Calculate the ${{S}^{2-}}$ ion concentration in a saturated solution $\left( 0.1M \right)$ of ${{H}_{2}}S$ whose $pH$ was adjusted to $2$ by the addition of $HCl$ . $\left( {{K}_{a}}=1.1\times {{10}^{-21}} \right)$
Answer
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Hint: $pH$ is the degree to measure acidity and basicity of an aqueous solution or a liquid. Acids have $pH$ value less than seven. And \[{{K}_{a}}\] is the dissociation constant of an acid, it is used to measure the strength of an acid.
Complete step by step answer:
Here, it is given that the dissociation constant of acid is $1.1\times {{10}^{-21}}$ ,
$pH$ of ${{H}_{2}}S$ is $2$,
And, ${{H}_{2}}S$ concentration is $0.1M$
$AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}$
${{K}_{a}}=\dfrac{[A][B]}{[AB]}$
where, \[{{K}_{a}}\] is the dissociation constant of an acid, $[A]$ and $[B]$ are the concentration of products and $[AB]$ is the concentration of reactant.
Now, let us see the reaction,
\[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\]
where, \[{{K}_{a}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]}\]
Now, substituting the values in the above formula, we get,
$1.1\times {{10}^{-21}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{0.1}$ – (1)
To calculate the concentration of ${{H}^{+}}$ ions, we have to use the formula of $pH$
$pH=-\log [{{H}^{+}}]$
where, $[{{H}^{+}}]$ is the concentration of ${{H}^{+}}$ ions.
Now, substituting the above value in this formula, we get,
$2=-\log [{{H}^{+}}]$
$\Rightarrow [{{H}^{+}}]=1\times {{10}^{-2}}M$
Now, the calculated concentration of $[{{H}^{+}}]$ ions is substituted in equation (1) and we get,
$1.1\times {{10}^{-21}}=\dfrac{{{[1\times {{10}^{-2}}]}^{2}}[{{S}^{2-}}]}{0.1}$
$\Rightarrow [{{S}^{2-}}]=\dfrac{1.1\times {{10}^{-21}}\times 0.1}{{{[1\times {{10}^{-2}}]}^{2}}}$
$\Rightarrow [{{S}^{2-}}]$=$1.1\times {{10}^{-18}}M$
Therefore, the concentration of ${{S}^{2-}}$ ions in a standard solution is $1.1\times {{10}^{-18}}M$ .
Additional information:
$pH$ is defined as a scale which is used to check acidity and basicity of an aqueous solution or liquid. The $pH$ less than seven considers an acidic condition, whereas $pH$ more than seven is a basic condition.
$pH=-\log [{{H}^{+}}]$
where, $[{{H}^{+}}]$ is the concentration of ${{H}^{+}}$ ions.
\[{{K}_{a}}\] is defined as the dissociation constant of an acid. If the value of the dissociation constant $({{K}_{a}})$ is high, then it is a strong acid and if the value of dissociation constant is low then, it is a weak acid.
\[{{K}_{a}}\] is defined as the ratio of concentration of products to the concentration of reactants.
Note: Here, ${{H}_{2}}S$ is a weak acid.
In the reaction, we get $2{{H}^{+}}$ ions, so therefore the concentration of the ${{H}^{+}}$ ions will get doubled.
In saturated solution, the maximum amount of solute is present in the solution that are capable of being dissolved.
Complete step by step answer:
Here, it is given that the dissociation constant of acid is $1.1\times {{10}^{-21}}$ ,
$pH$ of ${{H}_{2}}S$ is $2$,
And, ${{H}_{2}}S$ concentration is $0.1M$
$AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}$
${{K}_{a}}=\dfrac{[A][B]}{[AB]}$
where, \[{{K}_{a}}\] is the dissociation constant of an acid, $[A]$ and $[B]$ are the concentration of products and $[AB]$ is the concentration of reactant.
Now, let us see the reaction,
\[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\]
where, \[{{K}_{a}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]}\]
Now, substituting the values in the above formula, we get,
$1.1\times {{10}^{-21}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{0.1}$ – (1)
To calculate the concentration of ${{H}^{+}}$ ions, we have to use the formula of $pH$
$pH=-\log [{{H}^{+}}]$
where, $[{{H}^{+}}]$ is the concentration of ${{H}^{+}}$ ions.
Now, substituting the above value in this formula, we get,
$2=-\log [{{H}^{+}}]$
$\Rightarrow [{{H}^{+}}]=1\times {{10}^{-2}}M$
Now, the calculated concentration of $[{{H}^{+}}]$ ions is substituted in equation (1) and we get,
$1.1\times {{10}^{-21}}=\dfrac{{{[1\times {{10}^{-2}}]}^{2}}[{{S}^{2-}}]}{0.1}$
$\Rightarrow [{{S}^{2-}}]=\dfrac{1.1\times {{10}^{-21}}\times 0.1}{{{[1\times {{10}^{-2}}]}^{2}}}$
$\Rightarrow [{{S}^{2-}}]$=$1.1\times {{10}^{-18}}M$
Therefore, the concentration of ${{S}^{2-}}$ ions in a standard solution is $1.1\times {{10}^{-18}}M$ .
Additional information:
$pH$ is defined as a scale which is used to check acidity and basicity of an aqueous solution or liquid. The $pH$ less than seven considers an acidic condition, whereas $pH$ more than seven is a basic condition.
$pH=-\log [{{H}^{+}}]$
where, $[{{H}^{+}}]$ is the concentration of ${{H}^{+}}$ ions.
\[{{K}_{a}}\] is defined as the dissociation constant of an acid. If the value of the dissociation constant $({{K}_{a}})$ is high, then it is a strong acid and if the value of dissociation constant is low then, it is a weak acid.
\[{{K}_{a}}\] is defined as the ratio of concentration of products to the concentration of reactants.
Note: Here, ${{H}_{2}}S$ is a weak acid.
In the reaction, we get $2{{H}^{+}}$ ions, so therefore the concentration of the ${{H}^{+}}$ ions will get doubled.
In saturated solution, the maximum amount of solute is present in the solution that are capable of being dissolved.
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