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Chapter 5: Problem 18

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

Short Answer

Expert verified
Hence, the concentration of nitrate ions \([NO_{3}^{-}]\) in the solution will be \(0.255M\)

Step by step solution

01

Calculate the number of moles of nitrate from potassium nitrate

Number of moles can be calculated using the formula: \(Moles = Molarity \times Volume\). \nSo, for the first solution \(Moles = 0.283 mol/L \times 275 mL × (1 L/1000 mL) = 0.077825 mol\)
02

Calculate the number of moles of nitrate from magnesium nitrate

The same formula will be used for finding the moles in the second solution with small modification, since in each formula unit of magnesium nitrate, there are two nitrate ions - Mg(NO3)2 . Thus: \nMoles = 2 × Molarity × Volume. So, Moles = 2 × 0.421 mol/L × 328 mL × (1 L/1000 mL) = 0.275776 mol.
03

Find the total moles of nitrate

As we have the total moles from step 1 and step 2, we just need to add those: Total nitrate = 0.077825 Mol (from KNO3) + 0.275776 Mol (from Mg(NO3)2) = 0.353601 mol.
04

Calculate the total volume

The total volume of the resultant solution will be the sum of the volumes of its constituents, that is: Total volume = \(275 mL + 328 mL + 784 mL = 1387 mL = 1.387 L\).
05

Find the concentration of nitrate ions

The concentration or molarity can be found out by dividing the total number of moles of nitrate ions by the total volume of the solution. Thus: \([NO_{3}^{-}] = \frac{Total \, moles}{total \, volume} = \frac{0.353601 mol}{1.387 L} = 0.255 M\)

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry. It refers to the concentration of a solute in a solution. Defined as the number of moles of solute divided by liters of solution, it is expressed as moles per liter (mol/L).
To calculate molarity, you first need to know the quantity of solute (in moles) and the total volume of the solution (in liters). For example, if you have a solution containing 0.283 moles of potassium nitrate in 1 liter, its molarity is 0.283 M.
Understanding molarity helps predict how substances will interact in a solution and is key for preparing solutions with precise concentrations. This measurement is crucial for chemical reactions, where accurate solute concentrations are needed to ensure the desired reaction results.
Nitrate Ion Concentration
The concentration of nitrate ions in a solution can be determined by considering the contributions from different sources in the solution. In chemical solutions that contain nitrate like potassium nitrate ( KNO_3 ) and magnesium nitrate ( Mg(NO_3)_2 ), each compound releases nitrate ions once dissolved.

For KNO_3 , each molecule provides one nitrate ion, while Mg(NO_3)_2 yields two nitrate ions per formula unit.

  • Potassium Nitrate ( KNO_3 ): If you mix 275 mL of 0.283 M KNO_3 , it results in 0.077825 moles of nitrate.
  • Magnesium Nitrate ( Mg(NO_3)_2 ): Mixing 328 mL of 0.421 M Mg(NO_3)_2 gives 0.275776 moles of nitrate.

Understanding how to find these contributions is essential for accurately calculating the total nitrate ion concentration in any chemical solution.
Solution Volume Calculation
The total volume of a solution is important for calculating concentrations in chemistry. When preparing or combining solutions, volumes are often assumed to be additive. This means that the total volume is the sum of the individual volumes of the solutions mixed.
In this specific scenario, if you mix 275 mL of KNO_3 solution, 328 mL of Mg(NO_3)_2 solution, and 784 mL of water, the total volume of the final solution is 1387 mL or 1.387 liters.
  • This total volume is used to calculate the molarity of the resultant solution by dividing the total moles of the solute by the total volume in liters.
  • This approach provides a straightforward method to measure concentrations in mixed solutions, essential for consistent chemical reactions.

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Most popular questions from this chapter

Balance these equations for disproportionation reactions. (a) \(\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}\) (basic solution) (b) \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}\) (acidic solution)

Explain why these reactions cannot occur as written. (a) \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Balance these equations for redox reactions occurring in acidic solution. (a) \(\mathrm{P}_{4}(\mathrm{s})+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NO}(\mathrm{g})\) (b) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}+\mathrm{Mn}^{2+}\) (c) \(\mathrm{HS}^{-}+\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (d) \(\mathrm{Fe}^{3+}+\mathrm{NH}_{3} \mathrm{OH}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\)

What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)
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