the solution in liters. For instance, if you have 0.25 moles of NaNO3 in 1 liter of solution, the molar concentration is 0.25 M. This calculation is a backbone of many chemical calculations and provides a standard way to discuss the concentration of solutions.

When tackling exercises involving molar concentration, remember that accuracy in measuring volume and quantity of the solute is key. Also, consider the solute's molar mass to convert between grams and moles if needed.
Ion Dissociation
Ion dissociation is a process where ionic compounds separate into individual ions when dissolved in water. This phenomenon is observed in salts like NaCl, which, when in water, dissociate into Na鈦 and Cl鈦 ions. The ability to predict the degree of dissociation is essential, as it allows students to accurately determine the concentration of individual ions in a solution. For example, NaNO3 dissociates entirely into Na鈦 and NO3鈦 ions, each with a concentration of 0.25 M in a 0.25 M NaNO3 solution.

It's important to note that the dissociation process follows the stoichiometry of the compound. This means for every one formula unit of NaNO3 that dissociates, you will have one Na鈦 ion and one NO3鈦 ion in solution. This 1:1 relationship makes calculating the resulting ion concentrations straightforward.
Molecular Compounds
Molecular compounds, such as C鈧咹鈧佲倐O鈧 in our exercise, consist of molecules held together by covalent bonds, rather than ions. Unlike ionic compounds, they do not dissociate into ions in solution. Hence, when such compounds dissolve in water, the molar concentration of the solution refers to the intact molecules. For instance, a 0.0150 M solution of C鈧咹鈧佲倐O鈧 means that there are 0.0150 moles of glucose molecules in every liter of solution. These compounds are typically non-electrolytes and do not conduct electricity in solution because they lack free-moving ions.
Concentration of Mixtures
Mixtures are a combination of two or more substances. When dealing with the concentration of mixtures, as in the fourth part of the exercise, we need to calculate the molar concentration of each component in the final combined solution. This involves first determining the moles of each solute from their individual solutions and then dividing by the total volume of the mixed solution.

Example of Calculating the Concentration in Mixtures

In the mixture of NaCl and (NH鈧)鈧侰O鈧, the moles of each ion are calculated before they were mixed. Then, by adding the volumes of both solutions and assuming they are additive, the total volume is attained. The concentration of each ion is found by dividing its moles by the new total volume. Remember, the molar concentration changes when substances are mixed due to volume changes, even though the moles of the substance remain constant.

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Most popular questions from this chapter

Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), is a nonelectrolyte; hypochlorous acid, \(\mathrm{HClO}\), is a weak electrolyte; and ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a strong electrolyte. (a) What are the solute particles present in aqueous solutions of each compound? (b) If \(0.1 \mathrm{~mol}\) of each compound is dissolved in solution, which one contains \(0.2 \mathrm{~mol}\) of solute particles, which contains \(0.1 \mathrm{~mol}\) of solute particles, and which contains somewhere between \(0.1\) and \(0.2 \mathrm{~mol}\) of solute particles?

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution, (c) \(0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

(a) How would you prepare \(175.0 \mathrm{~mL}\) of \(0.150 \mathrm{MAgNO}_{3}\) solution starting with pure solute? (b) An experiment calls for you to use \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution. All you have available is a bottle of \(3.6 \mathrm{M} \mathrm{HNO}_{3}\). How would you prepare the desired solution?

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C}\), and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

Write a balanced molecular equation and a net ionic equation for the reaction that occurs when (a) solid \(\mathrm{CaCO}_{3}\) reacts with an aqueous solution of nitric acid; (b) solid iron(II) sulfide reacts with an aqueous solution of hydrobromic acid.

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