Answer:
[H₂CO₃(aq)] = 1.95x10ˉ⁶M,
[HCO₃ˉ(aq)] = 1.63x10ˉ⁴M, and
[CO₃ˉ²(aq)] = 1.27x10ˉ⁴M.
Step-by-step explanation:
Determine concentrations of carbonate ion species in Ag₂CO₃ solution with pH = 7.00.
Silver Carbonate => 3 carbonate species (H₂CO₃, HCO₃ˉ & CO₂²ˉ)
Given:
Ag₂CO₃(s) <=> 2Ag⁺(aq) + CO₃²ˉ(aq); Ksp = 8.2 x 10ˉ¹²
Ceq => ------ | 2x | x
Ksp = [Ag⁺]²[CO₃²ˉ] = (2x)²(x) = 3x³
=> x = [CO₃²ˉ] = CubeRt(Ksp/4) = CubeRt(8.2 x 10ˉ¹²/4)M = 1.27 x 10ˉ⁴M
Given:
H₂CO₃(aq) Ionization Constants => Kd1 = 4.3x10ˉ⁷; Kd2 = 4.8 x 10ˉ¹¹
Solubility of Ag₂CO₃(s) in neutral solution (pH = 7) delivers 1.27x10ˉ⁴M in CO₃²ˉ ions,
If the CO₃²ˉ ions undergo hydrolysis producing the bicarbonate and hydroxide ions in equal amounts.
Then:
CO₃²ˉ(aq) + H₂O(l) <=> HCO₃ˉ(aq) + OHˉ(aq)
Ceq => 1.27x10ˉ⁴M ------- x’ x’
Kb(CO₃²ˉ) = Kw/Kd2 = [HCO₃ˉ][OHˉ]/[CO₃²ˉ] = (1x10ˉ¹⁴/4.8x10ˉ¹¹) = (x’)²/1.27x10ˉ⁴
=> x’ = [HCO₃ˉ] = SqrRt[1.27(10 ˉ⁴)(10ˉ¹⁴)/(4.8x10ˉ¹¹)] = 1.63x10ˉ⁴M
The presence of 1.63x10ˉ⁴M HCO₃ˉ(aq) will under go hydrolysis according to the following rxn:
HCO₃ˉ(aq) + H₂O(l) <=> H₂CO₃(aq) + OHˉ(aq)
C(eq) => 1.63x10ˉ⁴M ----- x” x”
Kb(HCO₃²ˉ) = Kw/Kd1 = [H₂CO₃][OHˉ]/[HCO₃²ˉ] = (1x10ˉ¹⁴/4.8x10ˉ¹¹) = (x”)²/1.63x10ˉ⁴
=> x” = [HCO₃ˉ] = SqrRt[1.63(10 ˉ⁴)(10ˉ¹⁴)/(4.3x10ˉ⁷)] = 1.95x10ˉ⁶M
Therefore, the carbonate ion concentrations are:
[H₂CO₃(aq)] = 1.95x10ˉ⁶M,
[HCO₃ˉ(aq)] = 1.63x10ˉ⁴M, and
[CO₃ˉ²(aq)] = 1.27x10ˉ⁴M.
HCO₃⁻(aq) will have the higher concentration as the conjugate base CO₃²⁻(aq) is stronger in hydrolysis than the HCO₃⁻(aq) conjugate base of H₂CO₃(aq).