In a titration for the determination of the chloride ion, a solution of 0.1M agno3 is used against a solution, the volume of which is 100ml. At the completion of the titration, the residual concentrations of silver and chloride ions are equal and the total volume is now 125ml. What additional volume of the silver nitrate solution is needed to effect the precipitation of silver chromate (Ag2CrO4) if the molar concentration of the chromate ion is 1x10^-4.
ksp of AgCl- 1.6x10^-10
Ksp of Ag2CrO4- 9x10^-12
To determine the additional volume of silver nitrate solution needed to precipitate silver chromate (Ag2CrO4), we need to follow these steps:
1. Calculate the initial moles of chloride ions in the 100 ml solution:
Moles of chloride ions = Molarity × Volume
Moles of chloride ions = 0.1 M × 0.1 L = 0.01 moles
2. Since the residual concentrations of silver and chloride ions are equal, we have the same number of moles of silver ions present in the solution.
3. Calculate the moles of silver ions in 125 ml solution:
Moles of silver ions = Moles of chloride ions = 0.01 moles
4. Calculate the initial concentration of silver ions in the 100 ml solution:
Initial concentration of silver ions = Moles of silver ions / Initial volume
Initial concentration of silver ions = 0.01 moles / 0.1 L = 0.1 M
5. Determine the solubility product constant, Ksp, of Ag2CrO4:
Ksp of Ag2CrO4 = 9 x 10^-12
6. Calculate the molar concentration of chromate ions necessary to exceed Ksp and precipitate Ag2CrO4:
Concentration of chromate ions = Ksp of Ag2CrO4
7. Determine the moles of chromate ions required:
Moles of chromate ions = Concentration of chromate ions × Total volume
Moles of chromate ions = (9 x 10^-12) M × 0.125 L = 1.125 x 10^-12 moles
8. The ratio of silver to chromate ions in Ag2CrO4 is 2:1. Therefore, we need twice the moles of silver ions to precipitate the required moles of chromate ions:
Moles of silver ions for Ag2CrO4 = 2 × Moles of chromate ions
Moles of silver ions for Ag2CrO4 = 2 × 1.125 x 10^-12 moles = 2.25 x 10^-12 moles
9. Calculate the additional volume of silver nitrate solution needed:
Additional volume of silver nitrate solution = Moles of silver ions / Final concentration
Additional volume of silver nitrate solution = (2.25 x 10^-12 moles) / (0.1 M)
Additional volume of silver nitrate solution = 2.25 x 10^-11 L = 22.5 μL
Therefore, an additional volume of 22.5 μL of the silver nitrate solution is needed to precipitate the desired amount of silver chromate (Ag2CrO4).
To determine the additional volume of silver nitrate (AgNO3) solution needed to precipitate silver chromate (Ag2CrO4), we can use the solubility product constant (Ksp) equation for Ag2CrO4. The balanced equation for the precipitation reaction is:
2AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
First, let's find the initial number of moles of silver chloride (AgCl) and chloride ions (Cl-) in the 100 mL solution:
Number of moles of AgCl = (0.1 M) x (0.100 L) = 0.01 moles
Number of moles of Cl- = 0.01 moles
Since the residual concentrations of silver and chloride ions are equal at the end, the number of moles of both ions is 0.01 moles.
To determine the concentration of Ag+ ions required to precipitate Ag2CrO4, we need to find the concentration of Cl- ions from AgCl. We can use the Ksp equation for AgCl:
Ksp = [Ag+][Cl-]
Rearranging the equation to solve for [Cl-]:
[Cl-] = Ksp /[Ag+]
Substituting the given Ksp value for AgCl:
[Cl-] = (1.6 x 10^-10) /[Ag+]
[Cl-] = 1.6 x 10^-10 M
Since the concentrations of Ag+ and Cl- ions are equal, the concentration of Ag+ ions is also 1.6 x 10^-10 M.
To precipitate Ag2CrO4, we need to add chromate ions (CrO4^2-) from K2CrO4 to form a precipitate according to the balanced equation. The number of moles of Ag2CrO4 needs to be equal to the number of moles of AgCl:
Concentration of [Ag+] = 1.6 x 10^-10 M
Number of moles of Ag2CrO4 = 0.01 moles
Concentration of [Ag2CrO4] = (0.01 moles) / (0.125 L) = 0.08 M
Using the Ksp equation for Ag2CrO4:
Ksp = [Ag+]^2[CrO4^2-]
Rearranging and solving for [CrO4^2-]:
[CrO4^2-] = Ksp / [Ag+]^2
Substituting the given Ksp value for Ag2CrO4:
[CrO4^2-] = (9 x 10^-12) / (1.6 x 10^-10)^2
[CrO4^2-] = 3.515 x 10^-7 M
To reach this concentration, we need to calculate the additional volume of AgNO3 solution needed:
[Ag+] = (0.08 M) / (0.125/0.1 L) = 0.064 M
Now we can use the concentration and volume to calculate the number of moles:
Number of moles of Ag+ = (0.064 M) x (V L) = 0.064V moles
Using the balanced equation, the stoichiometry of AgNO3 and Ag2CrO4 is 2:1, so the number of moles of Ag2CrO4 required will be half of the number of moles of Ag+:
Number of moles of Ag2CrO4 = 0.064V / 2 = 0.032V moles
Since the concentration is given, we can find the volume (V) of AgNO3 solution needed:
[CrO4^2-] = (0.032V moles) / (V L) = 3.515 x 10^-7 M
Solving this equation for V:
V = (0.032V moles) / (3.515 x 10^-7 M)
Now, we can substitute the value of [CrO4^2-] and solve for V:
V = (0.032) / (3.515 x 10^-7) = 91.07 mL
Therefore, the additional volume of silver nitrate solution needed to precipitate silver chromate is 91.07 mL.