Questions: A 21.33 mL solution containing 1.607 g Mg(NO3)2 is mixed with a 32.75 mL solution containing 1.137 g NaOH. Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive. If a species fully precipitates, type 0. Be sure your answer has the correct number of significant digits. M Mg2+ M NO3- M Na+ M OH-

A 21.33 mL solution containing 1.607 g Mg(NO3)2 is mixed with a 32.75 mL solution containing 1.137 g NaOH. Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive. If a species fully precipitates, type 0. Be sure your answer has the correct number of significant digits. M Mg2+ M NO3- M Na+ M OH-
Transcript text: A 21.33 mL solution containing $1.607 \mathrm{~g} \mathrm{Mg}_{\left(\mathrm{NO}_{3}\right)_{2}}$ is mixed with a 32.75 mL solution containing 1.137 g NaOH . Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive. If a species fully precipitates, type 0. Be sure your answer has the correct number of significant digits. $\mathrm{M} \mathrm{Mg}^{2+}$ $\square$ 1 ×10 $\mathrm{M} \mathrm{NO}_{3}$ $\mathrm{M} \mathrm{Na}^{+}$ M OH
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Solution

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Solution Steps

Step 1: Determine the moles of each reactant

First, we need to calculate the moles of each reactant in the solutions.

For $\mathrm{Mg}(\mathrm{NO}_3)_2$: \[ \text{Molar mass of } \mathrm{Mg}(\mathrm{NO}_3)_2 = 24.305 + 2 \times (14.007 + 3 \times 16.00) = 148.315 \, \text{g/mol} \] \[ \text{Moles of } \mathrm{Mg}(\mathrm{NO}_3)_2 = \frac{1.607 \, \text{g}}{148.315 \, \text{g/mol}} = 0.01083 \, \text{mol} \]

For $\mathrm{NaOH}$: \[ \text{Molar mass of } \mathrm{NaOH} = 22.990 + 15.999 + 1.008 = 39.997 \, \text{g/mol} \] \[ \text{Moles of } \mathrm{NaOH} = \frac{1.137 \, \text{g}}{39.997 \, \text{g/mol}} = 0.02843 \, \text{mol} \]

Step 2: Write the balanced chemical equation

The balanced chemical equation for the reaction is: \[ \mathrm{Mg}(\mathrm{NO}_3)_2 + 2 \mathrm{NaOH} \rightarrow \mathrm{Mg}(\mathrm{OH})_2 \downarrow + 2 \mathrm{NaNO}_3 \]

Step 3: Determine the limiting reactant

From the balanced equation, 1 mole of $\mathrm{Mg}(\mathrm{NO}_3)_2$ reacts with 2 moles of $\mathrm{NaOH}$.

\[ \text{Moles of } \mathrm{NaOH} \text{ required} = 0.01083 \, \text{mol} \times 2 = 0.02166 \, \text{mol} \]

Since we have 0.02843 moles of $\mathrm{NaOH}$, which is more than 0.02166 moles, $\mathrm{Mg}(\mathrm{NO}_3)_2$ is the limiting reactant.

Step 4: Calculate the moles of products formed and remaining reactants

Since $\mathrm{Mg}(\mathrm{NO}_3)_2$ is the limiting reactant, it will completely react: \[ \text{Moles of } \mathrm{Mg}(\mathrm{NO}_3)_2 \text{ remaining} = 0 \] \[ \text{Moles of } \mathrm{NaOH} \text{ remaining} = 0.02843 \, \text{mol} - 0.02166 \, \text{mol} = 0.00677 \, \text{mol} \]

Step 5: Calculate the concentrations of ions in the final solution

The total volume of the solution after mixing is: \[ 21.33 \, \text{mL} + 32.75 \, \text{mL} = 54.08 \, \text{mL} = 0.05408 \, \text{L} \]

Concentration of $\mathrm{Mg}^{2+}$: \[ \text{Since } \mathrm{Mg}(\mathrm{OH})_2 \text{ precipitates completely, } [\mathrm{Mg}^{2+}] = 0 \]

Concentration of $\mathrm{NO}_3^{-}$: \[ \text{Moles of } \mathrm{NO}_3^{-} = 2 \times 0.01083 \, \text{mol} = 0.02166 \, \text{mol} \] \[ [\mathrm{NO}_3^{-}] = \frac{0.02166 \, \text{mol}}{0.05408 \, \text{L}} = 0.4005 \, \text{M} \]

Concentration of $\mathrm{Na}^{+}$: \[ \text{Moles of } \mathrm{Na}^{+} = 2 \times 0.01083 \, \text{mol} + 0.00677 \, \text{mol} = 0.02843 \, \text{mol} \] \[ [\mathrm{Na}^{+}] = \frac{0.02843 \, \text{mol}}{0.05408 \, \text{L}} = 0.5257 \, \text{M} \]

Concentration of $\mathrm{OH}^{-}$: \[ [\mathrm{OH}^{-}] = \frac{0.00677 \, \text{mol}}{0.05408 \, \text{L}} = 0.1252 \, \text{M} \]

Final Answer

\[ \boxed{[\mathrm{Mg}^{2+}] = 0} \] \[ \boxed{[\mathrm{NO}_3^{-}] = 0.4005 \, \text{M}} \] \[ \boxed{[\mathrm{Na}^{+}] = 0.5257 \, \text{M}} \] \[ \boxed{[\mathrm{OH}^{-}] = 0.1252 \, \text{M}} \]

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