Respuesta :
Answer : The concentration of silver ion is, [tex]3.8\times 10^{-3}M[/tex]
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,
[tex]Ag_2S(s)\rightleftharpoons 2Ag^+(aq)+S^{2-}(aq)[/tex]
The expression of [tex]K[/tex] will be,
[tex]K_{eq}=[Ag^+]^2[S^{2-}][/tex]
[tex]2.4\times 10^{-4}=(2.5\times 110^{-1})^2[S^{2-}][/tex]
[tex][S^{2-}]=3.8\times 10^{-3}M[/tex]
Therefore, the concentration of silver ion is, [tex]3.8\times 10^{-3}M[/tex]
3.84 x 10⁻³ M
Further explanation
Given:
- [tex]\boxed{K_{eq} = 2.4 \times 10^{-4}}[/tex]
- The equilibrium concentration of silver ion is [tex]\boxed{ \ [Ag^+] = 2.5 \times 10^{-1} \ M \ }[/tex].
Question:
What is [S²⁻] at equilibrium?
The Process:
For the reaction [tex]\boxed{ \ Ag_2S_{(s)} \rightleftharpoons 2Ag^+_{(aq)} + S^{2-}_{(aq)} \ }[/tex], we can observe that the substances on the right-hand side have a solution phase (aq) that is allowed into the equilibrium constant K.
Remember, pure solids (s) and liquids (l) are disregarded and kept at 1.
Therefore, from the initial formula [tex]\boxed{ \ K = \frac{ [Ag^+]^2.[S^{2-}] }{[Ag_2S]} \ }[/tex] we get the final result [tex]\boxed{ \ K = [Ag^+]^2.[S^{2-}] \ }[/tex].
Let us find out [S²⁻] at equilibrium.
[tex]\boxed{ \ K_{eq} = [Ag^+]^2.[S^{2-}] \ }[/tex]
[tex]\boxed{ \ 2.4 \times 10^{-4} = [2.5 \times 10^{-1}]^2.[S^{2-}] \ }[/tex]
[tex]\boxed{ \ 2.4 \times 10^{-4} = 6.25 \times 10^{-2} \cdot [S^{2-}] \ }[/tex]
[tex]\boxed{ \ [S^{2-}] = \frac{2.4 \times 10^{-4}}{6.25 \times 10^{-2}} \ }[/tex]
Thus we get [S²⁻] at equilibrium equal to [tex]\boxed{ \ [S^{2-}] = 3.84 \times 10^{-3} \ M \ }[/tex]
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Notes:
- [tex]\boxed{ \ K_c \ or \ K_{eq} \ }[/tex], denotes that the equilibrium constant is expressed using molar concentrations, i.e., [tex]\boxed{ \ mol.dm^{-3} \ or \ mol.L^{-1} \ \ }[/tex]. For this question, [tex]\boxed{ \ K_c \ }[/tex] means the same thing as [tex]\boxed{ \ K_{eq} \ }[/tex].
- The rIght-hand side of the equation on top, left-hand side of the equation on the bottom.
- The square brackets show concentrations in [tex]\boxed{ \ mol.dm^{-3} \ or \ mol.L^{-1} \ or \ M \ }[/tex].
- The indices are the numbers in front of each substance (or the coefficients) in the chemical equation.
- A heterogeneous balance consists of more than one phase. Typical examples include reactions involving solids and gases, or solids and liquids.
- Substances that can be included in formula K are substances with a gas phase (g) or aqueous phase (aq). Pure solids and liquids are unincluded in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1.
Learn more
- Write the equilibrium constant for the reaction https://brainly.com/question/10608589
- Write the equilibrium constant for the reaction a heterogeneous balance https://brainly.com/question/13026406
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