Problem 15 - Chemistry: Principles and Reactions

Question

A solution is prepared by mixing \(13.00 \mathrm{~mL}\) of \(0.0021 M\) aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) with \(25.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{HCl}\). Assume that volumes are additive. (a) Will precipitation occur? (b) Calculate \(\left[\mathrm{Hg}_{2}{\underline{\phantom{xx}}}^{2+}\right],\left[\mathrm{Cl}^{-}\right]\), and \(\left[\mathrm{NO}_{3}^{-}\right]\) after equilibrium is established.

Step-by-Step Solution

Answer

Provide a step-by-step solution for determining precipitation and calculating the equilibrium concentrations. Answer: Yes, precipitation will occur, forming the insoluble product \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\). The final equilibrium concentrations are: \(\left[\mathrm{Hg}_{2}{^{2+}}\right] = 0 M\), \(\left[\mathrm{Cl}^{-}\right] = 8.44\times10^{-3} M\), and \(\left[\mathrm{NO}_{3}^{-}\right] = 1.44\times10^{-3} M\). A step-by-step solution is provided in the explanation above.

Step 1: Determine the possible precipitation

For the possible precipitation reaction, the reactants are \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{HCl}\). The potential product is \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\), a mercury(I) chloride compound. We need to determine if this compound is soluble in water. According to solubility rules, most chlorides are soluble, but mercury(I) chlorides are insoluble. So, the \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\) is insoluble and precipitation will occur. #a_Step 2: Write the precipitation reaction and its equilibrium equation#

Step 2: Write the precipitation reaction and equilibrium equation

The balanced precipitation reaction is: $$\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(aq) + 2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) + 2\mathrm{HNO}_{3}(aq)$$ As the main product is insoluble, an equilibrium is established between the ions in the solution and the solid \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\). #a_Step 3: Determine the initial concentrations of the species in the mixture#

Step 3: Calculate the initial concentrations

Given \(\mathrm{Hg}_{2}{^{2+}}\) has a concentration of \(0.0021 M\) in \(13.00 \mathrm{~mL}\) and \(\mathrm{Cl^-}\) has a concentration of \(0.015 M\) in \(25.0 \mathrm{~mL}\). To find the initial concentration in the mixture, we need to calculate the moles of each ion and divide it by the total volume of the mixture (\(13.00\mathrm{~mL}+25.0\mathrm{~mL}\)). Initial moles of \(\mathrm{Hg}_{2}{^{2+}}\): \(0.0021 M \times 13.00 \mathrm{~mL} = 27.3\times10^{-3} \mathrm{~mol}\) Initial moles of \(\mathrm{Cl}^{-}\): \(0.015 M \times 25.0\mathrm{~mL} = 375\times10^{-3} \mathrm{~mol}\) Initial concentration of \(\mathrm{Hg}_{2}{^{2+}}\): \(\frac{27.3\times10^{-3}\mathrm{~mol}}{38\mathrm{~mL}} = 7.18\times10^{-4} M\) Initial concentration of \(\mathrm{Cl}^{-}\): \(\frac{375\times10^{-3}\mathrm{~mol}}{38\mathrm{~mL}} = 9.87\times10^{-3} M\) #a_Step 4: Calculate the equilibrium constant and Q#

Step 4: Calculate the equilibrium constant and Q

For the equilibrium reaction, we will first calculate the solubility product constant, \(K_{sp}\), which is given by: $$K_{sp} = [\mathrm{Hg}_{2}{^{2+}}][\mathrm{Cl}^{-}]^2$$ Next, we need to calculate the reaction quotient, Q, using the initial concentrations: $$Q = [\mathrm{Hg}_{2}{^{2+}}]_{initial}[\mathrm{Cl}^{-}]_{initial}^2 = (7.18\times10^{-4} M)(9.87\times10^{-3} M)^2$$ #a_Step 5: Calculate equilibrium concentration for each ion#

Step 5: Calculate equilibrium concentration

Since Q will be greater than \(K_{sp}\), the precipitation reaction will occur. Finally, we can calculate the equilibrium concentrations of each ion. For \(\mathrm{Hg}_{2}{^{2+}}\): Since one mole of \(\mathrm{Hg}_{2}{^{2+}}\) reacts with two moles of \(\mathrm{Cl}^{-}\), it's important to consider the limiting reactant when calculating the final concentration of \(\mathrm{Hg}_{2}{^{2+}}\). In this case, the initial moles of \(\mathrm{Hg}_{2}{^{2+}}\) are less than the half of the initial moles of \(\mathrm{Cl}^{-}\). Thus, all of the \(\mathrm{Hg}_{2}{^{2+}}\) will react: $$[\mathrm{Hg}_{2}{^{2+}}]_{equilibrium} = 0 M$$ For \(\mathrm{Cl}^{-}\): Since all the \(\mathrm{Hg}_{2}{^{2+}}\) reacted, we need to subtract the used moles of \(\mathrm{Cl}^{-}\). $$[\mathrm{Cl}^{-}]_{equilibrium} = [\mathrm{Cl}^{-}]_{initial} - 2\times[\mathrm{Hg}_{2}{^{2+}}]_{initial} = 9.87\times10^{-3} M - 2\times7.18\times10^{-4} M = 8.44\times10^{-3} M$$ For \(\mathrm{NO_{3}^{-}}\): As there are no changes in \(\mathrm{NO_{3}^{-}}\) concentration during the reaction, we can obtain the mole ratio of \(2\mathrm{NO_{3}^{-}}\) per \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\). Thus, \([\mathrm{NO_{3}^{-}}]_{initial} = 2\times[\mathrm{Hg}_{2}{^{2+}}]_{initial}\), and \([\mathrm{NO_{3}^{-}}]_{equilibrium} \\ = \frac{54.6\times10^{-3}\mathrm{~mol}}{38\mathrm{~mL}} = 1.44\times10^{-3} M\). The final equilibrium concentrations are: \(\left[\mathrm{Hg}_{2}{^{2+}}\right] = 0 M\), \(\left[\mathrm{Cl}^{-}\right] = 8.44\times10^{-3} M\), and \(\left[\mathrm{NO}_{3}^{-}\right] = 1.44\times10^{-3} M\).

Key Concepts

Solubility Rules

Solubility rules help us predict whether a compound will dissolve or form a precipitate in an aqueous solution. These rules are patterns based on observations to aid chemists in understanding reactions. For chlorides, like the compound in question, most are soluble in water. However, an exception stands out for mercury(I) chloride,

• Most chlorides are soluble, except for mercury(I), silver, and lead chlorides which are typically insoluble.
• Nitrates are generally soluble, with very few exceptions.

In this exercise, when mixing \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{HCl}\), mercury(I) chloride, \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\), is predicted to precipitate according to these rules. Understanding these guidelines allows us to predict the formation of a solid precipitate efficiently when mixing compounds.

Reaction Quotient (Q)

The reaction quotient, Q, is a measure used to determine the direction in which a reaction will proceed. It’s calculated in the same manner as the equilibrium constant, \(K\), but uses initial concentrations instead of equilibrium sales.

• When \(Q < K_{sp}\), the system will proceed towards forming more product until equilibrium is achieved.
• When \(Q = K_{sp}\), the system is at equilibrium, with no net change occurring.
• When \(Q > K_{sp}\), the reaction will proceed towards forming more reactants to achieve equilibrium, leading to precipitation.

In this exercise, after calculating \(Q\) with initial ion concentrations, we find that \(Q\) exceeds the given \(K_{sp}\). This condition indicates that the solution is supersaturated, leading to precipitation of \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}\). Hence, understanding \(Q\) helps us predict system shifts and the likelihood of reaching a precipitated state.

Equilibrium Concentrations

Equilibrium concentrations describe the amounts of ions remaining in solution once a reaction reaches equilibrium, particularly after precipitation occurs. Calculating these concentrations allows determination of what remains after most incompatible ions have formed the precipitate.To find the equilibrium concentrations in the given problem:

• Determine the limiting reactant, which is the ion that is completely consumed, restricting further reaction.
• Calculate the changes using stoichiometry, taking into account the initial concentrations and amount reacting.

As shown in this example, once the reaction of \(\mathrm{Hg}_{2}{\underline{\phantom{xx}}}^{2+}\) is complete with \(\mathrm{Cl}^{-}\), no \(\mathrm{Hg}_{2}{\underline{\phantom{xx}}}^{2+}\) ions remain, while the \(\mathrm{Cl}^{-}\) concentration is reduced according to the reaction ratio. Meanwhile, the \(\mathrm{NO}_{3}^{-}\) ions remain unaffected because they do not participate in forming the precipitate.Finding these concentrations requires a solid understanding of stoichiometry and limiting reactants to ensure accuracy and completeness in results.