Questions: Problem Set #9 3. (5 pts.) Silver nitrate and sodium iodide react to make a precipitate. a. When 50.0 mL of 5.0 g / L AgNO3 is added to a coffee-cup calorimeter containing 50.0 mL of 5.0 g / L NaI, with both solutions at 25.0 degrees C, how many moles of AgI forms? Include a reaction equation! b. Find ΔHrnn for this reaction from the balanced net ionic equation. NOTE: The ΔHf° for AgI(s) is -62.38 kJ / mol; the ΔHf° for Ag+(aq) is -105.9 kJ / mol, and the ΔHf° for I^- is 55.94 kJ / mol. c. What is the ΔTsoln (assuming that volumes are additive and the solution has the density and specific heat capacity of water)?

Solution Steps

The reaction between silver nitrate (\(\text{AgNO}_3\)) and sodium iodide (\(\text{NaI}\)) forms silver iodide (\(\text{AgI}\)) as a precipitate and sodium nitrate (\(\text{NaNO}_3\)) in solution. The balanced chemical equation is:

\[ \text{AgNO}_3(aq) + \text{NaI}(aq) \rightarrow \text{AgI}(s) + \text{NaNO}_3(aq) \]

The net ionic equation, which shows the formation of the precipitate, is:

\[ \text{Ag}^+(aq) + \text{I}^-(aq) \rightarrow \text{AgI}(s) \]

First, calculate the moles of \(\text{AgNO}_3\) and \(\text{NaI}\) in the solutions. Given concentrations are \(5.0 \, \text{g/L}\) for both solutions.

• Molar mass of \(\text{AgNO}_3\) is approximately \(169.87 \, \text{g/mol}\).
• Molar mass of \(\text{NaI}\) is approximately \(149.89 \, \text{g/mol}\).

Calculate moles of \(\text{AgNO}_3\):

\[ \text{Moles of } \text{AgNO}_3 = \frac{5.0 \, \text{g/L} \times 0.050 \, \text{L}}{169.87 \, \text{g/mol}} = 0.00147 \, \text{mol} \]

Calculate moles of \(\text{NaI}\):

\[ \text{Moles of } \text{NaI} = \frac{5.0 \, \text{g/L} \times 0.050 \, \text{L}}{149.89 \, \text{g/mol}} = 0.00167 \, \text{mol} \]

Since \(\text{AgNO}_3\) is the limiting reactant, the moles of \(\text{AgI}\) formed is equal to the moles of \(\text{AgNO}_3\):

\[ \text{Moles of } \text{AgI} = 0.00147 \, \text{mol} \]

Using Hess's Law and the given standard enthalpies of formation:

\[ \Delta H_{\text{rnn}} = \Delta H_f^\circ (\text{AgI}) - [\Delta H_f^\circ (\text{Ag}^+) + \Delta H_f^\circ (\text{I}^-)] \]

Substitute the given values:

\[ \Delta H_{\text{rnn}} = (-62.38 \, \text{kJ/mol}) - [(-105.9 \, \text{kJ/mol}) + 55.94 \, \text{kJ/mol}] \]

\[ \Delta H_{\text{rnn}} = -62.38 \, \text{kJ/mol} + 105.9 \, \text{kJ/mol} - 55.94 \, \text{kJ/mol} = -112.3 \, \text{kJ/mol} \]

Use the formula for heat transfer:

\[ q = m \cdot c \cdot \Delta T \]

Assume the density of the solution is \(1.00 \, \text{g/mL}\) and the specific heat capacity is \(4.18 \, \text{J/g}^\circ\text{C}\). The total mass of the solution is \(100.0 \, \text{g}\).

The heat released by the reaction is:

\[ q = \Delta H_{\text{rnn}} \times \text{moles of } \text{AgI} = -112.3 \, \text{kJ/mol} \times 0.00147 \, \text{mol} = -0.1651 \, \text{kJ} = -165.1 \, \text{J} \]

Solve for \(\Delta T\):

\[ \Delta T = \frac{q}{m \cdot c} = \frac{-165.1 \, \text{J}}{100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C}} = -0.395 \, ^\circ\text{C} \]

Final Answer