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Chapter 4: Problem 71

Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}_{3}\), (b) \(1.3 \times 10^{-2} \mathrm{M} \mathrm{MgSO}_{4}\), (c) \(0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (d) a mix- ture 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
In the given solutions, the concentration of each ion or molecule is as follows: (a) In 0.25 M NaNO鈧�: c(Na鈦�) = c(NO鈧冣伝) = 0.25 M (b) In \(1.3 \times 10^{-2} \mathrm{M}\) MgSO鈧�: c(Mg虏鈦�) = c(SO鈧劼测伝) = \(1.3 \times 10^{-2} \mathrm{M}\) (c) In 0.0150 M C鈧咹鈧佲倐O鈧�: c(C鈧咹鈧佲倐O鈧�) = 0.0150 M (d) In the mixed solution of 45.0 mL of 0.272 M NaCl and 65.0 mL of 0.0247 M (NH鈧�)鈧侰O鈧� (total volume = 110.0 mL): c(Na鈦�) = c(Cl鈦�) = \(\frac{45.0 \mathrm{~mL} \times 0.272\mathrm{M}}{110.0 \mathrm{~mL}}\) c(NH鈧勨伜) = \(2 \times \frac{65.0 \mathrm{~mL} \times 0.0247\mathrm{M}}{110.0 \mathrm{~mL}}\) c(CO鈧兟测伝) = \(\frac{65.0 \mathrm{~mL} \times 0.0247\mathrm{M}}{110.0 \mathrm{~mL}}\) Calculate the above expressions for the final concentrations in the mixed solution.

Step by step solution

01

(a) Determine the dissociation of NaNO3

NaNO3 is a strong electrolyte, which means it dissociates completely in water into its ions. In this case, NaNO3 dissociates into Na鈦� and NO鈧冣伝 ions. The balanced chemical equation for the dissociation of NaNO3 is: \[ \mathrm{NaNO}_3 \rightarrow \mathrm{Na^+} + \mathrm{NO_3^-} \]
02

(a) Get the concentration of Na鈦� and NO鈧冣伝 ions

The given concentration of NaNO3 is 0.25 M. Since one mole of NaNO3 dissociates into one mole of Na鈦� and one mole of NO鈧冣伝, the concentration of Na鈦� and NO鈧冣伝 ions will also be 0.25 M.
03

(b) Determine the dissociation of MgSO4

MgSO鈧� is an ionic compound, which means it dissociates completely in water into its ions. In this case, MgSO鈧� dissociates into Mg虏鈦� and SO鈧劼测伝 ions. The balanced chemical equation for the dissociation of MgSO鈧� is: \[ \mathrm{MgSO}_4 \rightarrow \mathrm{Mg^{2+}} + \mathrm{SO_4^{2-}} \]
04

(b) Get the concentration of Mg虏鈦� and SO鈧劼测伝 ions

The given concentration of MgSO鈧� is \(1.3 \times 10^{-2} \mathrm{M}\). Since one mole of MgSO鈧� dissociates into one mole of Mg虏鈦� and one mole of SO鈧劼测伝, the concentration of Mg虏鈦� and SO鈧劼测伝 ions will also be \(1.3 \times 10^{-2} \mathrm{M}\).
05

(c) Check if C6H12O6 dissociates

C鈧咹鈧佲倐O鈧� is a molecular compound and does not dissociate into ions in water. Therefore, its concentration remains at 0.0150 M in the solution.
06

(d) Calculate the final volume of the mixed solution

Given that the volume of 0.272 M NaCl is 45.0 mL and the volume of 0.0247 M (NH鈧�)鈧侰O鈧� is 65.0 mL, and the volumes are additive, the final volume of the mixed solution can be calculated by adding both volumes: \( V_{\mathrm{final}} = 45.0 \mathrm{~mL} + 65.0 \mathrm{~mL} = 110.0 \mathrm{~mL} \)
07

(d) Determine the dissociation of NaCl and (NH鈧�)鈧侰O鈧�

Both NaCl and (NH鈧�)鈧侰O鈧� dissociate completely in water: \[ \mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-} \] \[ \mathrm{(NH_4)_2CO_3} \rightarrow 2\mathrm{NH}_4^+ + \mathrm{CO_3^{2-}} \]
08

(d) Calculate the final concentration of each ion in the mixed solution

For Na鈦� ions (from NaCl): c(Na鈦�) = (Volume of NaCl solution 脳 Molarity of NaCl) / Final Volume = \((45.0 \mathrm{~mL} \times 0.272\mathrm{M}) / 110.0 \mathrm{~mL}\) For Cl鈦� ions (from NaCl): c(Cl鈦�) = (Volume of NaCl solution 脳 Molarity of NaCl) / Final Volume = \((45.0 \mathrm{~mL} \times 0.272\mathrm{M}) / 110.0 \mathrm{~mL}\) For NH鈧勨伜 ions (from (NH鈧�)鈧侰O鈧�): c(NH鈧勨伜) = 2 脳 (Volume of (NH鈧�)鈧侰O鈧� solution 脳 Molarity of (NH鈧�)鈧侰O鈧�) / Final Volume = \(2 \times (65.0 \mathrm{~mL} \times 0.0247\mathrm{M}) / 110.0 \mathrm{~mL}\) For CO鈧兟测伝 ions (from (NH鈧�)鈧侰O鈧�): c(CO鈧兟测伝) = (Volume of (NH鈧�)鈧侰O鈧� solution 脳 Molarity of (NH鈧�)鈧侰O鈧�) / Final Volume = \((65.0 \mathrm{~mL} \times 0.0247\mathrm{M}) / 110.0 \mathrm{~mL}\) Calculate the above expressions for the final concentrations.

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Key Concepts

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

Dissociation of Ionic Compounds
When ionic compounds dissolve in water, they dissociate into their component ions. This process involves breaking the ionic bonds within the compound.
For example, when sodium nitrate (\(\text{NaNO}_3\)), an ionic compound, is dissolved in water, it dissociates completely into sodium ions (\(\text{Na}^+\)) and nitrate ions (\(\text{NO}_3^-\)). The equation for this dissociation is: \[\mathrm{NaNO}_3 \rightarrow \mathrm{Na^+} + \mathrm{NO_3^-}\] Hence, each formula unit of sodium nitrate produces one sodium ion and one nitrate ion in solution.
Similar dissociation happens with magnesium sulfate (\(\text{MgSO}_4\)), producing magnesium ions (\(\text{Mg}^{2+}\)) and sulfate ions (\(\text{SO}_4^{2-}\)). This means for each mole of \(\text{MgSO}_4\), we get one mole of \(\text{Mg}^{2+}\) and one mole of \(\text{SO}_4^{2-}\). Understanding dissociation helps us calculate the concentration of each ion in solution.
Strong Electrolytes
Strong electrolytes are substances that completely dissociate into ions when dissolved in water.
They conduct electricity very well because of the free ions in the solution. Common examples include salts like sodium chloride (\(\text{NaCl}\)) and potassium nitrate (\(\text{KNO}_3\)).
The complete dissociation of these strong electrolytes means that the concentration of ions in the solution is equal to the concentration of the dissolved compound. For instance, if you dissolve 1 mole of \(\text{NaCl}\) in water, it will yield 1 mole of \(\text{Na}^+\) ions and 1 mole of \(\text{Cl}^-\) ions.
  • Each molecule of \(\text{NaCl}\) produces two types of ions.
  • This property is exploited in various applications, such as batteries and electrolysis processes.
Understanding strong electrolytes is fundamental in predicting the behavior of ionic solutions.
Calculating Ion Concentration
The concentration of ions in a solution can be calculated using the molarity of the dissolved ionic compound and its dissociation equation.
The molarity (M) indicates the number of moles of a solute per liter of solution. When an ionic compound dissociates in water, its molarity reflects the concentration of ions generated.For example:
If you have a 0.25 M solution of \(\text{NaNO}_3\), then the concentration of \(\text{Na}^+\) ions is also 0.25 M since each molecule dissociates into one ion each of \(\text{Na}^+\) and \(\text{NO}_3^-\). For more complex dissociation, like \(\text{(NH}_4)_2\text{CO}_3\), the ammonium nitrate contributes two \(\text{NH}_4^+\) ions per formula unit, so its molarity must be multiplied by two to get the concentration of \(\text{NH}_4^+\) ions.
  • Calculate individual ion concentrations using dissociation coefficients.
  • Adjust ion concentration calculations based on the volume and molarity of mixtures.
Knowing these calculations helps in understanding reactions and preparing solutions accurately.
Molecular Compounds
Molecular compounds, unlike ionic compounds, do not dissociate into ions in solution. They remain intact and dissolve as whole molecules.
This lack of ion formation means that molecular compounds do not conduct electricity when dissolved in water.
An example is glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) which dissolves in water but keeps its molecular form. Hence, in a 0.0150 M solution of glucose, the concentration of glucose remains unchanged at 0.0150 M, since it does not form ions.
  • Molecular compounds are often nonelectrolytes.
  • They interact with solvents through dipole interactions or hydrogen bonding.
Understanding the behavior of molecular compounds in solution enables better predictions about their solubility and reactivity.

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Most popular questions from this chapter

Formic acid, HCOOH, is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{HCOOH}\).

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), which reacts with sodium hydroxide in the following fashion: $$ \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow $$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of \(0.115 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00\) -qt sample of this vinegar?

A solution is made by mixing \(12.0 \mathrm{~g}\) of \(\mathrm{NaOH}\) and \(75.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

Can oxidation occur without accompanying reduction? Explain.

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or acid-base reactions. (a) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow\) $$ \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) $$ \mathrm{SrSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q) $$ (d) \(4 \mathrm{Zn}(s)+10 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow\) $$ 4 \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2} \mathrm{O}(g)+5 \mathrm{H}_{2} \mathrm{O}(l) $$
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