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Answer:
[Ag⁺] = 0.0666M
Explanation:
For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:
Ag⁺ + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]
As initial concentrations of Ag⁺ and CN⁻ are:
[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M
[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M
The equilibrium concentrations of each compound are:
[CN⁻] = 9.7x10⁻⁴M - x
[Ag⁺] = 0.0676M - x
[Ag(CN)₂⁻] = x
<em>Where x is reaction coordinate</em>
Replacing in Kf formula:
1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]
1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0
Solving for x:
X = 9.614x10⁻⁴M
Thus, equilibrium concentration of Ag⁺ is:
[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>
