1. if 10mL of 0.004M AgNO3 is added to 10mL of 0.0024M K2CrO4, is either Ag+ or CrO4(2-) in stoichiometric excess? if so, which is in excess?
2. the Ksp for BaCrO4 is 1.2x10-10. will BaCrO4 precipitate upon mixing 10 mL of 1x10-4M Ba(NO3)2 with 10mL of 1.0x10-4M K2CrO4?
If you made up a solution of NaOH by adding 0.010 mole of solid NaOH to 1.0 liter of distilled water, what would be the concentration of the OH-(aq)?
I want answers
To determine if either Ag+ or CrO4(2-) is in stoichiometric excess in the first question, we need to compare the number of moles of Ag+ and CrO4(2-) ions present in the solution after mixing.
1. For the silver nitrate (AgNO3) solution:
- Concentration: 0.004 M
- Volume: 10 mL
- Moles of Ag+ ions = concentration * volume (in liters)
= 0.004 M * 0.01 L
= 0.00004 moles
2. For the potassium chromate (K2CrO4) solution:
- Concentration: 0.0024 M
- Volume: 10 mL
- Moles of CrO4(2-) ions = concentration * volume (in liters)
= 0.0024 M * 0.01 L
= 0.000024 moles
Comparing the number of moles of Ag+ ions (0.00004 moles) to the number of moles of CrO4(2-) ions (0.000024 moles), we can see that Ag+ ions are in stoichiometric excess. This means that Ag+ ions are present in a higher amount than required for a complete reaction with CrO4(2-) ions.
Therefore, Ag+ ions are in stoichiometric excess in the given solution.
Moving on to the second question:
To determine if BaCrO4 will precipitate upon mixing the Ba(NO3)2 and K2CrO4 solutions, we need to compare the ion product (IP) with the solubility product constant (Ksp) for BaCrO4.
The ion product (IP) is calculated by multiplying the concentrations of the ions present in the solution at equilibrium.
1. For the Ba(NO3)2 solution:
- Concentration: 1x10-4 M
- Volume: 10 mL (0.01 L)
- Moles of Ba2+ ions = concentration * volume
= 1x10-4 M * 0.01 L
= 1x10-6 moles
2. For the K2CrO4 solution:
- Concentration: 1.0x10-4 M
- Volume: 10 mL (0.01 L)
- Moles of CrO4(2-) ions = concentration * volume
= 1.0x10-4 M * 0.01 L
= 1.0x10-6 moles
To calculate the ion product (IP), multiply the concentrations of Ba2+ and CrO4(2-) ions:
IP = [Ba2+] * [CrO4(2-)] = (1.0x10-6) * (1.0x10-6) = 1.0x10-12
Comparing the ion product (IP) to the solubility product constant (Ksp):
IP = 1.0x10-12
Ksp = 1.2x10-10
Since the ion product (IP) is smaller than the solubility product constant (Ksp), BaCrO4 will not precipitate. It means that the concentration of the ions in the solution is lower than what is required to exceed the solubility product constant and form a precipitate.
Therefore, BaCrO4 will not precipitate upon mixing the Ba(NO3)2 and K2CrO4 solutions.