Question

A solution is made up by mixing \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{AuNO}_{3}\) and \(225 \mathrm{~mL}\) of \(0.049 \mathrm{M} \mathrm{AgNO}_{3}\). Twenty-five \(\mathrm{mL}\) of a \(0.0100 \mathrm{M}\) solution of \(\mathrm{HCl}\) is then added. \(K_{\mathrm{sp}}\) of \(\mathrm{AuCl}=\) \(2.0 \times 10^{-13}\). When equilibrium is established, will there be \(\text{[ ]}\)no precipitate? \(\text{[ ]}\)a precipitate of \(\mathrm{AuCl}\) only? \(\text{[ ]}\)a precipitate of \(\mathrm{AgCl}\) only? \(\text{[ ]}\)a precipitate of both \(\mathrm{AgCl}\) and \(\mathrm{AuCl}\) ?

Short answer

Based on the comparison of reaction quotients with the \(K_{\mathrm{sp}}\) values, there will be a precipitate of both \(\mathrm{AgCl}\) and \(\mathrm{AuCl}\) when these solutions are mixed together.

Step-by-step solution

Calculate total volume

Since we have \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{AuNO}_{3}\), \(225 \mathrm{~mL}\) of \(0.049 \mathrm{M}\) \(\mathrm{AgNO}_{3}\), and \(25 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) \(\mathrm{HCl}\), we find the total volume by adding the individual volumes: Total Volume = \(125 + 225 + 25 = 375\) mL

Calculate initial concentrations

Make sure to convert the volumes to L and calculate the initial concentrations of \(\mathrm{Au^{+}}\), \(\mathrm{Ag^{+}}\), and \(\mathrm{Cl^{-}}\): \([\mathrm{Au^{+}}]_i = \frac{0.100 \mathrm{M} \cdot 0.125 \mathrm{L}}{0.375 \mathrm{L}} = 0.0333 \mathrm{M}\) \([\mathrm{Ag^{+}}]_i = \frac{0.049 \mathrm{M} \cdot 0.225 \mathrm{L}}{0.375 \mathrm{L}} = 0.0294 \mathrm{M}\) \([\mathrm{Cl^{-}}]_i = \frac{0.0100 \mathrm{M} \cdot 0.025 \mathrm{L}}{0.375 \mathrm{L}} = 0.000667 \mathrm{M}\)

Use \(K_{\mathrm{sp}}\) values to determine precipitates

The \(K_{\mathrm{sp}}\) of \(\mathrm{AuCl}\) is given: \(2.0 \times 10^{-13}\). Let's find the reaction quotient, \(Q\) for \(\mathrm{AuCl}\): \(Q_{\mathrm{AuCl}} = [\mathrm{Au^{+}}][\mathrm{Cl^{-}}] = (0.0333)(0.000667) = 2.22 \times 10^{-5}\) Now, let's compare \(Q_{\mathrm{AuCl}}\) with the \(K_{\mathrm{sp}}\) of \(\mathrm{AuCl}\). Since \(Q_{\mathrm{AuCl}} > K_{\mathrm{sp, AuCl}}\), a precipitate of \(\mathrm{AuCl}\) will form. Next, we need to find the \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\). The \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\) is \(1.8 \times 10^{-10}\). Let's find the reaction quotient, \(Q\) for \(\mathrm{AgCl}\): \(Q_{\mathrm{AgCl}} = [\mathrm{Ag^{+}}][\mathrm{Cl^{-}}] = (0.0294)(0.000667) = 1.96 \times 10^{-5}\) Compare \(Q_{\mathrm{AgCl}}\) with the \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\). As \(Q_{\mathrm{AgCl}} > K_{\mathrm{sp, AgCl}}\), a precipitate of \(\mathrm{AgCl}\) will also form.

Final result

Based on the calculations in Step 3, we have found that there will be a precipitate of both \(\mathrm{AgCl}\) and \(\mathrm{AuCl}\). So, the correct option is the following: \(\text{[ ]}\)no precipitate? \(\text{[ ]}\)a precipitate of \(\mathrm{AuCl}\) only? \(\text{[ ]}\)a precipitate of \(\mathrm{AgCl}\) only? \(\text{[X]}\)a precipitate of both \(\mathrm{AgCl}\) and \(\mathrm{AuCl}\) ?