Question

What is the pOH of $0.1$M KB (salt of weak acid and strong base) at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$? (Given ${\text{p}}{{\text{K}}_{\text{b}}}$of ${{\text{B}}^ - }$ = $7$)

Answer 1

We have to determine the pOH so, we will write the dissociation of the base in water. Then we will draw the ICE (initial, change, and equilibrium concentration) followed by the equilibrium constant expression. So, we can determine the concentration of base. Then by using the pOH formula we will determine the pOH.

Complete solution:
We will determine the ${{\text{K}}_{\text{b}}}$as follows:
${\text{p}}{{\text{K}}_{\text{b}}}\, = - {\text{log}}\,{{\text{K}}_{\text{b}}}$
Where,
${{\text{K}}_{\text{b}}}$is the base dissociation constant.
On substituting $7$ for ${\text{p}}{{\text{K}}_{\text{b}}}$.
$7\, = - \operatorname{lo} {\text{g}}\,\,{{\text{K}}_{\text{b}}}$
$\,{{\text{K}}_{\text{b}}}\, = \,1 \times {10^{ - 7}}\,$
KB is the salt of weak acid and strong base. The dissociation of base in water is shown as follows:
${{\text{B}}^ - }\, + \,{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{BH}}\,{\text{ + }}\,{\text{O}}{{\text{H}}^ - }$
Now we will write the ICE chart,
${{\text{B}}^ - }\, + \,{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{BH}}\,{\text{ + }}\,{\text{O}}{{\text{H}}^ - }$
Initial conc. $0.1\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,{\text{0}}\,\,\,{\text{ + }}\,\,\,\,0$
As water is in excess we can neglect the change in concentration of water.
Change $- {\text{x}}\,\,\, + 0\,\,\,\, \to \,\,\,\,\,\,\,{\text{x}}\,\,\,{\text{ + }}\,\,\,\,{\text{x}}$
Equi. Conc. $0.1 - {\text{x}}\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,{\text{x}}\,\,\,{\text{ + }}\,\,\,\,{\text{x}}$
Now we have to determine the concentration of hydroxide ions so, we will write the equilibrium constant for the above reaction,
${{\text{K}}_{\text{b}}}{\text{ = }}\,\dfrac{{\left[ {{\text{BH}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{B}}^ - }} \right]}}\,$
On substituting $1 \times {10^{ - 7}}\,$ for ${{\text{K}}_{\text{b}}}$, x for $\left[ {{\text{BH}}} \right]$and $\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ and $0.1 - {\text{x}}$ for $\left[ {{{\text{B}}^ - }} \right]$,
${\text{1}} \times {\text{1}}{{\text{0}}^{ - 7}}{\text{ = }}\,\dfrac{{\left[ {\text{x}} \right]\left[ {\text{x}} \right]}}{{\left[ {0.1 - {\text{x}}\,} \right]}}\,$
As the value of dissociation constant of base is very less means base dissociation is very small so, we can assume that $0.1 - {\text{x}} \approx \,{\text{0}}{\text{.1}}$ So,
${\text{1}} \times {\text{1}}{{\text{0}}^{ - 7}}{\text{ = }}\,\dfrac{{\left[ {\text{x}} \right]\left[ {\text{x}} \right]}}{{\left[ {0.1\,} \right]}}\,$
${{\text{x}}^2}\, = \,{\text{1}} \times {\text{1}}{{\text{0}}^{ - 7}} \times 0.1\,$
${{\text{x}}^2}\, = \,{\text{1}} \times {\text{1}}{{\text{0}}^{ - 8}}$
${\text{x}}\, = \sqrt {\,{\text{1}} \times {\text{1}}{{\text{0}}^{ - 8}}}$
${\text{x}}\, = {\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}$
So, the concentration of hydroxide ions is ${\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}$
The formula to calculate the pOH is as follows:
${\text{pOH}}\,\,{\text{ = }}\, - \,{\text{log }}\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
On substituting ${\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}$ for hydroxide ${\text{O}}{{\text{H}}^ - }$ concentration.
${\text{pOH}}\,\,{\text{ = }}\, - \,{\text{log }}\left[ {{\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}} \right]$
${\text{pOH}}\,\,{\text{ = }}\,4\,{\text{log}}\,{\text{10}}$
${\text{pOH}}\,\,{\text{ = }}\,4$

So, the pOH is$4$.

Note: We can also determine the pOH as follows:
We calculated in above solution that $\,{{\text{K}}_{\text{b}}}\, = \,1 \times {10^{ - 7}}\,$.
Now, the relation between acid dissociation constant and base dissociation constant is,
${{\text{K}}_{\text{a}}}{ \times}\,\,{{\text{K}}_{\text{b}}}\,{\text{ = }}\,{{\text{K}}_{\text{w}}}$
Where,
${{\text{K}}_{\text{a}}}$ is the acid dissociation constant
${{\text{K}}_{\text{b}}}$ is the base dissociation constant
${{\text{K}}_{\text{w}}}$ is the ionic product of water whose value is ${10^{ - 14}}$.

${{\text{K}}_{\text{a}}}{ \times }\,\,{{\text{K}}_{\text{b}}}\,{\text{ = }}\,{10^{ - {\text{14}}}}$
So,
${{\text{K}}_{\text{a}}}{ \times }\,\,1 \times {10^{ - 7}}\,{\text{ = }}\,{10^{ - {\text{14}}}}$
${{\text{K}}_{\text{a}}}\,{\text{ = }}\,\,1 \times {10^{ - 7}}$
The formula to calculate the hydroxide ion concentration from acid dissociation constant is as follows:
${\text{O}}{{\text{H}}^ - }\, = \,\sqrt {\dfrac{{{{\text{K}}_{\text{w}}}{\text{C}}}}{{{{\text{K}}_{\text{a}}}}}}$
Where,
C is the concentration of base.
On substituting $0.1$ M for C, $\,1 \times {10^{ - 7}}$ for ${{\text{K}}_{\text{a}}}$, and ${10^{ - {\text{14}}}}$ for ${{\text{K}}_{\text{w}}}$,
${\text{O}}{{\text{H}}^ - }\, = \,\sqrt {\dfrac{{\,{{10}^{ - 14}}\, \times 0.1}}{{\,1 \times {{10}^{ - 7}}}}}$
${\text{O}}{{\text{H}}^ - }\, = \,\sqrt {1 \times {{10}^{ - 8}}}$
${\text{O}}{{\text{H}}^ - }\, = \,1 \times {10^{ - 4}}$
On substituting ${\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}$ for hydroxide ${\text{O}}{{\text{H}}^ - }$ concentration.
${\text{pOH}}\,\,{\text{ = }}\, - \,{\text{log }}\left[ {{\text{1}} \times {\text{1}}{{\text{0}}^{ - 4}}} \right]$
${\text{pOH}}\,\,{\text{ = }}\,4$
So, the pOH is$4$.