Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mole of sodium phosphate in 10.0 mL of solution b. 0.300 mole of barium nitrate in 600.0 mL of solution c. 1.00 g of potassium chloride in 0.500 L of solution d. 132 g of ammonium sulfate in 1.50 L of solution

Short answer

a. Molarity of \(Na^+\) = 6.00 M, Molarity of \(PO_4^{3-}\) = 2.00 M b. Molarity of \(Ba^{2+}\) = 0.500 M, Molarity of \(NO_3^-\) = 1.00 M c. Molarity of \(K^+\) = 0.0268 M, Molarity of \(Cl^-\) = 0.0268 M d. Molarity of \(NH_4^+\) = 1.33 M, Molarity of \(SO_4^{2-}\) = 0.667 M

Step-by-step solution

a. 0.0200 mole of sodium phosphate in 10.0 mL of solution

Sodium phosphate dissociates as follows: \(Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}\) First, convert mL to L: 10.0 mL = 0.010 L Now, calculate the molarity for each ion: Molarity of \(Na^+\) = (0.0200 moles of Na_3PO_4 × 3 moles of \(Na^+\) per mole of Na_3PO_4) / 0.010 L = 6.00 M Molarity of \(PO_4^{3-}\) = (0.0200 moles of Na_3PO_4 × 1 mole of \(PO_4^{3-}\) per mole of Na_3PO_4) / 0.010 L = 2.00 M

b. 0.300 mole of barium nitrate in 600.0 mL of solution

Barium nitrate dissociates as follows: \(Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-\) First, convert mL to L: 600.0 mL = 0.600 L Now, calculate the molarity for each ion: Molarity of \(Ba^{2+}\) = (0.300 moles of Ba(NO_3)_2 × 1 mole of \(Ba^{2+}\) per mole of Ba(NO_3)_2) / 0.600 L = 0.500 M Molarity of \(NO_3^-\) = (0.300 moles of Ba(NO_3)_2 × 2 moles of \(NO_3^-\) per mole of Ba(NO_3)_2) / 0.600 L = 1.00 M

c. 1.00 g of potassium chloride in 0.500 L of solution

Potassium chloride dissociates as follows: \(KCl \rightarrow K^+ + Cl^-\) First, convert 1.00 g of potassium chloride to moles (molecular weight of KCl = 39.1 + 35.5 = 74.6 g/mol): 1.00 g KCl × (1 mol KCl / 74.6 g KCl) = 0.0134 moles of KCl Now, calculate the molarity for each ion: Molarity of \(K^+\) = (0.0134 moles of KCl × 1 mole of \(K^+\) per mole of KCl) / 0.500 L = 0.0268 M Molarity of \(Cl^-\) = (0.0134 moles of KCl × 1 mole of \(Cl^-\) per mole of KCl) / 0.500 L = 0.0268 M

d. 132 g of ammonium sulfate in 1.50 L of solution

Ammonium sulfate dissociates as follows: \((NH_4)_2SO_4 \rightarrow 2NH_4^+ + SO_4^{2-}\) First, convert 132 g of ammonium sulfate to moles (molecular weight of (NH_4)_2SO_4 = (2 × 18.0) + 32.1 + (4 × 16.0) = 132.1 g/mol): 132 g (NH_4)_2SO_4 × (1 mol (NH_4)_2SO_4 / 132.1 g (NH_4)_2SO_4) = 1.00 moles of (NH_4)_2SO_4 Now, calculate the molarity for each ion: Molarity of \(NH_4^+\) = (1.00 moles of (NH_4)_2SO_4 × 2 moles of \(NH_4^+\) per mole of (NH_4)_2SO_4) / 1.50 L = 1.33 M Molarity of \(SO_4^{2-}\) = (1.00 moles of (NH_4)_2SO_4 × 1 mole of \(SO_4^{2-}\) per mole of (NH_4)_2SO_4) / 1.50 L = 0.667 M

Key concepts

Strong Electrolytes

Strong electrolytes are substances that completely dissociate into ions when dissolved in water. This complete dissociation ensures that these substances conduct electricity well in a solution, as charged particles (ions) are responsible for the flow of electric current.
It's important to differentiate strong electrolytes from weak electrolytes, which only partially dissociate in solution.

• Examples of strong electrolytes include sodium chloride (\(NaCl\)), potassium chloride (\(KCl\)), and barium nitrate (\(Ba(NO_3)_2\)).
• These substances break into their respective cations and anions effortlessly when dissolved.

Understanding this concept is essential as it helps us predict and calculate the concentration of ions in solution, which is key to various chemical calculations.

Molarity Calculation

Molarity is a measure of the concentration of a solution, expressed in moles of solute per liter of solution (mol/L). It is represented by the symbol \(M\).
To calculate molarity, you can use the formula:

• \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]

Now, consider sodium phosphate, a strong electrolyte, which dissociates into ions completely.
For every mole of sodium phosphate (\(Na_3PO_4\)), three moles of \(Na^+\) and one mole of \(PO_4^{3-}\) are produced.
To find the concentration of each ion, convert the volume of the solution into liters and apply the dissociation ratio to the moles of the solute and plug it into the molarity formula.
Through these steps, you compute the exact molarity of each ion in the solution.

Ion Dissociation

Ion dissociation refers to the process where ionic compounds separate into their individual ions when dissolved in solution. This process is crucial because it allows the various ions to interact independently in a solution.
For example, barium nitrate (\(Ba(NO_3)_2\)) dissociates into one barium ion (\(Ba^{2+}\)) and two nitrate ions (\(NO_3^-\)).
Dissociation is often written in its simplified equation form:

• \(Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-\)

Knowing the dissociation process allows you to predict how each component of the solution will react in different scenarios.
Moreover, this understanding helps in calculating the concentration of each ion, which is determined by their dissociation from the parent compound.

Solution Preparation

Preparing a solution involves accurately measuring both solute and solvent to achieve the desired concentration. This process can start with weighing the solute or measuring the liquid volume to ensure precision.
For solid solutes like potassium chloride (\(KCl\)), you must convert the given mass to moles using its molar mass. This step is crucial because moles relate to the number of molecules and ions, which determine the resulting concentration.
Steps for solution preparation include:

• Finding the molar mass of the solute.
• Weighing the necessary amount of solute (in grams).
• Diluting the solute with a suitable volume of solvent to attain the required solution volume.

Through these careful measurements and calculations, you achieve the target molarity, ensuring the solution behaves as expected in experiments and practical applications.