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Chapter 19: Problem 69

Write the ion-product expressions for (a) calcium chromate; (b) silver cyanide; (c) silver phosphate.

Short Answer

Expert verified
(a) K_{sp} = [Ca^{2+}][CrO4^{2-}](b) K_{sp} = [Ag^{+}][CN^{-}](c) K_{sp} = [Ag^{+}]^3[PO4^{3-}]

Step by step solution

01

Understanding Ion-Product Expressions

The ion-product (Ksp) expression is an equation that shows the relationship between the ion concentrations in a saturated solution of a sparingly soluble compound. It's derived from the equilibrium constant for the solubility of the compound.
02

- Write the Dissociation Equation for Calcium Chromate

The first compound is calcium chromate, which dissociates as follows: CaCrO4(s) ⇌ Ca^{2+}(aq) + CrO4^{2-}(aq)
03

- Write the Ion-Product Expression for Calcium Chromate

The ion-product expression is written using the concentrations of the dissociated ions: K_{sp} = [Ca^{2+}][CrO4^{2-}]
04

- Write the Dissociation Equation for Silver Cyanide

The second compound is silver cyanide, which dissociates as follows: AgCN(s) ⇌ Ag^{+}(aq) + CN^{-}(aq)
05

- Write the Ion-Product Expression for Silver Cyanide

The ion-product expression is written using the concentrations of the dissociated ions: K_{sp} = [Ag^{+}][CN^{-}]
06

- Write the Dissociation Equation for Silver Phosphate

The third compound is silver phosphate, which dissociates as follows: Ag3PO4(s) ⇌ 3Ag^{+}(aq) + PO4^{3-}(aq)
07

- Write the Ion-Product Expression for Silver Phosphate

The ion-product expression is written using the concentrations of the dissociated ions: K_{sp} = [Ag^{+}]^3[PO4^{3-}]

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Key Concepts

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

calcium chromate dissociation
Calcium chromate is a sparingly soluble ionic compound. When it dissociates in water, it splits into calcium ions \(Ca^{2+}\) and chromate ions \(CrO4^{2-}\). The dissociation of calcium chromate can be represented by the following chemical equation:

\[ CaCrO4(s) \rightleftharpoons Ca^{2+}(aq) + CrO4^{2-}(aq) \]

To find the ion-product expression, also known as the solubility product constant \(K_{sp}\), we use the concentrations of the ions at equilibrium:

\[ K_{sp} = [Ca^{2+}][CrO4^{2-}] \]

This equation shows us how calcium chromate breaks apart in water and how to calculate the solubility product constant from the concentrations of its ions.
silver cyanide dissociation
Silver cyanide also is a sparingly soluble compound. It dissociates into silver ions \(Ag^{+}\) and cyanide ions \(CN^{-}\) in an aqueous solution. The dissociation equation for silver cyanide is:

\[ AgCN(s) \rightleftharpoons Ag^{+}(aq) + CN^{-}(aq) \]

The ion-product expression for silver cyanide, reflecting the equilibrium concentrations of the ions, is:

\[ K_{sp} = [Ag^{+}][CN^{-}] \]

This \(K_{sp}\) expression helps us determine how silver cyanide dissolves in water. By knowing the concentrations of the dissociated ions, we can understand its solubility behavior.
silver phosphate dissociation
Silver phosphate is another sparingly soluble compound. When it dissolves in water, it breaks apart into three silver ions \(Ag^{+}\) and one phosphate ion \(PO4^{3-}\). The dissociation equation for silver phosphate is:

\[ Ag3PO4(s) \rightleftharpoons 3Ag^{+}(aq) + PO4^{3-}(aq) \]

To find the ion-product expression for silver phosphate, we consider the equilibrium concentrations of its ions:

\[ K_{sp} = [Ag^{+}]^3[PO4^{3-}] \]

This expression shows the relationship between the dissolution of silver phosphate and the equilibrium concentrations of the ions. By using this equation, we can predict and calculate the solubility of silver phosphate in an aqueous solution.

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Most popular questions from this chapter

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a buffer that consists of \(0.20 M \mathrm{HF}\) and \(0.25 M \mathrm{KF}\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HF}=6.8 \times 10^{-4}\right) ?\)

A \(50.0-\mathrm{mL}\) volume of \(0.50 M \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is mixed with \(125 \mathrm{~mL}\) of \(0.25 M \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) (a) If aqueous \(\mathrm{NaOH}\) is added, which ion precipitates first? (See Appendix C.) (b) Describe how the metal ions can be separated using \(\mathrm{NaOH}\). (c) Calculate the \(\left[\mathrm{OH}^{-}\right]\) that will accomplish the separation.

In a gaseous equilibrium, the reverse reaction occurs when \(Q_{\mathrm{c}}>K_{\mathrm{c}} .\) What occurs in aqueous solution when \(Q_{\mathrm{sp}}>K_{\mathrm{sp}} ?\)

Calculate the molar solubility of \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad\left(K_{\mathrm{sp}}=\right.\) \(1.75 \times 10^{-13}\) ) in \(0.13 M \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\)

The normal \(\mathrm{pH}\) of blood is \(7.40 \pm 0.05\) and is controlled in part by the \(\mathrm{H}_{2} \mathrm{CO}_{3} / \mathrm{HCO}_{3}^{-}\) buffer system. (a) Assuming that the \(K_{\mathrm{a}}\) value for carbonic acid at \(25^{\circ} \mathrm{C}\) applies to blood, what is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in normal blood? (b) In a condition called acidosis, the blood is too acidic. What is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in a patient whose blood \(\mathrm{pH}\) is \(7.20 ?\)
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