Chapter 14, Problem 6QAP

Calculate K for the reactions in Question 2.

Interpretation Introduction

Interpretation:

The value of K needs to be calculated for the net ionic equation for reaction between aqueous solution of sodium acetate and nitric acid.

Concept Introduction :

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

  $\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

Square brackets represent the concentration.

Answer to Problem 6QAP

  $5.6\times {10}^4$

Explanation of Solution

The net ionic reaction between aqueous solution of sodium acetate and nitric acid is as follows:

  $H^{+}\left(aq\right)+C_2{H}_3{O}_2{}^{-}\left(aq\right)\to H{C}_2{H}_3{O}_2\left(aq\right)$

Since, for

  $H{C}_2{H}_3{O}_2\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C_2{H}_3{O}_2{}^{-}\left(aq\right)$ the value of acid dissociation constant, K_a is $1.8\times {10}^{-5}$ .

As the required reaction is inverse of the dissociation reaction of $H{C}_2{H}_3{O}_2$ so, the expression of K for this reaction is as follows:

  $K=\frac{1}{K_a(H{C}_2{H}_3{O}_2)}$

Substitute the value in the above expression as follows:

  $K=\frac{1}{K_{a}H{C}_2{H}_3{O}_2}$

  $\begin{array}{l}=\frac{1}{1.8\times {10}^{-5}}\hfill \\ =5.6\times {10}^4\hfill \end{array}$

Therefore, the value of K is $5.6\times {10}^4$ .

Interpretation Introduction

Interpretation:

The value of K needs to be calculated for the net ionic equation for reaction between aqueous solution of hydrobromic acid and strontium hydroxide.

Concept Introduction :

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

  $\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

Square brackets represent the concentration.

Answer to Problem 6QAP

  $4.8\times {10}^3$

Explanation of Solution

The net ionic equation for the reaction between hydrobromic acid and strontium hydroxide is as follows:

  $H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to H_{2}O\left(l\right)$

Since, for

  $H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$ the value of acid dissociation constant, K_w is $1.0\times {10}^{-14}$ .

As the required reaction is inverse of the dissociation reaction of $H_{2}O$ so, the expression of K for this reaction is as follows:

  $K=\frac{1}{K_w}$

Substituting the value in the above expression as below:

  $K=\frac{1}{K_w}$

  $\begin{array}{l}=\frac{1}{1.0\times {10}^{-14}}\hfill \\ =1.0\times {10}^{14}\hfill \\ \hfill \end{array}$

Therefore, the value of K is $1.0\times {10}^{-14}$ .

Interpretation Introduction

Interpretation:

The value of K needs to be calculated for the net ionic equation for reaction between aqueous solution of hypochlorous acid and sodium cyanide.

Concept Introduction :

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

  $\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

Square brackets represent the concentration.

Answer to Problem 6QAP

48

Explanation of Solution

The net ionic equation for the reaction between Hypochlorous acid and sodium cyanide is as below:

  $\begin{array}{l}HOCl\left(aq\right)+C{N}^{-}\left(aq\right)\to HCN\left(aq\right)+OC{l}^{-}\left(aq\right)\hfill \\ HOCl\left(aq\right)\to H^{+}\left(aq\right)+OC{l}^{-}\left(aq\right)\hfill \\ H^{+}\left(aq\right)+C{N}^{-}\left(aq\right)\to HCN\left(aq\right)\hfill \end{array}$

Since, for

  $HCN\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{N}^{-}\left(aq\right)$ the value of acid dissociation constant, K_a is $5.8\times {10}^{-10}$ . And for $HOCl\left(aq\right)\to H^{+}\left(aq\right)+OC{l}^{-}\left(aq\right)$ is $2.8\times {10}^{-8}$ .

As the required reaction of HCN is inverse of the dissociation reaction of HCN so, the expression of K for this reaction is as follows:

  ${\text{K = K}}_{\text{a}}\text{(HOCl)}\frac{\text{1}}{{\text{ K}}_{\text{a}}\text{(HCN)}}$

Substituting the value in the above expression as follows:

  ${\text{K = K}}_{\text{a}}\text{(HOCl)}\frac{\text{1}}{{\text{ K}}_{\text{a}}\text{(HCN)}}$

  = $2.8\times {10}^{-8}\times \left(\frac{1}{5.8\times {10}^{-10}}\right)$

The K value is 48.

Interpretation Introduction

Interpretation:

The value of K needs to be calculated for the net ionic equation for reaction between aqueous solution of sodium hydroxide and nitrous acid.

Concept Introduction :

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

  $\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

Square brackets represent the concentration.

Answer to Problem 6QAP

  $6.0\times {10}^{10}$

Explanation of Solution

The net ionic equation for the reaction between sodium hydroxide and nitrous acid is as below:

  $\begin{array}{l}HN{O}_2\left(aq\right)+O{H}^{-}\left(aq\right)\to N{O}_2{}^{-}\left(aq\right)+H_{2}O\left(l\right)\hfill \\ \hfill \end{array}$

The expression of K for this reaction is:

  $\text{K =}\frac{\text{1}}{{\text{ K}}_{\text{b}}{\text{(NO}}_{{}_2}^{-}\text{)}}$

The K_b value for NO₂- is $1.7\times {10}^{-11}$ , thus, substitute this value in the above expression as follows:

  $\begin{array}{l}\text{K =}\frac{\text{1}}{{\text{ K}}_{\text{b}}{\text{(NO}}_{{}_2}^{-}\text{)}}\\ =\frac{1}{1.7\times {10}^{-11}}\\ =6.0\times {10}^{10}\end{array}$

The K value is $6.0\times {10}^{10}$