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Which ions remain in solution, unreacted, after each of the following pairs of solutions is mixed? \begin{equation} \begin{array}{l}{\text { (a) potassium carbonate and magnesium sulfate }} \\\ {\text { (b) lead nitrate and lithium sulfide }} \\ {\text { (c) ammonium phosphate and calcium chloride }}\end{array} \end{equation}

Short Answer

Expert verified
For the given pairs of solutions, the ions that remain unreacted in the solution are: (a) K鈦 and SO鈧劼测伝 (b) Li鈦 and NO鈧冣伝 (c) NH鈧勨伜 and Cl鈦

Step by step solution

01

Identifying possible reactions for part (a)

For the pair potassium carbonate (K2CO3) and magnesium sulfate (MgSO4), first write down the dissociated ions as K鈦, CO鈧兟测伝, Mg虏鈦, and SO鈧劼测伝. Now, consider all possible combinations of cations and anions: K鈦 with SO鈧劼测伝, yielding potassium sulfate (K2SO4), and Mg虏鈦 with CO鈧兟测伝, yielding magnesium carbonate (MgCO3).
02

Determine solubility for part (a)

Using solubility rules, we can determine that potassium sulfate (K2SO4) is soluble, while magnesium carbonate (MgCO3) is not. Therefore, magnesium carbonate will precipitate out of the solution.
03

Identify unreacted ions for part (a)

Since potassium sulfate remains soluble, the K鈦 and SO鈧劼测伝 ions remain unreacted in the solution.
04

Identifying possible reactions for part (b)

For the pair lead nitrate (Pb(NO鈧)鈧) and lithium sulfide (Li鈧係), first write down the dissociated ions as Pb虏鈦, NO鈧冣伝, Li鈦, and S虏鈦. Now, consider all possible combinations of cations and anions: Pb虏鈦 with S虏鈦, yielding lead sulfide (PbS), and Li鈦 with NO鈧冣伝, yielding lithium nitrate (LiNO鈧).
05

Determine solubility for part (b)

Using solubility rules, we can determine that lead sulfide (PbS) is not soluble, while lithium nitrate (LiNO鈧) is. Therefore, lead sulfide will precipitate out of the solution.
06

Identify unreacted ions for part (b)

Since lithium nitrate remains soluble, the Li鈦 and NO鈧冣伝 ions remain unreacted in the solution.
07

Identifying possible reactions for part (c)

For the pair ammonium phosphate ((NH鈧)鈧働O鈧) and calcium chloride (CaCl鈧), first write down the dissociated ions as NH鈧勨伜, PO鈧劼斥伝, Ca虏鈦, and Cl鈦. Now, consider all possible combinations of cations and anions: NH鈧勨伜 with Cl鈦, yielding ammonium chloride (NH鈧凜l), and Ca虏鈦 with PO鈧劼斥伝, yielding calcium phosphate (Ca鈧(PO鈧)鈧).
08

Determine solubility for part (c)

Using solubility rules, we can determine that ammonium chloride (NH鈧凜l) is soluble, while calcium phosphate (Ca鈧(PO鈧)鈧) is not. Therefore, calcium phosphate will precipitate out of the solution.
09

Identify unreacted ions for part (c)

Since ammonium chloride remains soluble, the NH鈧勨伜 and Cl鈦 ions remain unreacted in the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
Precipitation reactions occur when two solutions containing soluble salts are mixed, leading to the formation of an insoluble compound. These reactions are essential to many processes in chemistry and are often depicted in chemical equations.

For instance, when mixing solutions of potassium carbonate and magnesium sulfate, we observe the outcome of a precipitation reaction. Two new compounds can form: potassium sulfate (K2SO4) and magnesium carbonate (MgCO3). However, the solubility rules tell us that magnesium carbonate is insoluble in water and therefore precipitates out of the solution as a solid.

Applying these principles helps identify which ions will remain in solution and which will form the precipitate. Soluble ions, like the potassium (K鈦) and sulfate (SO鈧劼测伝) ions in our example, will not participate in the precipitate formation and will remain in the solution as spectator ions.
Ionic Equations
Ionic equations provide a deeper understanding of the reactions happening in a solution by showing which ions are involved and their states. The process begins by writing the complete ionic equation, which lists all ions in their dissociated form. For example, lead nitrate and lithium sulfide dissociate into their constituent ions Pb虏鈦, NO鈧冣伝, Li鈦, and S虏鈦 when dissolved in water.

In an ionic equation, the ions that do not participate in the formation of the precipitate are termed 'spectator ions'. These ions do not change state and are not part of the net ionic equation. For instance, when lead nitrate reacts with lithium sulfide, the net ionic equation would only include the ions that form the lead sulfide precipitate, not the spectator ions Li鈦 and NO鈧冣伝 which are soluble and remain in the solution.
Soluble and Insoluble Compounds
Understanding the concept of solubility is essential for predicting the outcome of reactions in aqueous solutions. Solubility rules are guidelines that allow us to predict whether a compound will dissolve in water (soluble) or form a precipitate (insoluble).

Compounds containing alkali metal ions and the ammonium ion are typically soluble. Hence, potassium sulfate and ammonium chloride remain in solution in the given examples. On the other hand, many carbonates, phosphates, and sulfides, such as magnesium carbonate, calcium phosphate, and lead sulfide, are generally insoluble, except when paired with specific ions that can make them soluble.

These solubility rules come in handy when predicting which ions will remain unreacted in a solution, such as Ca虏鈦 and PO鈧劼斥伝 ions forming the insoluble compound calcium phosphate, whereas NH鈧勨伜 and Cl鈦 remain in solution as they form the soluble compound ammonium chloride.

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Most popular questions from this chapter

(a) How many milliliters of 0.120 \(\mathrm{M}\) HCl are needed to completely neutralize 50.0 \(\mathrm{mL}\) of 0.101 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of 0.125 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize 0.200 \(\mathrm{g}\) of NaOH? \((\mathrm{c})\) If 55.8 \(\mathrm{mL}\) of a BaCl \(_{2}\) solution is needed to precipitate all the sulfate ion in a 752 -mg sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the BaCl\(_{2}\) solution?

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