1. You have 88.8 mL of 2.50 M solution of Na2CrO4 (aq). You have 125 mL of a 2.50 M solution of AgNO3 (aq). Calculate the concetration of Na+ when the two solutions are added together.
2. You have 75.0 mL of 2.50 M solution of Na2CrO4 (aq). You have 125 mL of a 1.84 M solution of AgNO3 (aq). Calculate the concetration of CrO4^2- when the two solutions are added together.
umm DrBob222 i still don't understand wat u did.... lik is it possible for a step-by-step help?
1.1.769 M of sodium ions
To find the concentration of Na+ when the Na2CrO4 and AgNO3 solutions are added together, you can use the concept of stoichiometry and the idea that the total volume remains constant. Here's the step-by-step process:
1. Convert the given volumes of solution into liters:
- 88.8 mL = 88.8/1000 = 0.0888 L (volume of Na2CrO4 solution)
- 125 mL = 125/1000 = 0.125 L (volume of AgNO3 solution)
2. Calculate the number of moles of Na+ and the total volume of the mixture after combining the solutions:
- Moles of Na+ from Na2CrO4 solution: (0.0888 L)(2.50 mol/L) = 0.222 mol
- Moles of Na+ from AgNO3 solution: (0.125 L)(2.50 mol/L) = 0.3125 mol
- Total volume of the mixture: 0.0888 L + 0.125 L = 0.2138 L
3. Find the concentration of Na+ in the final mixture:
- Concentration of Na+ (final) = (Total moles of Na+) / (Total volume of the mixture)
- Concentration of Na+ = (0.222 mol + 0.3125 mol) / 0.2138 L
- Concentration of Na+ = 2.431 M
Therefore, the concentration of Na+ in the final mixture is 2.431 M.
To calculate the concentration of CrO4^2- when the Na2CrO4 and AgNO3 solutions are mixed, you can follow a similar approach:
1. Convert the given volumes of solution into liters:
- 75.0 mL = 75.0/1000 = 0.075 L (volume of Na2CrO4 solution)
- 125 mL = 125/1000 = 0.125 L (volume of AgNO3 solution)
2. Calculate the number of moles of CrO4^2- and the total volume of the mixture after combining the solutions:
- Moles of CrO4^2- from Na2CrO4 solution: (0.075 L)(2.50 mol/L) = 0.1875 mol
- Moles of CrO4^2- from AgNO3 solution: (0.125 L)(1.84 mol/L) = 0.230 mol
- Total volume of the mixture: 0.075 L + 0.125 L = 0.2 L
3. Find the concentration of CrO4^2- in the final mixture:
- Concentration of CrO4^2- (final) = (Total moles of CrO4^2-) / (Total volume of the mixture)
- Concentration of CrO4^2- = (0.1875 mol + 0.230 mol) / 0.2 L
- Concentration of CrO4^2- = 2.085 M
Therefore, the concentration of CrO4^2- in the final mixture is 2.085 M.
1. You have 2.50M x 0.0.0888 L = ?? moles Na2CrO4. (Na^+) is twice that.
2. Write the equation. Balance it. Ag2CrO4 will ppt. Determine moles Na2CrO4 and moles AgNO3, determine which material is in excess and how much ppt is formed, from that find CrO4^-
Post your work if you get stuck.