Chapter 18, Problem 18.95P

(a)

Interpretation Introduction

Interpretation:

The higher $\text{pH}$ of the solution has to be identified from the given pairs, $\text{ 0}{\text{.1 M NiCl}}_{\text{2}}\text{ or 0}\text{.1 M NaCl}$.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

The reaction of any base B with water is written as,

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

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

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

An equilibrium constant $\left(K\right)$ with subscript $\text{b}$ indicates that it is an equilibrium constant of the base in water.

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_a$ value is calculating by using following formula,

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

Acids strength is mainly depending on the dissociation of ions, strong acids dissociates completely and weak acid dissociate slightly.

The acid strength is depending on the ${\text{K}}_{\text{a}}$ value, if the ${\text{K}}_{\text{a}}$ value is larger the stronger the acid and it is lower $\text{pH}$. if the ${\text{K}}_{\text{a}}$ value is small the weaker the acid and it is higher $\text{pH}$.

Explanation of Solution

The $\text{pH}$ of the $\text{0}\text{.1 M NaCl}$ is seven because it is formed from strong hydrochloric acid and strong sodium hydroxide.

The $\text{pH}$ of the $\text{0}{\text{.1 M NiCl}}_{\text{2}}$ is given below,

The ${\text{K}}_{\text{a}}$ value of ${\text{NiCl}}_{\text{2}}$ is ${\text{1×10}}^{\text{–10}}$

The ${\text{K}}_{\text{b}}$ is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{1.0\times {10}^{-10}}\\ {\text{K}}_{\text{b}}\text{ = 1}\text{.0}\times {10}^{-4}\end{array}$

The hydrolysis equation is given below,

  ${\text{2Cl}}^{-}\text{+}2{\text{H}}_{\text{2}}\text{O}\underset{}{\rightleftharpoons }2\text{HCl (aq) }\text{+}{\text{ 2OH}}^{-}$

Therefore,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[{\text{Cl}}^{\text{-}}\right]}^2}\\ therefore,\\ 1\times {10}^{-4}\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[0.1\right]}^2}\\ x^4=1\times {10}^{-6}\\ x=3.16\times {10}^{-2}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}{\text{(OH}}^{-1}\text{)}\\ \text{pOH}\text{=}\text{-}\text{log}\text{(}3.16\times {10}^{-2}\text{)}\\ \text{pOH=}1.5\end{array}$

$\text{pH}$ calculation is given below,

  $\begin{array}{l}\text{pH}+\text{pOH}\text{=}14\\ \text{pH}+1.5\text{=}14\\ \text{pH}=12.5\end{array}$

Therefore, $\text{0}{\text{.1 M NiCl}}_{\text{2}}$ is higher $\text{pH}$, less acidic.

(b)

Interpretation Introduction

Interpretation:

The higher $\text{pH}$ of the solution has to be identified in the given $\text{ 0}{\text{.1 M Sn(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{ or 0}{\text{.1 M Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ pairs.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

The reaction of any base B with water is written as,

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

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

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

An equilibrium constant $\left(K\right)$ with subscript $\text{b}$ indicates that it is an equilibrium constant of the base in water.

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_a$ value is calculating by using following formula,

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

Acids strength is mainly depending on the dissociation of ions, strong acids dissociates completely and weak acid dissociate slightly.

The acid strength is depending on the ${\text{K}}_{\text{a}}$ value, if the ${\text{K}}_{\text{a}}$ value is larger the stronger the acid and it is lower $\text{pH}$. if the ${\text{K}}_{\text{a}}$ value is small the weaker the acid and it is higher $\text{pH}$.

Explanation of Solution

The $\text{pH}$ of the ${\text{Sn(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is given below,

The ${\text{K}}_{\text{a}}$ value of ${\text{Sn(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is ${\text{4×10}}^{\text{–4}}$

The ${\text{K}}_{\text{b}}$ is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{{\text{4×10}}^{\text{–4}}}\\ {\text{K}}_{\text{b}}\text{ = 2}\text{.5}\times {10}^{-11}\end{array}$

The hydrolysis equation is given below,

  ${\text{2NO}}_{\text{3}}{}^{-}\text{+}2{\text{H}}_{\text{2}}\text{O}\underset{}{\rightleftharpoons }2{\text{HNO}}_3\text{ (aq) }\text{+}{\text{ 2OH}}^{-}$

Therefore,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[N{O}_3{}^{\text{-}}\right]}^2}\\ therefore,\\ 2.5\times {10}^{-11}\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[0.1\right]}^2}\\ x^4=2.5\times {10}^{-13}\\ x=7.07\times {10}^{-4}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}{\text{(OH}}^{-}\text{)}\\ \text{pOH}\text{=}\text{-}\text{log}\text{(}7.07\times {10}^{-4}\text{)}\\ \text{pOH=}3.15\end{array}$

$\text{pH}$ calculation is given below,

  $\begin{array}{l}\text{pH}+\text{pOH}\text{=}14\\ \text{pH}+3.15\text{=}14\\ \text{pH}=10.84\end{array}$

Therefore, ${\text{Sn(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is 11.34.

(ii) The $\text{pH}$ of the ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is given below,

The ${\text{K}}_{\text{a}}$ value of ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is ${\text{2×10}}^{\text{–10}}$

The ${\text{K}}_{\text{b}}$ of acetate ion is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{{\text{2×10}}^{\text{–10}}}\\ {\text{K}}_{\text{b}}\text{ = 5}\times {10}^{-5}\end{array}$

The hydrolysis equation is given below,

  ${\text{2NO}}_{\text{3}}{}^{-}\text{+}2{\text{H}}_{\text{2}}\text{O}\underset{}{\rightleftharpoons }2{\text{HNO}}_3\text{ (aq) }\text{+}{\text{ 2OH}}^{-}$

Therefore,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[N{O}_3{}^{\text{-}}\right]}^2}\\ therefore,\\ 5\times {10}^{-5}\text{ =}\frac{{\left[x\right]}^2{\left[x\right]}^2}{{\left[0.1\right]}^2}\\ x^4=5\times {10}^{-7}\\ x=2.66\times {10}^{-2}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}{\text{(OH}}^{-1}\text{)}\\ \text{pOH}\text{=}\text{-}\text{log}\text{(}2.66\times {10}^{-2}\text{)}\\ \text{pOH=}1.60\end{array}$

$\text{pH}$ calculation is given below,

  $\begin{array}{l}\text{pH}+\text{pOH}\text{=}14\\ \text{pH}+1.60\text{=}14\\ \text{pH}=12.39\end{array}$

Therefore, ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is 12.39, hence ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is more basic.