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Chapter 4: Problem 40

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.

Short Answer

Expert verified
After mixing the solutions, the concentration of each ion is as follows: \( Co^{2+} = 0.1667 \mathrm{M} \), \( Ni^{2+} = 0.1167 \mathrm{M} \), and \( Cl^- = 0.5667 \mathrm{M} \).

Step by step solution

01

Determine initial moles of ions in each solution

To find the initial moles of ions in each solution, use the following formula: moles = molarity × volume (in liters) - For CoCl2 solution: moles of Co^2+ = 0.250 M × 0.050 L = 0.0125 mol moles of Cl^- = (0.250 M × 0.050 L) × 2 = 0.025 mol (since there are 2 moles of Cl^- for each mole of CoCl2) - For NiCl2 solution: moles of Ni^2+ = 0.350 M × 0.025 L = 0.00875 mol moles of Cl^- = (0.350 M × 0.025 L) × 2 = 0.0175 mol (since there are 2 moles of Cl^- for each mole of NiCl2)
02

Calculate the final volume of the mixed solution

To find the final volume of the mixed solution, just add the individual volumes: final volume = volume of CoCl2 + volume of NiCl2 final volume = 50.0 mL + 25.0 mL = 75.0 mL Now, convert the final volume to liters: final volume = 75.0 mL × \( \frac{1 L}{1000 mL} \) = 0.075 L
03

Calculate the final concentration of ions

To find the final concentration of each ion, use the formula: molarity = \( \frac{moles}{final~volume~in~liters} \) - For Co^2+ ion: final concentration of Co^2+ = \( \frac{0.0125~mol}{0.075~L} \) = 0.1667 M - For Ni^2+ ion: final concentration of Ni^2+ = \( \frac{0.00875~mol}{0.075~L} \) = 0.1167 M - For Cl^- ion: final concentration of Cl^- = \( \frac{(0.025 + 0.0175)~mol}{0.075~L} \) = \( \frac{0.0425~mol}{0.075~L} \) = 0.5667 M Therefore, the concentration of each ion after mixing is 0.1667 M for Co^2+, 0.1167 M for Ni^2+, and 0.5667 M for Cl^-.

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Most popular questions from this chapter

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

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