Chapter 15, Problem 15.88QE

(a)

Interpretation Introduction

Interpretation:

The solution of $0.150M{\text{ NaHSO}}_4$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

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

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

Here,

$K_{\text{w}}$ denotes the ionic product of water.

$K_{\text{a}}$ denotes the ionization constant of the weak acid.

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

Answer to Problem 15.88QE

$0.150M{\text{ NaHSO}}_4$ is weakly basic and its $\text{pH}$ is estimated to be 8.05.

Explanation of Solution

Since ${\text{HSO}}_4{}^{-}$ is a stronger acid than water it donates its proton to water and therefore the ionization of ${\text{NaHSO}}_4$ salt occurs as follows:

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right){\text{+Na}}^{+}\left(\text{aq}\right)$

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

  $K_{\text{a}}=\frac{\left[{\text{SO}}_4^{2-}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{HSO}}_4{}^{-}\right]}$        (1)

Here,

$\left[{\text{SO}}_4^{2-}\right]$ denotes the concentration of ${\text{HSO}}_4{}^{-}$ ion.

$\left[{\text{H}}_3{\text{O}}^{+}\right]$ denotes the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions.

$\left[{\text{HSO}}_4{}^{-}\right]$ denotes the concentration of ${\text{HSO}}_4{}^{-}$ .

$K_{\text{a}}$ denotes ionization constant of ${\text{HSO}}_4{}^{-}$ ion.

Refer table 15.6 for the $K_{\text{b}}$ value of ${\text{HSO}}_4{}^{-}$.

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

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

Substitute $1.0\times {10}^{-2}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $K_{\text{a}}\left(1.0\times {10}^{-2}\right)=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{a}}=\frac{1\times {10}^{-14}}{1.0\times {10}^{-2}}\\ =1\times {10}^{-12}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{HSO}}_4{}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}& +& {\text{SO}}_4^{2-}\\ \text{Initial, }M& 0.15& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.15-x\right)& & & & x& & x\end{array}$

Substitute $0.15-x$ for $\left[{\text{HSO}}_4{}^{-}\right]$, $x$ for $\left[{\text{SO}}_4^{2-}\right]$, $x$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $1\times {10}^{-12}$ for $K_{\text{a}}$ in equation (1).

  $1\times {10}^{-12}=\frac{\left(x\right)\left(x\right)}{\left(0.15-x\right)}$

Assume $x<<0.15M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(1\times {10}^{-12}\right)\left(0.15\right)\\ x=\sqrt{0.15\times {10}^{-12}}\\ =0.38729\times {10}^{-6}\end{array}$

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

  $\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$        (2)

Substitute $0.38729\times {10}^{-6}M$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (2).

  $\begin{array}{c}\text{pH}=-\text{log}\left(0.38729\times {10}^{-6}\right)\\ =5.588\end{array}$

(b)

Interpretation Introduction

Interpretation:

The solution of $0.050M{\text{ Na}}_3{\text{PO}}_{\text{4}}$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

The ionization of a hypothetical weak acid $\left(\text{HA}\right)$ is given as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{A}}^{-}\left(\text{aq}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)$

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

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

Here,

$K_{\text{w}}$ denotes the ionic product of water.

$K_{\text{a}}$ denotes the ionization constant of weak acid.

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

Answer to Problem 15.88QE

$0.050M{\text{ Na}}_3{\text{PO}}_{\text{4}}$ is strongly basic and its $\text{pH}$ is estimated to be 11.3218.

Explanation of Solution

Since ${\text{Na}}_3{\text{PO}}_{\text{4}}$ is a salt formed from stronger acid ${\text{H}}_3{\text{PO}}_3$ hence the anion ${\text{PO}}_{\text{4}}^{\text{3}-}$ is a weak conjugate base so the ionization of ${\text{Na}}_3{\text{PO}}_{\text{4}}$ salt occurs as follows:

  ${\text{3Na}}^{+}\left(\text{aq}\right)+{\text{PO}}_{\text{4}}^{\text{3}-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons {\text{HPO}}_4^{-}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right){\text{+Na}}^{+}\left(\text{aq}\right)$

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

  $K_{\text{b}}=\frac{\left[{\text{HPO}}_4^{-}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{PO}}_{\text{4}}^{\text{3}-}\right]}$        (3)

Here,

$\left[{\text{PO}}_{\text{4}}^{\text{3}-}\right]$ denotes the concentration of ${\text{PO}}_{\text{4}}^{\text{3}-}$ ion.

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

$\left[{\text{HPO}}_4^{-}\right]$ denotes the concentration of ${\text{HPO}}_4^{-}$ .

$K_{\text{b}}$ denotes ionization constant of ${\text{PO}}_{\text{4}}^{\text{3}-}$ ion.

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

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

Substitute $2.2\times {10}^{-13}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (3).

  $\left(2.2\times {10}^{-13}\right)K_{\text{b}}=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{2.2\times {10}^{-13}}\\ =4.545\times {10}^{-2}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{PO}}_{\text{4}}^{\text{3}-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& {\text{HPO}}_4^{-}\\ \text{Initial, }M& 0.05& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.05-x\right)& & & & x& & x\end{array}$

Substitute $0.05-x$ for $\left[{\text{PO}}_{\text{4}}^{\text{3}-}\right]$, $x$ for $\left[{\text{HPO}}_4^{-}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $4.545\times {10}^{-2}$ for $K_{\text{a}}$ in equation (1).

  $4.545\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{\left(0.05-x\right)}$

Assume $x<<0.05M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(4.545\times {10}^{-2}\right)\left(0.05\right)\\ x=\sqrt{0.22725\times {10}^{-2}}\\ =4.767\times {10}^{-2}\end{array}$

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

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (2)

Substitute $4.767\times {10}^{-2}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (2).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(4.767\times {10}^{-2}\right)\\ =2.6782\end{array}$

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

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

Substitute 2.6782 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-2.6782\\ =11.3218\end{array}$

(c)

Interpretation Introduction

Interpretation:

Solution of $0.10M\text{ KBr}$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.88QE

$0.10M\text{ KBr}$ is weakly acidic and its $\text{pH}$ is estimated to be7.

Explanation of Solution

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

$\text{KBr}\left(\text{aq}\right)\rightleftharpoons {\text{K}}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

${\text{Br}}^{-}$ the conjugate base of strong acid $\text{HBr}$ hence it is weak and not able to accept a proton from water. Similarly, ${\text{K}}^{+}$ conjugate acid of strong base potassium hydroxide and hence it is also not strong enough to undergo any reaction with water. ${\text{K}}^{+}$ acts as a spectator ion and does not alter the $\text{pH}$. Therefore $\text{KBr}$ salt is neutral as there is no reaction with water and thus its $\text{pH}$ is estimated to be 7.