Chapter 18, Problem 1E

Write $K_{10}$ expressions for the following equilibria. For example, for the reaction $\text{AgCl}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{\text{4}}\left(\text{aq}\right)+{\text{Cl}}^{\text{-}}\left(\text{aq}\right)$ , $K_{10}=\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]$ .

1. ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)\rightleftharpoons 2{\text{Ag}}^{\text{4}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}^{\text{2-}}\left(\text{aq}\right)$
2. $Ra{\left({\text{la}}_2\right)}_2\left(\text{s}\right)\rightleftharpoons {\text{Ra}}^{\text{2+}}\left(\text{aq}\right)+2{\text{lO}}_2^{-}\left(\text{aq}\right)$
3. ${\text{Ni}}_{\text{2}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons 3\text{Ni}\left(\text{aq}\right)+2{\text{PO}}_{\text{4}}^{\text{2-}}$
4. ${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{PuO}}_{\text{2}}^{\text{2+}}\left(\text{aq}\right)+{\text{CO}}_{\text{2}}^{\text{2-}}\left(\text{aq}\right)$

Interpretation Introduction

(a)

Interpretation:

The K_sp expression for the following equilibria should be determined:

${\text{Ag}}_{\text{2}}{\text{SO}}_4\text{(s)}\rightleftharpoons {\text{2Ag}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Concept introduction:

A constant that defines the solubility equilibrium that exists between a compound in its solid state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\\ {\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}}^{\text{]}}\end{array}$

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 1E

${\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{2+}}{\text{]}}^2{\text{[SO}}_4^{2-}\text{]}$

Explanation of Solution

The dissociation reaction of ${\text{Ag}}_{\text{2}}{\text{SO}}_4\text{(s)}$ is:

${\text{Ag}}_{\text{2}}{\text{SO}}_4\text{(s)}\rightleftharpoons {\text{2Ag}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Thus, the products formed after dissociation of ${\text{Ag}}_{\text{2}}{\text{SO}}_4\text{(s)}$ are ${\text{Ag}}^{\text{+}}\text{(aq)}$ and ${\text{SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$.

So, the expression for K_sp of ${\text{Ag}}_{\text{2}}{\text{SO}}_4\text{(s)}$ is:

${\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{2+}}{\text{]}}^2{\text{[SO}}_4^{2-}\text{]}$

Interpretation Introduction

(b)

Interpretation:

The K_sp expression for the following equilibria should be determined:

${\text{Ra(IO}}_{\text{3}}\text{)}\text{(s)}\rightleftharpoons {\text{Ra}}^{\text{2+}}{\text{(aq)+2IO}}_{\text{3}}^{\text{-}}\text{(aq)}$

Concept introduction:

A constant that defines the solubility equilibrium that exists between a compound in its solid state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\\ {\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}}^{\text{]}}\end{array}$

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 1E

${\text{K}}_{\text{sp}}{\text{= [Ra}}^{\text{2+}}{\text{][IO}}_3^{-}{\text{]}}^2$

Explanation of Solution

The dissociation reaction of ${\text{Ra(IO}}_{\text{3}}\text{)}\text{(s)}$ is:

${\text{Ra(IO}}_{\text{3}}\text{)}\text{(s)}\rightleftharpoons {\text{Ra}}^{\text{2+}}{\text{(aq)+2IO}}_{\text{3}}^{\text{-}}\text{(aq)}$

Thus, the products formed after dissociation of ${\text{Ra(IO}}_{\text{3}}\text{)}\text{(s)}$ are ${\text{Ra}}^{\text{2+}}\text{(aq)}$ and ${\text{IO}}_{\text{3}}^{\text{-}}\text{(aq)}$.

So, the expression for K_sp of ${\text{Ra(IO}}_{\text{3}}\text{)}\text{(s)}$ is:

${\text{K}}_{\text{sp}}{\text{= [Ra}}^{\text{2+}}{\text{][IO}}_3^{-}{\text{]}}^2$

Interpretation Introduction

(c)

Interpretation:

The K_sp expression for the following equilibria should be determined:

${\text{Ni}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{(s)}\rightleftharpoons {\text{3Ni}}^{\text{2+}}{\text{(aq)+2PO}}_{\text{4}}^{\text{3-}}\text{(aq)}$

Concept introduction:

A constant that defines the solubility equilibrium that exists between a compound in its solid state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\\ {\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}}^{\text{]}}\end{array}$

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 1E

${\text{K}}_{\text{sp}}{\text{= [Ni}}^{\text{2+}}{\text{]}}^3{{\text{[PO}}_4^{3-}\text{]}}^2$

Explanation of Solution

The dissociation reaction of ${\text{Ni}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{(s)}$ is:

${\text{Ni}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{(s)}\rightleftharpoons {\text{3Ni}}^{\text{2+}}{\text{(aq)+2PO}}_{\text{4}}^{\text{3-}}\text{(aq)}$

Thus, the products formed after dissociation of ${\text{Ni}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{(s)}$ are ${\text{Ni}}^{\text{2+}}\text{(aq)}$ and ${\text{PO}}_{\text{4}}^{\text{3-}}\text{(aq)}$.

So, the expression for K_sp of ${\text{Ni}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{(s)}$ is:

${\text{K}}_{\text{sp}}{\text{= [Ni}}^{\text{2+}}{\text{]}}^3{{\text{[PO}}_4^{3-}\text{]}}^2$

Interpretation Introduction

(d)

Interpretation:

The K_sp expression for the following equilibria should be determined:

${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s)}\rightleftharpoons {\text{PuO}}_{\text{2}}^{\text{2-}}{\text{(aq)+CO}}_{\text{3}}^{\text{2-}}\text{(aq)}$

Concept introduction:

A constant that defines the solubility equilibrium that exists between a compound in its solid state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\\ {\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}}^{\text{]}}\end{array}$

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 1E

${\text{K}}_{\text{sp}}{\text{= [PuO}}_{\text{2}}^{\text{2-}}{\text{][CO}}_{\text{3}}^{\text{2-}}\text{]}$

Explanation of Solution

The dissociation reaction of ${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s)}$ is:

${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s)}\rightleftharpoons {\text{PuO}}_{\text{2}}^{\text{2-}}{\text{(aq)+CO}}_{\text{3}}^{\text{2-}}\text{(aq)}$

Thus, the products formed after dissociation of ${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s)}$ are ${\text{PuO}}_{\text{2}}^{\text{2-}}\text{(aq)}$ and ${\text{CO}}_{\text{3}}^{\text{2-}}\text{(aq)}$.

So, the expression for K_sp of ${\text{PuO}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s)}$ is:

${\text{K}}_{\text{sp}}{\text{= [PuO}}_{\text{2}}^{\text{2-}}{\text{][CO}}_{\text{3}}^{\text{2-}}\text{]}$