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You are asked to prepare \(0.250 \mathrm{~L}\) of a solution that is \(0.500 M\) in nitrate ion. Your only source of nitrate ion is a bottle of \(1.00 M\) calcium nitrate. What volume (mL) of the calcium nitrate solution must you use?

Short Answer

Expert verified
You need 62.5 mL of the calcium nitrate solution.

Step by step solution

01

Understand the Problem

You need to determine the volume of a solution containing calcium nitrate that will provide a specific concentration of nitrate ions in a different solution. The target solution needs to have a volume of \(0.250\) L and a concentration of \(0.500\, M\) in nitrate ions. The available solution contains calcium nitrate with a concentration of \(1.00\, M\).
02

Determine Moles of Nitrate Required

Calculate the moles of nitrate ions needed using the target concentration and volume of the nitrate solution. Using the formula for molarity, \( \text{moles} = M \times V = 0.500 \times 0.250 = 0.125 \text{ moles} \).
03

Factor in the Stoichiometry

Calcium nitrate has the chemical formula \( \text{Ca(NO}_3)_2 \). Since each formula unit provides two nitrate ions, the moles of calcium nitrate needed is half the moles of nitrate ions. Thus, the moles of calcium nitrate needed is \( \frac{0.125}{2} = 0.0625 \text{ moles} \).
04

Calculate the Required Volume of Calcium Nitrate Solution

Use the moles of calcium nitrate needed to find the volume of the \(1.00\, M\) calcium nitrate solution using the formula \( \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} \). Thus, \( \text{Volume (L)} = \frac{0.0625}{1.00} = 0.0625 \text{ L} \).
05

Convert the Volume to Milliliters

Convert the volume from liters to milliliters. Since \(1\,L = 1000\,mL\), \(0.0625\,L = 62.5\,mL\).

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. It's like a recipe for a cake, where each ingredient (or chemical reactant) has a specific amount to contribute to make the perfect cake (or the desired product). In this exercise, stoichiometry helps us determine the volume of calcium nitrate solution needed to obtain the desired nitrate ion concentration. Since calcium nitrate dissociates to produce two nitrate ions for each formula unit, it's essential to adjust our calculations based on this stoichiometric relationship. To solve the problem, we first calculated the total moles of nitrate ions needed for our target solution, then used the stoichiometric factor of 1:2 (one mole of calcium nitrate gives two moles of nitrate ions) to find the moles of calcium nitrate required. Understanding stoichiometry ensures that our calculations accurately account for the formation of nitrate ions from calcium nitrate.
Concentration of Ions
When dealing with solutions, the concentration of ions is a crucial aspect to understand. It's the measure of how much of a given ion is present in a solution. Molarity, a common unit for expressing concentration, is defined as moles of solute per liter of solution. In our exercise, we needed a final solution that was 0.500 M in nitrate ions. This meant for every liter of solution, there should be 0.500 moles of nitrate ions. However, our calcium nitrate solution was 1.00 M, so we had to calculate the exact amount to mix. Calculating ion concentration helps us maintain the desired properties of the solution. In many applications, like in this exercise, getting the concentration right is imperative for the solution's intended purpose. By knowing the moles of nitrate required and the stoichiometry involved, we were able to accurately determine the volume of the calcium nitrate solution needed.
Chemical Formula Interpretation
Chemical formulas are like the blueprints of molecules, giving insight into the composition and the number of each type of atom in a molecule. Interpreting these correctly is key to working effectively in chemistry, particularly in stoichiometric calculations.For the calcium nitrate in our exercise, the formula is \( \text{Ca(NO}_3)_2 \). This tells us that each unit of calcium nitrate contains one calcium ion and two nitrate ions. This interpretation plays a crucial role in determining how many moles of calcium nitrate are required to achieve the desired moles of nitrate ions in a solution. When working with chemical formulas, remember:
  • Subscripts indicate the quantity of atoms or groups.
  • The formula reflects the smallest whole number ratio of elements in the compound.
  • Understanding formulas is essential for converting between chemical quantities and solving stoichiometry problems.
Properly interpreting chemical formulas is fundamental for accurate problem-solving in chemistry, as it allows you to precisely calculate the quantities needed for any chemical process.

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Most popular questions from this chapter

4.52 The following reactions occur in aqueous solution. Complete and balance the molecular equations using phase labels. Then write the net ionic equations. $$ \begin{array}{l} \mathrm{(a)} \mathrm{BaCO}_{3}+\mathrm{HNO}_{3} \longrightarrow \\ \mathrm{(b)} \mathrm{K}_{2} \mathrm{~S}+\mathrm{HCl} \longrightarrow \\ \mathrm{(c)} \mathrm{CaSO}_{3}(s)+\mathrm{HI} \longrightarrow \end{array} $$

A 1.345 -g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give \(2.012 \mathrm{~g}\) of barium chromate, \(\mathrm{BaCrO}_{4} .\) What is the formula of the compound?

Describe in words how you would do each of the following preparations. Then give the molecular equation for each preparation. (a) \(\mathrm{CuCl}_{2}(s)\) from \(\mathrm{CuSO}_{4}(s)\) (b) \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(s)\) from \(\mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{NaNO}_{3}(s)\) from \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) (d) \(\mathrm{MgCl}_{2}(s)\) from \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

A \(50.0-\mathrm{mL}\) volume of \(\mathrm{AgNO}_{3}\) solution contains \(0.0345 \mathrm{~mol} \mathrm{AgNO}_{3}\) (silver nitrate). What is the molarity of the solution?

A metal, \(\mathrm{M}\), was converted to the chloride \(\mathrm{MCl}_{2}\). Then a solution of the chloride was treated with silver nitrate to give silver chloride crystals, which were filtered from the solution. \(\mathrm{MCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow\) $$ \mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{AgCl}(s) $$ If \(2.434 \mathrm{~g}\) of the metal gave \(7.964 \mathrm{~g}\) of silver chloride, what is the atomic weight of the metal? What is the metal?

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