Question

Calculate the number of moles of each ion present in \(2.00 \times 10^{2} \mathrm{~cm}^{3}\) of (a) \(0.200 \mathrm{M} \mathrm{NaCl}\), (b) \(0.350 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4}\) (c) \(1.44 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Short answer

In summary, the number of moles of each ion present in the given solutions are: (a) Na+: 0.0400 moles, Cl-: 0.0400 moles (b) K+: 0.210 moles, PO43-: 0.0700 moles (c) Al3+: 0.288 moles, NO3-: 0.864 moles

Step-by-step solution

Find the moles for each solution

Firstly, we have to find the total moles of each compound using the formula: Moles = Volume × Molarity Note that the given volume is 2.00 × 10^2 cm³, which we need to convert to liters because Molarity is in moles per liter. \(1 L = 1000 cm^{3}\), so \((2.00 \times 10^{2} cm^{3}) = 0.200 L\). Now let's find the moles for each compound: (a) Moles of NaCl = Molarity × Volume = 0.200 M × 0.200 L = 0.0400 moles (b) Moles of K3PO4 = Molarity × Volume = 0.350 M × 0.200 L = 0.0700 moles (c) Moles of Al(NO3)3 = Molarity × Volume = 1.44 M × 0.200 L = 0.288 moles

Calculate moles for the ions of each compound

Now we will break each compound into its constituent ions to find the number of moles for each ion: (a) 1 mole of NaCl gives 1 mole of Na+ and 1 mole of Cl-. So, Moles of Na+: 0.0400 moles Moles of Cl-: 0.0400 moles (b) 1 mole of K3PO4 gives 3 moles of K+ and 1 mole of PO43-. So, Moles of K+: 3 × 0.0700 moles = 0.210 moles Moles of PO43-: 0.0700 moles (c) 1 mole of Al(NO3)3 gives 1 mole of Al3+ and 3 moles of NO3-. So, Moles of Al3+: 0.288 moles Moles of NO3-: 3 × 0.288 moles = 0.864 moles In summary, the number of moles of each ion present in the given solutions are: (a) Na+: 0.0400 moles, Cl-: 0.0400 moles (b) K+: 0.210 moles, PO43-: 0.0700 moles (c) Al3+: 0.288 moles, NO3-: 0.864 moles

Key concepts

Molarity

Molarity is a common way to express the concentration of a solution. It tells us how many moles of a substance are present in one liter of solution. The formula to calculate molarity is simple and vital for various calculations. It is defined as:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]When you know the molarity and volume of a solution, you can find the number of moles using the equation:

• Moles = Molarity × Volume

For example, a 0.200 M NaCl solution tells us that there are 0.200 moles of NaCl in every liter.It's important to ensure the volume is in liters when you plug values into these equations. This makes sure your calculations are correct, as molarity is defined in terms of liters.

Ionic Compounds

Ionic compounds are substances made up of positively and negatively charged ions. They are held together by ionic bonds, which occur due to the attraction between oppositely charged ions. These compounds often dissolve in water, disassociating into their respective ions.In a solution like NaCl, it separates into sodium ions (\( \text{Na}^+ \)) and chloride ions (\( \text{Cl}^- \)) when dissolved. Similarly, \( \text{K}_3\text{PO}_4 \) separates into potassium ions (\( \text{K}^+ \)) and phosphate ions (\( \text{PO}_4^{3-} \)).

The ability to separate into ions is crucial in many chemical reactions and calculations. Understanding how many ions result from dissolving a compound helps in precisely calculating the stoichiometry of solutions.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. It's all about the relations between the different participants in a reaction. This involves the use of balanced chemical equations, which shows the ratios of substances involved in the reaction.For example, knowing 1 mole of \( \text{NaCl} \) yields 1 mole of \( \text{Na}^+ \) and 1 mole of \( \text{Cl}^- \), we can use stoichiometry to determine the amount of product from a known amount of reactant. This concept helps us quantify the ions in a solution after a compound has dissociated.

• In \( \text{K}_3\text{PO}_4 \), stoichiometry tells us that for every mole, 3 moles of \( \text{K}^+ \) are produced.
• Thus, if you have 0.0700 moles of \( \text{K}_3\text{PO}_4 \), you will have 0.210 moles of \( \text{K}^+ \).

Having a grasp of stoichiometry helps in understanding and performing these calculations effectively.

Volume Conversion

Volume conversion is an essential skill in chemistry for correctly solving problems involving solutions and concentrations. Normally, volume in chemistry is expressed in liters (L), but sometimes measurements might start off in milliliters (mL) or cubic centimeters (\( \text{cm}^3 \)).Knowing the right conversions is essential:

• 1 L = 1000 mL
• 1 mL = 1 \( \text{cm}^3 \)

Therefore, to convert from \( \text{cm}^3 \) to L, you divide by 1000. For instance, 200 \( \text{cm}^3 \) is 0.200 L.

Correct volume conversions allow for precise calculations of moles in a solution, since molarity is given in terms of liters. Ensuring accuracy in your volume measurements ensures accuracy in your overall computations for chemical concentrations.