- Calculate the initial moles of \(\text{K}_3\text{PO}_4\): \(0.48\text{ M} \times 0.035\text{ L} = 0.0168\text{ mol}\).
- Determine the moles of \(\text{KNO}_3\) produced from stoichiometry: \(3 \times 0.0168\text{ mol} = 0.0504\text{ mol}\).
- Divide the produced moles of \(\text{KNO}_3\) by the final total volume of \(0.060\text{ L}\) to find the molarity: \(\frac{0.0504\text{ mol}}{0.060\text{ L}} = 0.84\text{ M}\).
- The final molarity of the \(\text{KNO}_3\) solution is \(\boxed{0.84\text{ M}}\).
Explanation
- Analysis of the Reaction and Data
To determine the molarity of the \(\text{KNO}_3\) solution, we first identify the reactants and the stoichiometric relationship from the balanced chemical equation:
\[3\text{AgNO}_3(aq) + \text{K}_3\text{PO}_4(aq) \rightarrow \text{Ag}_3\text{PO}_4(s) + 3\text{KNO}_3(aq)\]
We are given that \(\text{AgNO}_3\) is in excess, meaning all of the \(\text{K}_3\text{PO}_4\) provided will react completely. We have \(35\text{ ml}\) (which is \(0.035\text{ L}\)) of \(0.48\text{ M K}_3\text{PO}_4\). The total volume of the resulting mixture is \(60\text{ ml}\) (which is \(0.060\text{ L}\)).
- Determining the Moles of Product
We calculate the moles of \(\text{K}_3\text{PO}_4\) initially present using the formula \(\text{moles} = \text{Molarity} \times \text{Volume (in Liters)}\):
\[n_{\text{K}_3\text{PO}_4} = 0.48\text{ M} \times 0.035\text{ L} = 0.0168\text{ moles}\]
According to the balanced chemical equation, \(1\text{ mole}\) of \(\text{K}_3\text{PO}_4\) reacts to form \(3\text{ moles}\) of \(\text{KNO}_3\). Therefore, the number of moles of \(\text{KNO}_3\) produced is:
\[n_{\text{KNO}_3} = 3 \times n_{\text{K}_3\text{PO}_4} = 3 \times 0.0168\text{ moles} = 0.0504\text{ moles}\]
- Calculating the Molarity of \(\text{KNO}_3\)
Now that we have the total moles of \(\text{KNO}_3\) produced (\(0.0504\text{ moles}\)) and the final total volume of the solution (\(0.060\text{ L}\)), we can find the molarity of the \(\text{KNO}_3\) solution using the formula \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\):
\[M_{\text{KNO}_3} = \frac{0.0504\text{ moles}}{0.060\text{ L}} = 0.84\text{ M}\]
The molarity of the resulting \(\text{KNO}_3\) solution is \(\boxed{0.84\text{ M}}\).
Examples
In a chemistry laboratory, preparing specific concentrations of solutions is essential for accurate titration or synthesis. For instance, when creating a fertilizer component like potassium nitrate (\(\text{KNO}_3\)), chemists must account for the stoichiometry of the reaction and the final volume to ensure the desired concentration is achieved for crop application. This calculation methodology is fundamental in industries like agriculture, pharmacology, and material science, where precise chemical concentrations dictate the safety and effectiveness of the final product.