Chapter 6, Problem 6D.14E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous sodium hydroxide solution has to be calculated.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base -10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6D.14E

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous sodium hydroxide solution is 9.0.

Explanation of Solution

The ${\text{OH}}^{-}$ ion concentration is equal to the concentration of sodium hydroxide, because it is a strong base therefore, it dissociates completely and the following equation is given below.

  ${\text{NaOH+H}}_{\text{2}}\text{O}\to {\text{Na}}^{\text{+}}{\text{+OH}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

If consider, $\text{1}{\text{.0×10}}^{\text{-5}}\text{M NaOH}$=$\text{1}{\text{.0×10}}^{\text{-5}}{\text{M [OH}}^{-}\text{]}$, using the information we can calculate $\text{pOH}$ of the solution.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}\\ \\ \text{=}{\text{-log(1×10}}^{\text{-5}}\text{)}\\ \\ \text{=}\text{5}\text{.0}\end{array}$

Therefore, the $\text{pOH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous sodium hydroxide solution is 5.0.  From this value, the pH of sodium hydroxide solution is calculated.

  $\begin{array}{l}\text{pH+pOH=14}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-5}\text{.0}\\ \\ \text{=}\text{9}\text{.0}\end{array}$

Therefore, the $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous sodium hydroxide solution is 9.0

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.20}{\text{M NaNO}}_{\text{2}}$ aqueous solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.14E

The $\text{pH}$ of $\text{0}\text{.20}{\text{M NaNO}}_{\text{2}}$ aqueous solution is 8.33.

Explanation of Solution

The equilibrium reaction of sodium nitrite is given below.

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

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[HNO}}_{\text{2}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[NO}}_{\text{2}}{}^{\text{-}}\text{]}}$

	${\text{NO}}_2{}^{-}$	${\text{HNO}}_{\text{2}}$	${\text{OH}}^{-}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.20-x}}$

Nitrous acid ${\text{K}}_{\text{a}}$ value is $\text{4}{\text{.3×10}}^{\text{-4}}$ which is given in table 6C.1 and using this value ${\text{K}}_{\text{b}}$ is calculated and the respective formula is given below,

$\begin{array}{l}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{= K}}_{\text{w}}\\ \\ {\text{ K}}_{\text{b}}\text{=}\text{}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ \text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{4}{\text{.3×10}}^{\text{-4}}}\\ \\ \text{=}\text{2}{\text{.33×10}}^{\text{-11}}\end{array}$

Therefore, the ${\text{K}}_{\text{b}}$ value of sodium nitrite is $\text{2}{\text{.33×10}}^{\text{-11}}$.

The obtained ${\text{K}}_{\text{b}}$ value is substitute in below the equilibrium reaction.

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{= }\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\\ \\ \text{2}{\text{.33×10}}^{\text{-11}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\end{array}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{2}{\text{.33×10}}^{\text{-11}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(2}{\text{.33×10}}^{\text{-11}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.20×(2}{\text{.33×10}}^{\text{-11}}\text{)}}\\ \\ \text{=}2.16\times {10}^{-6}\end{array}$

Therefore, the pOH concentration of sodium nitrite is $\text{2}{\text{.16×10}}^{\text{-6}}$.

Now, the $\text{pOH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}\\ \\ \text{=}\text{-log(2}{\text{.16×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.67}\end{array}$

Therefore, the $\text{pOH}$ of $\text{0}\text{.20}\text{M}$ aqueous sodium nitrite solution is 5.67.  From this value, the pH of sodium nitrite solution is calculated.

  $\begin{array}{l}\text{pH+pOH=14}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-5}\text{.67}\\ \\ \text{=}8.33\end{array}$

Therefore, the $\text{pH}$ of 0.20 M aqueous sodium nitrite solution is 8.33.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of aqueous ammonia solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.14E

The $\text{pH}$ of $\text{0}\text{.20}{\text{M NH}}_3$ aqueous solution is 11.28.

Explanation of Solution

Ammonia is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{NH}}_{\text{3(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}$

	${\text{NH}}_3$	${\text{NH}}_4{}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.20-x}}$

Ammonia ${\text{K}}_{\text{b}}$ value is $\text{1}{\text{.8×10}}^{\text{-5}}$ which is given in table 6C.2 and, the obtained ${\text{K}}_{\text{b}}$ value is substitute in above equation.

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{= }\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\\ \\ \text{1}{\text{.8×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\end{array}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(1}{\text{.8×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.20×(1}{\text{.8×10}}^{\text{-5}}\text{)}}\\ \\ \text{=}1.9\times {10}^{-3}\end{array}$

Therefore, the pOH concentration of ammonia is $\text{1}{\text{.9×10}}^{\text{-3}}$.

Now, the $\text{pOH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}\\ \\ \text{=}\text{-log(1}{\text{.9×10}}^{\text{-3}}\text{)}\\ \\ \text{=}2.72\end{array}$

Therefore, the $\text{pOH}$ of $\text{0}\text{.20}\text{M}$ aqueous ammonia solution is 2.72.  From this value, the pH of ammonia solution is calculated.

  $\begin{array}{l}\text{pH+pOH=14}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-2}\text{.72}\\ \\ \text{=}11.28\end{array}$

Therefore, $\text{pH}$ of 0.20 M aqueous ammonia solution is 11.28.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.20}\text{M NaCN}$ aqueous solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.14E

The $\text{pH}$ of $\text{0}\text{.20}\text{M NaCN}$ aqueous solution is 9.31.

Explanation of Solution

Sodium cyanide is a strong base when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{CN}}^{-}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HCN}}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{[HCN]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[CN}}^{-}\text{]}}$

	${\text{CN}}^{-}$	$\text{HCN}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.20-x}}$

Hydrocyanic acid ${\text{K}}_{\text{a}}$ value is $4.9\times {10}^{-10}$ which is given in table 6C.1 and using this value ${\text{K}}_{\text{b}}$ is calculated and the respective formula is given below,

$\begin{array}{l}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{= K}}_{\text{w}}\\ \\ {\text{ K}}_{\text{b}}\text{=}\text{}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ \text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{4}{\text{.9×10}}^{\text{-10}}}\\ \\ \text{=}\text{2}{\text{.04×10}}^{\text{-5}}\end{array}$

Therefore, the ${\text{K}}_{\text{b}}$ value of sodium nitrite is $\text{2}{\text{.04×10}}^{\text{-5}}$.

The obtained ${\text{K}}_{\text{b}}$ value is substitute in below the equilibrium reaction.

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{= }\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\\ \\ \text{2}{\text{.04×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\end{array}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{2}{\text{.04×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(2}{\text{.04×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.20×(2}{\text{.04×10}}^{\text{-5}}\text{)}}\\ \\ \text{=}2.02\times {10}^{-3}\end{array}$

Therefore, the pOH concentration of sodium cyanide is $\text{2}{\text{.02×10}}^{\text{-3}}$.

Now, the $\text{pOH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}\\ \\ \text{=}\text{-log(2}{\text{.04×10}}^{\text{-5}}\text{)}\\ \\ \text{=}4.69\end{array}$

Therefore, the $\text{pOH}$ of $\text{0}\text{.20}\text{M}$ aqueous sodium cyanide solution is 4.69.  From this value, the pH of sodium cyanide solution is calculated.

  $\begin{array}{l}\text{pH+pOH=14}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-4}\text{.69}\\ \\ \text{=}9.31\end{array}$

Therefore, the $\text{pH}$ of the aqueous sodium cyanide solution is 9.31.

The solutions are ranked in the order of increasing pH and it is given below.

$\text{0}\text{.20}\text{M}{\text{NaNO}}_{\text{2}}\text{<}\text{1}{\text{.0×10}}^{\text{-5}}\text{M}\text{NaOH}\text{<}\text{0}\text{.20}\text{M}\text{NaCN}\text{<}\text{0}\text{.20}\text{M}{\text{NH}}_{\text{3}}$