
Calculate the ${K_b}$ of a base whose $0.1M$ solution has $pH$ of 10.5.
Answer
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Hint: The compound can broadly be classified as either neutral, as the base of an acid. The acidic strength is related to the production of hydroxyl ions while the strength of the base is determined by the hydronium ion concentration. These are the ions that are formed when the ions interact with the water molecule.
Complete step by step answer:
The base can be represented as $BOH$ Which dissociates as
Given the initial concentration of the base as $0.1M$ so the initial concentration can be represented as
The initial concentration of BOH is 0.1
After dissociation, the concentration of the base can be represented as
$BOH \rightleftharpoons {B^ + } + O{H^ - }$
Thus writing the equation of ${K_b}$ we get
${K_b} = \dfrac{{(0.1\alpha )(0.1\alpha )}}{{0.1(1 - \alpha )}}$
Since the $pH$ of the base is given as 10.5 so we can calculate the concentration of hydroxyl ion as
$ - \log [{H^ + }] = pH$
$ \Rightarrow - \log [{H^ + }] = 10.5$
$ \Rightarrow [{H^ + }] = 3.16 \times {10^{ - 11}}M$
As we know that $[{H^ + }][O{H^ - }] = {10^{ - 14}}$ so
$[O{H^ - }] = 3.16 \times {10^{ - 4}}M = 0.1\alpha $
Therefore $\alpha = 3.16 \times {10^{ - 3}}$
Putting the value of $\alpha $d in the equation of ${K_b}$ we get,
$\alpha < < 1{{ so, 1 - }}\alpha \approx 1$
So ${K_b} = 0.1{\alpha ^2} = {10^{ - 6}}$.
Note: The dissociation of the base is represented by ${K_b}$ what is called the base dissociation constant. It is a measure of how completely the base dissociates into its constituent ions.
${K_W} = {K_a} \times {K_b}$ , here the dissociation constant of water in a solution is the same as the product of the dissociation constant of the base and the acid that is present in the solution.
The $pH$ of the base is usually above 7 since 7 is the $pH$ of neutral compounds. For bases, it ranges from 7 to 14 which is the end of the scale, while for the acids it is lower than 7.
The dissociation constant of water is fixed so if the concentration of any one of the bases or acid is known the concentration of the counterpart can be figured of easily through this constant.
Complete step by step answer:
The base can be represented as $BOH$ Which dissociates as
Given the initial concentration of the base as $0.1M$ so the initial concentration can be represented as
The initial concentration of BOH is 0.1
After dissociation, the concentration of the base can be represented as
| BOH | ${B^ + }$ | $O{H^ - }$ | |
| INITIAL | 0.1 | 0 | 0 |
| FINAL | ${{0}}{{.1(1 - }}\alpha {{)}}$ | ${{0}}{{.1}}\alpha $ | ${{0}}{{.1}}\alpha $ |
$BOH \rightleftharpoons {B^ + } + O{H^ - }$
Thus writing the equation of ${K_b}$ we get
${K_b} = \dfrac{{(0.1\alpha )(0.1\alpha )}}{{0.1(1 - \alpha )}}$
Since the $pH$ of the base is given as 10.5 so we can calculate the concentration of hydroxyl ion as
$ - \log [{H^ + }] = pH$
$ \Rightarrow - \log [{H^ + }] = 10.5$
$ \Rightarrow [{H^ + }] = 3.16 \times {10^{ - 11}}M$
As we know that $[{H^ + }][O{H^ - }] = {10^{ - 14}}$ so
$[O{H^ - }] = 3.16 \times {10^{ - 4}}M = 0.1\alpha $
Therefore $\alpha = 3.16 \times {10^{ - 3}}$
Putting the value of $\alpha $d in the equation of ${K_b}$ we get,
$\alpha < < 1{{ so, 1 - }}\alpha \approx 1$
So ${K_b} = 0.1{\alpha ^2} = {10^{ - 6}}$.
Note: The dissociation of the base is represented by ${K_b}$ what is called the base dissociation constant. It is a measure of how completely the base dissociates into its constituent ions.
${K_W} = {K_a} \times {K_b}$ , here the dissociation constant of water in a solution is the same as the product of the dissociation constant of the base and the acid that is present in the solution.
The $pH$ of the base is usually above 7 since 7 is the $pH$ of neutral compounds. For bases, it ranges from 7 to 14 which is the end of the scale, while for the acids it is lower than 7.
The dissociation constant of water is fixed so if the concentration of any one of the bases or acid is known the concentration of the counterpart can be figured of easily through this constant.
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