
What is the pH of \[0.1~M~\]$NaHC{{O}_{3}}$? \[{{K}_{1}}~=~5~\times ~{{10}^{-7}}\]and \[{{K}_{2}}~=~5~\times ~{{10}^{-11}}\] for carbonic acids.
Answer
573.9k+ views
Hint: First we will calculate the base dissociation constant using the given values from the question and using that base dissociation constant we will find the value of pOH followed by the value of pH which is asked in the question.
Complete step by step answer:
First we will write the balanced chemical equation according to question:
$NaHC{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}+NaOH$
We know a relationship between ${{K}_{a}}$,${{K}_{b}}$and ${{K}_{w}}$:
${{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}$ …….(i)
${{K}_{b}}=$Base dissociation constant
${{K}_{a}}$= Acid dissociation constant
${{K}_{w}}=$Dissociation constant for water
Here ${{K}_{a}}={{K}_{1}}=4.5\times {{10}^{-7}}$
${{K}_{w}}={{10}^{-14}}$
Using the above two given values in (i) we will get value of base dissociation constant which is ${{K}_{b}}$
${{K}_{b}}=\dfrac{{{10}^{-14}}}{4.5\times {{10}^{-7}}}=2.22\times {{10}^{-8}}$
Dissociation is low. So,
$\Rightarrow 2.22\times {{10}^{-8}}=\dfrac{{{[O{{H}^{-}}]}^{2}}}{0.1}$
On solving above equation we will get the value of $[O{{H}^{-}}]$equal to $4.7\times {{10}^{-5}}$ ………(ii)
We know pOH = -log${{[OH]}^{-}}$ ……..(iii)
Using (ii) and (iii) we will get pOH = $4.32$
$pH=14-4.32=9.68$
So the pH of $0.1M$ $NaHC{{O}_{3}}$ is $9.68$.
Note:
Some examples of weak bases are ammonium hydroxide, hydrazine (somewhat structurally related with the anterior one), phosphine, zinc hydroxide (besides the fact that this hydroxide is an amphoteric one) and, amongst the organic compounds, there is much more examples, as: methylamine, dymethylamine, Pyridine,etc.
Some weak acid examples are acetic acid, oxalic acid, formic acid, benzoic acid, hydrofluoric acid, nitrous acid, etc.
some examples of strong acid are sulphuric acid, hydrochloric acid, nitric acid, perchloric acid, chloric acid,etc
Some examples of strong base are potassium hydroxide, sodium hydroxide, barium hydroxide, calcium hydroxide and lithium hydroxide.
Complete step by step answer:
First we will write the balanced chemical equation according to question:
$NaHC{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}+NaOH$
We know a relationship between ${{K}_{a}}$,${{K}_{b}}$and ${{K}_{w}}$:
${{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}$ …….(i)
${{K}_{b}}=$Base dissociation constant
${{K}_{a}}$= Acid dissociation constant
${{K}_{w}}=$Dissociation constant for water
Here ${{K}_{a}}={{K}_{1}}=4.5\times {{10}^{-7}}$
${{K}_{w}}={{10}^{-14}}$
Using the above two given values in (i) we will get value of base dissociation constant which is ${{K}_{b}}$
${{K}_{b}}=\dfrac{{{10}^{-14}}}{4.5\times {{10}^{-7}}}=2.22\times {{10}^{-8}}$
Dissociation is low. So,
$\Rightarrow 2.22\times {{10}^{-8}}=\dfrac{{{[O{{H}^{-}}]}^{2}}}{0.1}$
On solving above equation we will get the value of $[O{{H}^{-}}]$equal to $4.7\times {{10}^{-5}}$ ………(ii)
We know pOH = -log${{[OH]}^{-}}$ ……..(iii)
Using (ii) and (iii) we will get pOH = $4.32$
$pH=14-4.32=9.68$
So the pH of $0.1M$ $NaHC{{O}_{3}}$ is $9.68$.
Note:
Some examples of weak bases are ammonium hydroxide, hydrazine (somewhat structurally related with the anterior one), phosphine, zinc hydroxide (besides the fact that this hydroxide is an amphoteric one) and, amongst the organic compounds, there is much more examples, as: methylamine, dymethylamine, Pyridine,etc.
Some weak acid examples are acetic acid, oxalic acid, formic acid, benzoic acid, hydrofluoric acid, nitrous acid, etc.
some examples of strong acid are sulphuric acid, hydrochloric acid, nitric acid, perchloric acid, chloric acid,etc
Some examples of strong base are potassium hydroxide, sodium hydroxide, barium hydroxide, calcium hydroxide and lithium hydroxide.
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