Answer:
Concentration of [tex]Zn^{+2}[/tex] = 1.9 * [tex]10^{-19}[/tex]
Explanation:
Find the provided attachment for solution.
The equilibrium concentration of uncomplexed Zn²⁺ ions if the concentration of cyanide ions at equilibrium is 0.50 M, is 19.18 × 10⁻²² M.
Equilibrium constant of any reaction will be calculated as the ration of the concentration of products to the concentration of reactants with raise to to respective coefficients at the equilibrium condition.
Moles of Zn²⁺ & Zinc cyanide complex are equal, so the concentration of 5×10⁻³ moles of Zinc cyanide complex in 1 liter of the solution is 5×10⁻³ M.
Chemical reaction for the formation of given complex with ICE table is:
Zn²⁺ + 4(cyanide ions) ⇄ Zinc cyanide complex
Initial: 0 0.50 5×10⁻³
Change: +x +4x -x
Equilibrium: x 0.50+4x 5×10⁻³-x
Equilibrium constant for the above reaction will be calculated as:
Kf = Zinc cyanide complex / [Zn²⁺][cyanide ion]⁴
Given value of Kf = 4.17 × 10¹⁹
On putting all values on the above equation we get,
4.17 × 10¹⁹ = [5×10⁻³-x] / [x].[0.50+4x]⁴
x is very small as compared to the 5×10⁻³ and 0.50, so we can neglect x and equation becomes:
4.17 × 10¹⁹ = [5×10⁻³] / [x].[0.50]⁴
x = 19.18 × 10⁻²² M
Hence, required concentration of Zn²⁺ is 19.18 × 10⁻²² M.
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