Solveeit Logo
Login

Question

Question: Calculate the \({{S}^{2-}}\) ion concentration in a saturated solution \(\left( 0.1M \right)\) of \(...

Calculate the S2−{{S}^{2-}} ion concentration in a saturated solution (0.1M)\left( 0.1M \right) of H2S{{H}_{2}}S whose pHpH was adjusted to 22 by the addition of HClHCl . (Ka=1.1×10−21)\left( {{K}_{a}}=1.1\times {{10}^{-21}} \right)

Explanation

Solution

pHpH is the degree to measure acidity and basicity of an aqueous solution or a liquid. Acids have pHpH value less than seven. And Ka{{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 1.1×10−211.1\times {{10}^{-21}} ,
pHpH of H2S{{H}_{2}}S is 22,
And, H2S{{H}_{2}}S concentration is 0.1M0.1M
AB⇌A++B−AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}
Ka=[A][B][AB]{{K}_{a}}=\dfrac{[A][B]}{[AB]}
where, Ka{{K}_{a}} is the dissociation constant of an acid, [A][A] and [B][B] are the concentration of products and [AB][AB] is the concentration of reactant.
Now, let us see the reaction,
H2S⇌2H++S2−{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}
where, Ka=[H+]2[S2−][H2S]{{K}_{a}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]}
Now, substituting the values in the above formula, we get,
1.1×10−21=[H+]2[S2−]0.11.1\times {{10}^{-21}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{0.1} – (1)
To calculate the concentration of H+{{H}^{+}} ions, we have to use the formula of pHpH
pH=−log⁡[H+]pH=-\log [{{H}^{+}}]
where, [H+][{{H}^{+}}] is the concentration of H+{{H}^{+}} ions.
Now, substituting the above value in this formula, we get,
2=−log⁡[H+]2=-\log [{{H}^{+}}]
⇒[H+]=1×10−2M\Rightarrow [{{H}^{+}}]=1\times {{10}^{-2}}M
Now, the calculated concentration of [H+][{{H}^{+}}] ions is substituted in equation (1) and we get,
1.1×10−21=[1×10−2]2[S2−]0.11.1\times {{10}^{-21}}=\dfrac{{{[1\times {{10}^{-2}}]}^{2}}[{{S}^{2-}}]}{0.1}
⇒[S2−]=1.1×10−21×0.1[1×10−2]2\Rightarrow [{{S}^{2-}}]=\dfrac{1.1\times {{10}^{-21}}\times 0.1}{{{[1\times {{10}^{-2}}]}^{2}}}
⇒[S2−]\Rightarrow [{{S}^{2-}}]=1.1×10−18M1.1\times {{10}^{-18}}M

Therefore, the concentration of S2−{{S}^{2-}} ions in a standard solution is 1.1×10−18M1.1\times {{10}^{-18}}M .

Additional information:
pHpH is defined as a scale which is used to check acidity and basicity of an aqueous solution or liquid. The pHpH less than seven considers an acidic condition, whereas pHpH more than seven is a basic condition.
pH=−log⁡[H+]pH=-\log [{{H}^{+}}]
where, [H+][{{H}^{+}}] is the concentration of H+{{H}^{+}} ions.
Ka{{K}_{a}} is defined as the dissociation constant of an acid. If the value of the dissociation constant (Ka)({{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.
Ka{{K}_{a}} is defined as the ratio of concentration of products to the concentration of reactants.

Note: Here, H2S{{H}_{2}}S is a weak acid.
In the reaction, we get 2H+2{{H}^{+}} ions, so therefore the concentration of the H+{{H}^{+}} ions will get doubled.
In saturated solution, the maximum amount of solute is present in the solution that are capable of being dissolved.