Chapter 13, Problem 67QAP

Using the equilibrium constants listed in Table 13.2, arrange the following 0.1 M aqueous solutions in order of increasing pH (from lowest to highest).

(a) NaNO₂

(b) HCl

(c) NaF

(d) Zn(H₂O)₃(OH)(NO₃)

Interpretation Introduction

Interpretation: The order of increasing pH value needs to be determined.

Concept Introduction: The dissociation reaction of a weak base is represented as follows:

  $\text{BOH}\rightleftarrows {\text{B}}^{+}+{\text{OH}}^{-}$

The expression for the base dissociation constant will be as follows:

  $K_b=\frac{\left[{\text{B}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{BOH}\right]}$

Here, $\left[{\text{B}}^{+}\right]$ is equilibrium concentration of conjugate acid, $\left[{\text{OH}}^{-}\right]$ is equilibrium concentration of hydroxide ion and $\left[\text{BOH}\right]$ is the equilibrium concentration of weak base.

From hydroxide ion concentration, pOH of the solution can be calculated as follows:

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

From pOH, pH of the solution can be calculated as follows:

  $\text{pH}=14-\text{pOH}$

Answer to Problem 67QAP

0.1 M $\text{HCl}$<0.1 M $\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)$<0.1 M $\text{NaF}$<0.1 M ${\text{NaNO}}_{\text{2}}$

Explanation of Solution

The given aqueous solutions are as follows:

M ${\text{NaNO}}_{\text{2}}$ , 0.1 M $\text{HCl}$ , 0.1 M $\text{NaF}$ , 0.1 M $\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)$

Here, HCl is a strong acid thus, it completely dissociates in the solution.

From the table 13.2, the equilibrium constants for ${\text{NO}}_2^{-}$ is $1.7\times {10}^{-11}$ , ${\text{F}}^{-}$ is $1.4\times {10}^{-11}$ and $\mathrm{Zn}{\left(H_{2}O\right)}_3{\left(OH\right)}^{+}$ is $3.0\times {10}^{-5}$ .

(a)

The dissociation reaction of ${\text{NaNO}}_{\text{2}}$ is represented as follows:

  $\begin{array}{l}{\text{NaNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{HNO}}_{\text{2}}+\text{NaOH}\\ \text{0}\text{.1 - - -}\\ \text{0}\text{.1-x - x x}\end{array}$

The equilibrium constant can be calculated as follows:

  $K_b=\frac{\left[{\text{HNO}}_{\text{2}}\right]\left[\text{NaOH}\right]}{\left[{\text{NaNO}}_{\text{2}}\right]}$

Putting the values,

  $1.7\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.1-x\right)}$

Thus,

  $x=-1.304\times {10}^{-6},+1.304\times {10}^{-6}$

Therefore, the value of $\left[\text{NaOH}\right]$ is $1.304\times {10}^{-6}$ .

Thus,

  $\left[{\text{OH}}^{-}\right]=1.304\times {10}^{-6}$

The pOH of the solution can be calculated as follows:

  $\text{pOH=}-\log \left[{\mathrm{OH}}^{-}\right]$

Putting the values,

  $\text{pOH=}-\log \left(1.304\times {10}^{-6}\right)=5.88$

The pH of the solution can be calculated as follows:

  $\text{pH}=14-\text{pOH}$

Putting the values,

  $\text{pH}=14-\text{5}\text{.88=8}\text{.12}$

(b)

HCl is a strong acid, the dissociation can be represented as follows:

  $\text{HCl}\to {\text{H}}^{+}+{\text{Cl}}^{-}$

Since, molarity of HCl is 0.1 M thus, concentration of hydrogen ion is 0.1 M.

The pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Putting the value,

  $\text{pH}=-\log \left(0.1\right)=1$

(c)

0.1 M $\text{NaF}$ solution.

The dissociation reaction can be represented as follows:

  $\begin{array}{l}{\text{NaF+H}}_{\text{2}}\text{O}\rightleftharpoons \text{HF+NaOH}\\ \text{0}\text{.1 - - -}\\ \text{0}\text{.1-x - x x}\end{array}$

The equilibrium constant for the reaction will be:

  $k_b=\frac{\left[\text{HF}\right]\left[\text{NaOH}\right]}{\left[\text{NaF}\right]}$

Putting the values,

  $1.4\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.1-x\right)}$

Thus,

  $x=-1.18\times {10}^{-6},+1.18\times {10}^{-6}$

The value of x cannot be negative thus, it is equal to $1.18\times {10}^{-6}$ .

Thus,

  $\left[\text{NaOH}\right]=\left[{\text{OH}}^{-}\right]=1.18\times {10}^{-6}$

The pOH of the solution can be calculated as follows:

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Putting the value,

  $\text{pOH}=-\log \left(1.18\times {10}^{-6}\right)=5.93$

The pH of the solution can be calculated as follows:

  $\text{pH}=14-\text{pOH}$

Putting the value,

  $\text{pH}=14-\text{5}\text{.93=8}\text{.07}$

(d)

0.1 M $\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)$

The decomposition reaction can be represented as follows:

  $\begin{array}{l}\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \mathrm{Zn}{\left(H_{2}O\right)}_3{\left(OH\right)}_2+{\text{HNO}}_3\\ 0.1\text{ - - -}\\ \text{0}\text{.1-}x\text{ - x x}\end{array}$

The equilibrium constant for the reaction will be:

  $k_b=\frac{\left[\mathrm{Zn}{\left(H_{2}O\right)}_3{\left(OH\right)}_2\right]\left[{\text{HNO}}_3\right]}{\left[\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)\right]}$

Putting the values,

  $3.0\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.1-x\right)}$

Thus,

  $x=-0.00174,+0.00712$

Since, the value of x cannot be negative thus, it is equal to $0.00712$ .

  $\left[{\text{HNO}}_3\right]=0.00712$

Since, ${\text{HNO}}_3$ is a strong acid thus, $\left[{\text{H}}^{+}\right]=0.00712$

The pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Putting the values,

  $\text{pH}=-\log \left(0.00712\right)=2.15$

The order of increasing pH value can be represented as follows:

0.1 M $\text{HCl}$<0.1 M $\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)$<0.1 M $\text{NaF}$<0.1 M ${\text{NaNO}}_{\text{2}}$

Conclusion

Thus, the order of increasing pH value can be represented as follows:

0.1 M $\text{HCl}$<0.1 M $\mathrm{Zn}{\left(H_{2}O\right)}_3\left(OH\right)\left(N{O}_3\right)$<0.1 M $\text{NaF}$<0.1 M ${\text{NaNO}}_{\text{2}}$