Question

Suppose \(50.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

Short answer

The concentration of the ions in the mixed solution is 0.1667 M for Co\(^{2+}\), 0.1167 M for Ni\(^{2+}\), and 0.5667 M for Cl\(^{-}\).

Step-by-step solution

Determine the total volume of the mixed solutions

Since the volumes are additive, we simply sum the volumes of the two solutions to get the total volume: Total volume = \(50.0 \mathrm{~mL} + 25.0 \mathrm{~mL} = 75.0 \mathrm{~mL}\)

Calculate initial moles of each ion

We will find the initial moles of each ion in the separate solutions by multiplying the volume (in liters) by the concentration (in moles per liter) for each solution. For Co\(^{2+}\) ion from \(\mathrm{CoCl}_{2}\): Initial moles of Co\(^{2+}\) = Initial volume (in L) × Initial concentration = (\(50.0 \mathrm{~mL}\) × \(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}) × 0.250 \mathrm{M} = 0.0125 \mathrm{~mol}\) For Cl\(^{-}\) ion from \(\mathrm{CoCl}_{2}\) (remembering that there are 2 Cl\(^{-}\) ions for every Co\(^{2+}\) ion, because of the chemical formula \(\mathrm{CoCl}_{2}\)): Initial moles of Cl\(^{-}\) = 2 × Initial moles of Co\(^{2+}\) = 2 × 0.0125 \mathrm{~mol} = 0.0250 \mathrm{~mol}$ For Ni\(^{2+}\) ion from \(\mathrm{NiCl}_{2}\): Initial moles of Ni\(^{2+}\) = Initial volume (in L) × Initial concentration = (\(25.0 \mathrm{~mL}\) × \(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}) × 0.350 \mathrm{M} = 0.00875 \mathrm{~mol}\) For Cl\(^{-}\) ion from \(\mathrm{NiCl}_{2}\) (since there are 2 Cl\(^{-}\) ions for every Ni\(^{2+}\) ion): Initial moles of Cl\(^{-}\) from \(\mathrm{NiCl}_{2}\) = 2 × Initial moles of Ni\(^{2+}\) = 2 × 0.00875 \mathrm{~mol} = 0.0175 \mathrm{~mol}$

Calculate total moles of each ion

Adding the moles of each ion from their respective solutions will give us the total moles of each ion after mixing. Total moles of Co\(^{2+}\) = 0.0125 \mathrm{~mol} (since it comes only from \(\mathrm{CoCl}_{2}\)) Total moles of Ni\(^{2+}\) = 0.00875 \mathrm{~mol} (since it comes only from \(\mathrm{NiCl}_{2}\)) Total moles of Cl\(^{-}\) = 0.0250 \mathrm{~mol} + 0.0175 \mathrm{~mol} = 0.0425 \mathrm{~mol}$

Calculate final concentrations of each ion

Now, to find the final concentration of each ion in the mixed solution, we will divide the total moles of each ion by the total volume (in liters). Final concentration of Co\(^{2+}\): = \(\frac{0.0125 \mathrm{~mol}}{75.0 \mathrm{~mL} × \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}} = 0.1667 \mathrm{M}\) Final concentration of Ni\(^{2+}\): = \(\frac{0.00875 \mathrm{~mol}}{75.0 \mathrm{~mL} × \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}} = 0.1167 \mathrm{M}\) Final concentration of Cl\(^{-}\): = \(\frac{0.0425 \mathrm{~mol}}{75.0 \mathrm{~mL} × \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}} = 0.5667 \mathrm{M}\) The concentration of the ions in the mixed solution is 0.1667 M for Co\(^{2+}\), 0.1167 M for Ni\(^{2+}\), and 0.5667 M for Cl\(^{-}\).