The concentration of Ag+ after dissociation of Ag(CN)2- is 2.91 × 10^-22 M.
here is the calculation of the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of AgNO3 and 10.0 g of KCN in sufficient water to make 1.00 L of solution:
Given:
Mass of AgNO3 (m_AgNO3) = 1.00 g
Molar mass of AgNO3 (M_AgNO3) = 169.87 g/mol
Mass of KCN (m_KCN) = 10.00 g
Molar mass of KCN (M_KCN) = 65.11 g/mol
Volume of solution (V) = 1.00 L
Formation constant of Ag(CN)2- complex (K_f) = 1.0 × 10²¹
Calculations:
1. Calculate the moles of AgNO3 and KCN:
moles of AgNO3 = m_AgNO3 / M_AgNO3 = 1.00 g / 169.87 g/mol = 0.0058868546535586035 mol
moles of KCN = m_KCN / M_KCN = 10.00 g / 65.11 g/mol = 0.15384615384615385 mol
2. Assume complete reaction of AgNO3 with KCN to form Ag(CN)2-:
moles of Ag(CN)2- = moles of AgNO3 = 0.0058868546535586035 mol
3. Calculate the initial concentration of Ag+:
initial concentration of Ag+ = moles of AgNO3 / V = 0.0058868546535586035 mol / 1.00 L = 0.0058868546535586035 M
4. Calculate the concentration of Ag+ after dissociation of Ag(CN)2-:
dissociation constant = 1/K_f = 1/1.0 × 10²¹ = 1.0 × 10^-21
Concentration of Ag(CN)2- = moles of Ag(CN)2- / V = 0.0058868546535586035 mol / 1.00 L = 0.0058868546535586035 M
Let x be the concentration of Ag+ after dissociation. Then, we can set up the following equation:
x² + (2 ×0.0058868546535586035) ×x - 0.0058868546535586035 = 0
Solving for x using the quadratic formula, we get:
x = 2.91 × 10^-22 M
Therefore, the concentration of Ag+ after dissociation of Ag(CN)2- is 2.91 × 10^-22 M.