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

First, we need to determine the moles of each ion in the \(32.0 \mathrm{mL}\) of \(0.30 \mathrm{M} \ \mathrm{KMnO}_{4}\) solution. Since molarity (M) = moles of solute/volume of solution in L, we can calculate the moles of KMnO4. Moles of KMnO4 = Molarity × Volume = \(0.30 \mathrm{M} \times 32.0 \mathrm{mL}\) However, we need to convert the volume from mL to L. Moles of KMnO4 = \(0.30 \mathrm{M} \times 0.032 \mathrm{L} = 0.0096 \ \mathrm{mol}\) Now we will find the moles of ions in the solution. KMnO4 dissociates into K+ and MnO4- ions. Moles of K+ = moles of KMnO4 = \(0.0096 \ \mathrm{mol}\) Moles of MnO4- = moles of KMnO4 = \(0.0096 \ \mathrm{mol}\)

Now, let's find the moles of each ion in the mixture of solutions. Moles of K+ in the mixture = Moles of K+ in 32.0 mL of 0.30 M KMnO4 + moles of K+ in 15.0 mL of 0.60 M KMnO4 Moles of K+ in the mixture = \(0.0096 \mathrm{mol} + 0.60 \mathrm{M} \times 0.015 \mathrm{L}\) Moles of K+ in the mixture = \(0.0096 \mathrm{mol} + 0.0090 \mathrm{mol} = 0.0186 \mathrm{mol}\) Moles of MnO4- in the mixture = Moles of MnO4- in 32.0 mL of 0.30 M KMnO4 + moles of MnO4- in 15.0 mL of 0.60 M KMnO4 Moles of MnO4- in the mixture = \(0.0096 \mathrm{mol} + 0.60 \mathrm{M} \times 0.015 \mathrm{L}\) Moles of MnO4- in the mixture = \(0.0096 \mathrm{mol} + 0.0090 \mathrm{mol} = 0.0186 \mathrm{mol}\)

Now that we have the moles of each ion in the mixture, we can find their final concentrations. The total volume of the mixture is \(32.0 \mathrm{mL} + 15.0 \mathrm{mL} = 47.0 \mathrm{mL}\). Concentration of K+ = Moles of K+ in the mixture / Total volume of the mixture = \(0.0186 \ \mathrm{mol} / 0.047 \ \mathrm{L} = 0.396 \ \mathrm{M}\) Concentration of MnO4- = Moles of MnO4- in the mixture / Total volume of the mixture = \(0.0186 \ \mathrm{mol} / 0.047 \ \mathrm{L} = 0.396 \ \mathrm{M}\) The final concentrations in the mixture are: K+ = \(0.396 \ \mathrm{M}\) MnO4- = \(0.396 \ \mathrm{M}\)

For part (b), we have to mix \(60.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \ \mathrm{ZnCl}_{2}^{+}\) solution and \(5.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \ \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) solution. For both solutions, we have multiple ions to consider. First, we need to calculate the moles of each ion in both solutions. For ZnCl2: Moles of Zn2+ = Molarity × Volume = \(0.100 \mathrm{M} \times 0.060 \mathrm{L} = 0.0060 \ \mathrm{mol}\) Moles of Cl- = 2 × Moles of Zn2+ (since there are two Cl- ions in ZnCl2) = 2 × \(0.0060 \ \mathrm{mol} = 0.0120 \ \mathrm{mol}\) For Zn(NO3)2: Moles of Zn2+ = Molarity × Volume = \(0.200 \mathrm{M} \times 0.005 \mathrm{L} = 0.0010 \ \mathrm{mol}\) Moles of NO3- = 2 × Moles of Zn2+ (since there are two NO3- ions in Zn(NO3)2) = 2 × \(0.0010 \ \mathrm{mol} = 0.0020 \ \mathrm{mol}\) Now we sum up the moles of each ion in the mixture: Moles of Zn2+ in the mixture = Moles of Zn2+ in ZnCl2 + Moles of Zn2+ in Zn(NO3)2 = \(0.0060 \mathrm{mol} + 0.0010 \mathrm{mol} = 0.0070 \mathrm{mol}\) Moles of Cl- in the mixture = Moles of Cl- in ZnCl2 = \(0.0120 \ \mathrm{mol}\) Moles of NO3- in the mixture = Moles of NO3- in Zn(NO3)2 = \(0.0020 \ \mathrm{mol}\) Now we calculate the concentration of each ion in the mixture. The total volume of the mixture is \(60.0 \mathrm{mL} + 5.0 \mathrm{mL} = 65.0 \mathrm{mL}\). Concentration of Zn2+ = Moles of Zn2+ in the mixture / Total volume of the mixture = \(0.0070 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.108 \ \mathrm{M}\) Concentration of Cl- = Moles of Cl- in the mixture / Total volume of the mixture = \(0.0120 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.185 \ \mathrm{M}\) Concentration of NO3- = Moles of NO3- in the mixture / Total volume of the mixture = \(0.0020 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.031 \ \mathrm{M}\) The final concentrations in the mixture are: Zn2+ = \(0.108 \ \mathrm{M}\) Cl- = \(0.185 \ \mathrm{M}\) NO3- = \(0.031 \ \mathrm{M}\)

For part (c), we have to mix \(4.2 \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) and \(150.0 \mathrm{mL}\) of \(0.02 \mathrm{M} \ \mathrm{KCl}\). First, let's find the moles of CaCl2: Moles of CaCl2 = mass / molar mass = \(4.2 \mathrm{g} / 110.98 \mathrm{g/mol} = 0.0378 \ \mathrm{mol}\) Now we find the moles of ions in the CaCl2: Moles of Ca2+ = moles of CaCl2 = \(0.0378 \ \mathrm{mol}\) Moles of Cl- in CaCl2 = 2 × moles of Ca2+ (since there are two Cl- ions in CaCl2) = 2 × \(0.0378 \ \mathrm{mol} = 0.0756 \mathrm{mol}\) Next, let's find the moles of ions in the KCl solution: Moles of K+ = Molarity × Volume = \(0.02 \mathrm{M} \times 0.150 \mathrm{L} = 0.0030 \ \mathrm{mol}\) Moles of Cl- in KCl solution = Moles of K+ (since there is one Cl- ion in KCl) = \(0.0030 \ \mathrm{mol}\) Now we sum up the moles of each ion in the mixture: Moles of Ca2+ in the mixture = Moles of Ca2+ = \(0.0378 \ \mathrm{mol}\) Moles of K+ in the mixture = Moles of K+ = \(0.0030 \ \mathrm{mol}\) Moles of Cl- in the mixture = Moles of Cl- in CaCl2 + Moles of Cl- in KCl = \(0.0756 \ \mathrm{mol} + 0.0030 \ \mathrm{mol} = 0.0786 \ \mathrm{mol}\) Now, calculate the concentration of each ion in the mixture. The total volume of the mixture is \(4.2 \mathrm{g} \times 2 \mathrm{mL/g} + 150.0 \mathrm{mL} = 158.4 \mathrm{mL}\). Concentration of Ca2+ = Moles of Ca2+ in the mixture / Total volume of the mixture = \(0.0378 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.239 \ \mathrm{M}\) Concentration of K+ = Moles of K+ in the mixture / Total volume of the mixture = \(0.0030 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.019 \ \mathrm{M}\) Concentration of Cl- = Moles of Cl- in the mixture / Total volume of the mixture = \(0.0786 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.496 \ \mathrm{M}\) The final concentrations in the mixture are: Ca2+ = \(0.239 \ \mathrm{M}\) K+ = \(0.019 \ \mathrm{M}\) Cl- = \(0.496 \ \mathrm{M}\)