Question

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}\) with \(5.0 \mathrm{~mL}\) of \(0.200 \mathrm{MZn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.

Short answer

(a) K⁺ and MnO₄⁻: 0.396 M; (b) Zn²⁺: 0.108 M, Cl⁻: 0.185 M, NO₃⁻: 0.0308 M; (c) Ca²⁺: 0.252 M, Cl⁻: 0.524 M.

Step-by-step solution

Calculate moles of each substance

First, we calculate the moles of each solute in the given solutions. For each ion, we'll apply the formula: \[\text{moles} = \text{concentration (M)} \times \text{volume (L)}\](a) For \(32.0 \, \text{mL}\) of \(0.30 \, \text{M} \, \text{KMnO}_4\), the moles are:\[0.30 \, \text{mol/L} \times 0.0320 \, \text{L} = 0.0096 \, \text{mol}\]For \(15.0 \, \text{mL}\) of \(0.60 \, \text{M} \, \text{KMnO}_4\), the moles are:\[0.60 \, \text{mol/L} \times 0.0150 \, \text{L} = 0.0090 \, \text{mol}\](b) For \(60.0 \, \text{mL}\) of \(0.100 \, \text{M} \, \text{ZnCl}_2\), the moles are:\[0.100 \, \text{mol/L} \times 0.0600 \, \text{L} = 0.0060 \, \text{mol}\]For \(5.0 \, \text{mL}\) of \(0.200 \, \text{M} \, \text{Zn(NO}_3)_2\), the moles are:\[0.200 \, \text{mol/L} \times 0.0050 \, \text{L} = 0.0010 \, \text{mol}\](c) The moles of \(\text{CaCl}_2\):\[\text{mass} = 4.2 \, \text{g}, \text{molar mass of CaCl}_2 = 111 \, \text{g/mol}\]\[\text{moles of CaCl}_2 = \frac{4.2 \, \text{g}}{111 \, \text{g/mol}} = 0.0378 \, \text{mol}\]

Calculate total volume of mixtures

Assume the volumes are additive:(a) The total volume after mixing is:\[32.0 \, \text{mL} + 15.0 \, \text{mL} = 47.0 \, \text{mL} = 0.0470 \, \text{L}\](b) The total volume after mixing is:\[60.0 \, \text{mL} + 5.0 \, \text{mL} = 65.0 \, \text{mL} = 0.0650 \, \text{L}\](c) The total volume is:\[150.0 \, \text{mL} = 0.150 \, \text{L}\] (No need to add, already given as total volume with \(\text{KCl}\)).

Calculate concentration of ions

Using the moles calculated and the total volume, we find the concentration of each ion:(a) Combine moles of \(\text{KMnO}_4\):\[\text{total moles} = 0.0096 + 0.0090 = 0.0186 \, \text{mol}\]\[\text{Concentration of } \text{K}^+\text{ and } \text{MnO}_4^- = \frac{0.0186 \, \text{mol}}{0.0470 \, \text{L}} = 0.396 \, \text{M}\](b) For \(\text{ZnCl}_2\), \(\text{Zn}^{2+}\) is:\[\text{total } \text{Zn}^{2+} = 0.0060 \, \text{mol} + 0.0010 \, \text{mol} = 0.0070 \, \text{mol}\]\[\text{Concentration of } \text{Zn}^{2+} = \frac{0.0070 \, \text{mol}}{0.0650 \, \text{L}} = 0.108 \, \text{M}\]For \(\text{Cl}^-\), from \(\text{ZnCl}_2\) only:\[\text{moles of } \text{Cl}^- = 2 \times 0.0060 = 0.0120 \, \text{mol}\]\[\text{Concentration of } \text{Cl}^- = \frac{0.0120 \, \text{mol}}{0.0650 \, \text{L}} = 0.185 \, \text{M}\]For \(\text{NO}_3^-\), from \(\text{Zn(NO}_3)_2\) only:\[\text{moles of } \text{NO}_3^- = 2 \times 0.0010 = 0.0020 \, \text{mol}\]\[\text{Concentration of } \text{NO}_3^- = \frac{0.0020 \, \text{mol}}{0.0650 \, \text{L}} = 0.0308 \, \text{M}\](c) For \(\text{CaCl}_2\), \(\text{Ca}^{2+}\) is:\[\text{Concentration of } \text{Ca}^{2+} = \frac{0.0378 \, \text{mol}}{0.150 \, \text{L}} = 0.252 \, \text{M}\]For \(\text{Cl}^-\) from both sources (\(\text{CaCl}_2\) and \(\text{KCl}\)):\[\text{From } \text{CaCl}_2: 2 \times 0.0378 = 0.0756 \, \text{mol}\]\[\text{From } \text{KCl}: 0.02 \, \text{mol/L} \times 0.150 \, \text{L} = 0.003 \, \text{mol}\]\[\text{Total moles } \text{Cl}^- = 0.0756 + 0.003 = 0.0786 \, \text{mol}\]\[\text{Concentration of } \text{Cl}^- = \frac{0.0786 \, \text{mol}}{0.150 \, \text{L}} = 0.524 \, \text{M}\]

Final Step: Conclusion

After finding all concentrations for each part:- (a) \(\text{K}^+\) and \(\text{MnO}_4^-\) are both at \(0.396 \, \text{M}\).- (b) \(\text{Zn}^{2+}\) is at \(0.108 \, \text{M}\), \(\text{Cl}^-\) at \(0.185 \, \text{M}\), and \(\text{NO}_3^-\) at \(0.0308 \, \text{M}\).- (c) \(\text{Ca}^{2+}\) is at \(0.252 \, \text{M}\) and \(\text{Cl}^-\) at \(0.524 \, \text{M}\).

Key concepts

Understanding Molarity

Molarity is a fundamental concept in chemistry that helps us understand the concentration of a solution. It tells us how many moles of a solute are present in one liter of solution, measured in mol/L or simply "M". This measurement is crucial because it allows chemists to precisely determine the amounts of chemicals involved in reactions. To calculate molarity, you use the formula:\[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}}\]For example, if you dissolve 1 mole of sodium chloride in 1 liter of water, you have a solution with a molarity of 1 M. Molarity facilitates the ease of using chemical solutions by helping predict how substances will react when combined.When diluting or mixing solutions, it's important to remember that molarity can change depending on the total volume of the solution. However, the total moles of the solute remain the same, making molarity a convenient way to control concentrations in lab experiments.

Chemical Calculations Made Simple

Chemical calculations can seem daunting at first, but they become manageable when broken down into simple steps using basic algebraic principles. One of the key components in chemical calculations is the conversion between mass, moles, and number of particles using Avogadro's number. Here are some basic steps for performing chemical calculations:

• Understand the problem: Determine what is being asked, and identify which chemical quantities you know (mass, moles, volume, etc.).
• Convert units if necessary: Often, you'll need to convert grams to moles using the substance's molar mass, which is its mass per mole.
• Set up the calculation: Use the appropriate formula or equation, ensuring units cancel where necessary.
• Perform the calculation: Solve using algebra, ensuring each step is clear and logical.
• Verify the result: Check if the result makes sense chemically and mathematically.

Performing accurate chemical calculations is vital in having proper experimental results and predictions when working with laboratory reactions and is a skill developed with practice.

Ionic Compounds and Their Roles

Ionic compounds are formed by the electrostatic attraction between oppositely charged ions. They are usually composed of metals bonded to non-metals. When dissolved in water, these compounds dissociate, or break apart, into their constituent ions, which are charged particles that conduct electricity.For example, when zinc chloride (\(\text{ZnCl}_2\)) is dissolved in water, it separates into \(\text{Zn}^{2+}\) and \(2\text{Cl}^-\) ions. Similarly, calcium chloride (\(\text{CaCl}_2\)) dissociates into \(\text{Ca}^{2+}\) and \(2\text{Cl}^-\) ions. The ability to conduct electricity makes these aqueous solutions known as electrolytes, which are vital in many biological and chemical processes.Understanding how ionic compounds dissociate helps in calculating the concentration of individual ions in a solution, as shown in the following example:- For a solution of \(\text{KMnO}_4\), the molarity of the solution is the same for \(\text{K}^+\) and \(\text{MnO}_4^-\) because each formula unit of \(\text{KMnO}_4\) contributes one \(\text{K}^+\) and one \(\text{MnO}_4^-\) to the solution.Ionic compounds have significant roles in both nature and industry due to their distinct properties and widespread applications.