Question

Write the solubility product expression, \(K_{\mathrm{sp}}\), for these substances: (a) \(\mathrm{AgCl}\) (c) \(\mathrm{Zn}(\mathrm{OH})_{2}\) (b) \(\mathrm{PbCrO}_{4}\) (d) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Short answer

The solubility product expressions are: \(\mathrm{AgCl}: K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\); \(\mathrm{PbCrO}_{4}: K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{CrO}_{4}^{2-}]\), \(\mathrm{Zn} (\mathrm{OH})_{2}: K_{\mathrm{sp}} = [\mathrm{Zn}^{2+}][\mathrm{OH}^{-}]^{2}\), \(\mathrm{Ca}_{3} (\mathrm{PO}_{4})_{2}: K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\)

Step-by-step solution

- Write the dissociation equation

For each substance, write the balanced chemical equation for the dissociation in water. This will help identify the ions that are produced.

Step 1(a) - \(\mathrm{AgCl}\)

The dissociation equation for \(\mathrm{AgCl}\) is: \[ \mathrm{AgCl} (s) \leftrightarrow \mathrm{Ag}^{+} (aq) + \mathrm{Cl}^{-} (aq) \]

Step 2(a) - Write \(K_{\mathrm{sp}}\) expression for \(\mathrm{AgCl}\)

Based on the dissociation equation, the solubility product expression is: \[ K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]

Step 1(b) - \(\mathrm{PbCrO}_{4}\)

The dissociation equation for \(\mathrm{PbCrO}_{4}\) is: \[ \mathrm{PbCrO}_{4} (s) \leftrightarrow \mathrm{Pb}^{2+} (aq) + \mathrm{CrO}_{4}^{2-} (aq) \]

Step 2(b) - Write \(K_{\mathrm{sp}}\) expression for \(\mathrm{PbCrO}_{4}\)

Based on the dissociation equation, the solubility product expression is: \[ K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{CrO}_{4}^{2-}] \]

Step 1(c) - \(\mathrm{Zn} (\mathrm{OH})_{2}\)

The dissociation equation for \(\mathrm{Zn} (\mathrm{OH})_{2}\) is: \[ \mathrm{Zn} (\mathrm{OH})_{2} (s) \leftrightarrow \mathrm{Zn}^{2+} (aq) + 2 \mathrm{OH}^{-} (aq) \]

Step 2(c) - Write \(K_{\mathrm{sp}}\) expression for \(\mathrm{Zn} (\mathrm{OH})_{2}\)

Based on the dissociation equation, the solubility product expression is: \[ K_{\mathrm{sp}} = [\mathrm{Zn}^{2+}][\mathrm{OH}^{-}]^{2} \]

Step 1(d) - \(\mathrm{Ca}_{3} (\mathrm{PO}_{4})_{2}\)

The dissociation equation for \(\mathrm{Ca}_{3} (\mathrm{PO}_{4})_{2}\) is: \[ \mathrm{Ca}_{3} (\mathrm{PO}_{4})_{2} (s) \leftrightarrow 3 \mathrm{Ca}^{2+} (aq) + 2 \mathrm{PO}_{4}^{3-} (aq) \]

Step 2(d) - Write \(K_{\mathrm{sp}}\) expression for \(\mathrm{Ca}_{3} (\mathrm{PO}_{4})_{2}\)

Based on the dissociation equation, the solubility product expression is: \[ K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2} \]

Key concepts

dissociation equation

Understanding the dissociation equation is crucial when dealing with the solubility of ionic compounds. These equations help us determine how compounds break down into ions in a solvent, usually water. For example, when we consider \(\text{AgCl}\); its dissociation equation is: \[ \text{AgCl} (s) \leftrightarrow \text{Ag}^{+} (aq) + \text{Cl}^{-} (aq) \]. This shows that solid silver chloride splits into silver ions and chloride ions in the aqueous phase.

Each dissociation equation must be balanced. This means that the number of each kind of atom on both sides of the equation must be equal. Balancing ensures that we adhere to the principle of conservation of mass.

So, for \(PbCrO_4\), the dissociation is: \[PbCrO_4(s) \leftrightarrow Pb^{2+} (aq) + CrO_4^{2-} (aq) \]. This follows the same pattern, where lead chromate dissociates into lead and chromate ions.

solubility product expression

The solubility product constant, \(K_{\mathrm{sp}}\), is a useful way to express the solubility of sparsely soluble ionic compounds. The \(K_{\mathrm{sp}}\) value is derived from the concentrations of the ions in a saturated solution.The general format of the \(K_{\mathrm{sp}}\) expression depends on the balanced dissociation equation.

For the dissociation of \(Zn(\mathrm{OH})_2\): \[Zn(\mathrm{OH})_2 (s) \leftrightarrow Zn^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \], the \(K_{\mathrm{sp}}\) is: \[K_{\mathrm{sp}} = [Zn^{2+}][\mathrm{OH}^{-}]^{2} \].

Similarly, for \(Ca_3(PO_4)_2\): \[Ca_3(PO_4)_2(s) \leftrightarrow 3Ca^{2+}(aq) + 2PO_4^{3-}(aq) \], the \(K_{\mathrm{sp}}\) is: \[K_{\mathrm{sp}} = [Ca^{2+}]^{3}[PO_{4}^{3-}]^{2} \].

Each ion's concentration is raised to the power of its coefficient from the balanced equation. This multiplicativity and exponentiation reflect the stoichiometry of the dissociation reaction.

ionic compounds

Ionic compounds are composed of positively charged ions (cations) and negatively charged ions (anions). They form a crystal lattice structure where each cation is surrounded by anions and vice versa. When these compounds dissolve in water, they dissociate into their constituent ions.

For example, \(PbCrO_4\) consists of \(Pb^{2+}\) and \(CrO_4^{2-}\). Upon dissociation in water, these ions separate, forming a solution.

Understanding the behavior of ionic compounds is key in predicting the solubility of substances. Factors such as lattice energy and hydration energy play significant roles. The higher the lattice energy, the harder it is for the compound to dissociate.

Compounds such as \(Ca_3(PO_4)_2\) demonstrate the complexity of ionic interactions, dissociating into multiple ions that further affect the equilibrium in the solution.

chemical equilibrium

In the context of solubility, chemical equilibrium refers to a state where the rate of dissolution of a substance equals the rate of precipitation. For ionic compounds, this involves the dynamic balance between the solid form of the compound and its dissolved ions in solution.

Consider \(AgCl\): \[\text{AgCl} (s) \leftrightarrow \text{Ag}^{+} (aq) + \text{Cl}^{-} (aq) \]. At equilibrium, the concentrations of \(Ag^{+} \) and \(Cl^{-}\) remain constant. The solubility product constant, \(K_{\mathrm{sp}}\), quantifies this balance.

When the ion product (the product of ion concentrations) exceeds \(K_{\mathrm{sp}}\), the solution becomes supersaturated. This can lead to precipitation until equilibrium is restored.

Conversely, if the ion product is less than \(K_{\mathrm{sp}}\), the solution can still dissolve more of the ionic compound until equilibrium is achieved. This foundational concept helps predict and explain solubility phenomena in various chemical contexts.