Chapter 15, Problem 15.92QE

(a)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ solution of $0.25M{\text{ KNO}}_2$ has to be calculated.

Concept Introduction:

The ionization of a hypothetical weak base is given as follows:

  $\text{B}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

Here,

$\left[\text{B}\right]$ denotes the concentration of hypothetical weak base.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{BH}}^{+}\right]$ denotes the concentration of ${\text{BH}}^{+}$ ions.

$K_{\text{b}}$ denotes ionization constant of weak base.

Answer to Problem 15.92QE

The value of $\text{pH}$ of $0.25M{\text{ KNO}}_2$ is 8.325.

Explanation of Solution

Refer table 15.6 for $K_{\text{a}}$ values of acids.

The relation among $K_{\text{a}}$, ${\text{K}}_{\text{b}}$ and $K_{\text{w}}$ is given as follows:

  $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}$        (1)

Substitute $5.6\times {10}^{-4}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $\left(5.6\times {10}^{-4}\right)K_{\text{b}}=1\times {10}^{-14}$

Rearrange to obtain the value of $K_{\text{b}}$.

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{5.6\times {10}^{-4}}\\ =1.7857\times {10}^{-11}\end{array}$

The ionization of ${\text{KNO}}_2$ salt occurs as follows:

  ${\text{K}}^{+}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{HNO}}_2\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{K}}^{+}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NO}}_2^{-}\right]}$        (2)

Here,

$\left[{\text{HNO}}_2\right]$ denotes the concentration of ${\text{HNO}}_2$.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{NO}}_2^{-}\right]$ denotes the concentration of ${\text{NO}}_2^{-}$ .

$K_{\text{b}}$ denotes equilibrium constant for this ionization.

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{NO}}_2^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& {\text{HNO}}_2\\ \text{Initial, }M& 0.25& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.25-x\right)& & & & x& & x\end{array}$

Substitute $0.25-x$ for $\left[{\text{NO}}_2^{-}\right]$, $x$ for $\left[{\text{HNO}}_2\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $1.7857\times {10}^{-11}$ for $K_{\text{b}}$ in equation (2).

  $1.7857\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}$

Assume $x<<0.25M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(1.7857\times {10}^{-11}\right)\left(0.25\right)\\ x=\sqrt{4.46425\times {10}^{-12}}\\ =2.1128\times {10}^{-6}\end{array}$

The formula to calculate $\text{pOH}$ is given as follows:

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (3)

Substitute $2.1128\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(2.1128\times {10}^{-6}\right)\\ =5.675\end{array}$

The formula to calculate $\text{pH}$ from $\text{pOH}$ is given as follows:

  $\text{pH}=14-\text{pOH}$        (4)

Substitute 5.675 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-5.675\\ =8.325\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ solution $0.50M$ of sodium formate has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.92QE

The value of $\text{pH}$ the solution $0.50M$ of sodium formate is 8.7219.

Explanation of Solution

The ionization of sodium formate salt occurs as follows:

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{HCOO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons \text{HCOOH}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{Na}}^{+}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[\text{HCOOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HCOO}}^{-}\right]}$        (5)

Here,

$\left[\text{HCOOH}\right]$ denotes the concentration of $\text{HCOOH}$.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{HCOO}}^{-}\right]$ denotes the concentration of ${\text{HCOO}}^{-}$ .

$K_{\text{b}}$ denotes equilibrium constant for this ionization.

Substitute $1.8\times {10}^{-4}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $\left(1.8\times {10}^{-4}\right)K_{\text{b}}=1\times {10}^{-14}$

Rearrange to obtain the value of $K_{\text{b}}$.

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-4}}\\ =5.55\times {10}^{-11}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{HCOO}}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& \text{HCOOH}\\ \text{Initial, }M& 0.50& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.50-x\right)& & & & x& & x\end{array}$

Substitute $0.50-x$ for $\left[{\text{HCOO}}^{-}\right]$, $x$ for $\left[\text{HCOOH}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $5.55\times {10}^{-11}$ for $K_{\text{b}}$ in equation (5).

  $5.55\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.50-x\right)}$

Assume $x<<0.50M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(5.55\times {10}^{-11}\right)\left(0.50\right)\\ x=\sqrt{27.775\times {10}^{-12}}\\ =5.27\times {10}^{-6}\end{array}$

Substitute $5.27\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(5.27\times {10}^{-6}\right)\\ =5.2781\end{array}$

Substitute 5.2781 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-5.2781\\ =8.7219\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ of the solution $0.015M$ of sodium fluoride has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.92QE

The value of $\text{pH}$ of the solution $0.015M$ of sodium fluoride is 8.189.

Explanation of Solution

The ionization of $\text{NaF}$ salt occurs as follows:

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{F}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons \text{HF}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right){\text{+Na}}^{+}\left(\text{aq}\right)$

The expression of ${\text{K}}_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[\text{HF}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{F}}^{-}\right]}$        (6)

Here,

$\left[{\text{F}}^{-}\right]$ denotes the concentration of fluoride ion.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[\text{HF}\right]$ denotes the concentration of $\text{HF}$ .

$K_{\text{b}}$ denotes ionization constant of fluoride ion.

Substitute $6.3\times {10}^{-5}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $\left(6.3\times {10}^{-5}\right)K_{\text{b}}=1\times {10}^{-14}$

Rearrange to obtain the value of ${\text{K}}_{\text{b}}$.

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{6.3\times {10}^{-5}}\\ =1.587\times {10}^{-10}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{F}}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & \text{HF}& +& {\text{OH}}^{-}\\ \text{Initial, }M& 0.015& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.015-x\right)& & & & x& & x\end{array}$

Substitute $0.015-x$ for $\left[{\text{F}}^{-}\right]$, $x$ for $\left[\text{HF}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $1.587\times {10}^{-10}$ for $K_{\text{b}}$ in equation (6).

  $1.587\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.015-x\right)}$

Assume $x<<0.015M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(1.587\times {10}^{-10}\right)\left(0.015\right)\\ x=\sqrt{0.023805\times {10}^{-10}}\\ =1.5428\times {10}^{-6}\end{array}$

Substitute $1.5428\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(1.5428\times {10}^{-6}\right)\\ =5.81169\end{array}$

Substitute 5.81169 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-5.81169\\ =8.18831\\ \approx 8.189\end{array}$