1. **State the problem:** We need to find the concentrations of $\mathrm{Cu^{2+}}$ and $\mathrm{NO_3^-}$ ions in solution given the initial concentration of $\mathrm{Cu(NO_3)_2}$ and additional moles of ions. 2. **Formula and explanation:** Concentration is moles of solute per liter of solution. For ionic compounds, use stoichiometry to find moles of each ion from moles of compound. 3. **Calculate $C_{\mathrm{Cu^{2+}}}$:** - Initial moles of $\mathrm{Cu(NO_3)_2}$ per liter: $0.1$ mol/L - Each mole of $\mathrm{Cu(NO_3)_2}$ gives 1 mole of $\mathrm{Cu^{2+}}$ - Additional moles of $\mathrm{Cu^{2+}}$: $4.405 \times 10^{-3}$ mol $$C_{\mathrm{Cu^{2+}}} = 0.1 \times \frac{1}{1} + 4.405 \times 10^{-3} = 0.104405 \text{ mol/L}$$ 4. **Calculate $C_{\mathrm{NO_3^-}}$:** - Initial moles of $\mathrm{Cu(NO_3)_2}$ per liter: $0.1$ mol/L - Each mole of $\mathrm{Cu(NO_3)_2}$ gives 2 moles of $\mathrm{NO_3^-}$ - Subtract moles of $\mathrm{NO_3^-}$ lost: $0.02226$ mol $$C_{\mathrm{NO_3^-}} = 0.1 \times \frac{2}{1} - 0.02226 = 0.17774 \text{ mol/L}$$ 5. **Final concentrations:** - $C_{\mathrm{Cu^{2+}}} = 0.104405$ mol/L - $C_{\mathrm{NO_3^-}} = 0.17774$ mol/L These calculations use stoichiometric ratios and account for additional or lost moles to find the final ion concentrations.