You have 75.0 mL of a 2.50 M solution of Na2CrO4 (aq). You also have 125 mL of a 2.15 M solution of AgNO3 (aq). Calculate the concentration of NO3 when the two solutions are added together.
Well, when the solutions are mixed together, it's like a chemistry version of a mixer party! Na2CrO4 and AgNO3 are both guests, and they'll be mingling in a single container.
To find the concentration of NO3 in the final solution, we need to consider the total volume of the solution after mixing. So let's do some math magic:
The initial volume of Na2CrO4 solution is 75.0 mL, and the initial concentration is 2.50 M. So, we have:
Initial moles of Na2CrO4 = initial volume (in L) × initial concentration
= 75.0 mL ÷ 1000 mL/L × 2.50 M
Similarly, for AgNO3, we have:
Initial moles of AgNO3 = initial volume (in L) × initial concentration
= 125 mL ÷ 1000 mL/L × 2.15 M
After mixing, the total volume will be 75.0 mL + 125 mL = 200 mL. Let's convert that to liters:
Total volume of solution = 200 mL ÷ 1000 mL/L
Now, we can use the fact that the number of moles remains constant before and after mixing. The number of moles of NO3 will be the same as the number of moles of AgNO3:
Final moles of NO3 = Initial moles of AgNO3
Finally, we can calculate the concentration of NO3 in the final solution:
Final concentration of NO3 = Final moles of NO3 ÷ Total volume of solution
So, get your calculator ready and perform the necessary calculations!
To calculate the resulting concentration of NO3 when the two solutions are mixed together, we need to find the moles of NO3 in each solution and then add them together.
Step 1: Calculate the moles of Na2CrO4 in the first solution.
Moles of Na2CrO4 = (concentration of Na2CrO4) x (volume of Na2CrO4)
Moles of Na2CrO4 = 2.50 M x 0.075 L
Moles of Na2CrO4 = 0.1875 mol
Step 2: Calculate the moles of AgNO3 in the second solution.
Moles of AgNO3 = (concentration of AgNO3) x (volume of AgNO3)
Moles of AgNO3 = 2.15 M x 0.125 L
Moles of AgNO3 = 0.26875 mol
Step 3: Add the moles of NO3 together.
Total moles of NO3 = moles of Na2CrO4 + moles of AgNO3
Total moles of NO3 = 0.1875 mol + 0.26875 mol
Total moles of NO3 = 0.45625 mol
Step 4: Determine the final volume of the solution.
Final volume = volume of Na2CrO4 + volume of AgNO3
Final volume = 75.0 mL + 125 mL
Final volume = 200 mL
Step 5: Calculate the concentration of NO3 in the final solution.
Concentration of NO3 = (total moles of NO3) / (final volume)
Concentration of NO3 = 0.45625 mol / 0.200 L
Concentration of NO3 = 2.28125 M
Therefore, the concentration of NO3 in the final solution is 2.28125 M.
To calculate the concentration of NO3 when the two solutions are mixed together, we first need to determine the number of moles of Na2CrO4 and AgNO3 that are present in each solution.
For the Na2CrO4 solution:
Given volume of the solution = 75.0 mL
Given molarity (concentration) of the solution = 2.50 M
Using the formula:
moles = volume (in liters) × molarity
Converting the volume to liters:
75.0 mL = 75.0 / 1000 L = 0.075 L
Calculating the moles of Na2CrO4:
moles of Na2CrO4 = 0.075 L × 2.50 mol/L = 0.1875 mol
For the AgNO3 solution:
Given volume of the solution = 125 mL
Given molarity (concentration) of the solution = 2.15 M
Converting the volume to liters:
125 mL = 125 / 1000 L = 0.125 L
Calculating the moles of AgNO3:
moles of AgNO3 = 0.125 L × 2.15 mol/L = 0.26875 mol
Since the compound Na2CrO4 dissociates into two moles of Na+ ions and one mole of CrO4^2- ions, the number of moles of NO3 ions can be determined from the ratio of Na2CrO4 to AgNO3.
From the balanced equation:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
The ratio of moles of Na2CrO4 to NO3 ions is 1:2.
Therefore, the moles of NO3 ions is twice the number of moles of Na2CrO4.
Calculating the moles of NO3 ions:
moles of NO3 = 2 × 0.1875 mol = 0.375 mol
Finally, to find the concentration (molarity) of NO3 ions when the two solutions are mixed, we divide the moles of NO3 ions by the total volume of the solution.
Total volume of the solution = volume of Na2CrO4 solution + volume of AgNO3 solution
= 75.0 mL + 125 mL
= 200 mL = 200 / 1000 L = 0.20 L
Concentration of NO3 = moles of NO3 / total volume of the solution
Concentration of NO3 = 0.375 mol / 0.20 L = 1.875 M
Therefore, the concentration of NO3 when the two solutions are mixed together is 1.875 M.
The Ag^+ and the CrO4^2- react to form Ag2CrO4 but the NO3^- is not involved in the reaction (it is a spectator ion); therefore, it's concentration is not affected. It will be diluted, of course, due to the added volume from the Na2CrO4.
2.15 M x [(125/(125+75)] = ? M.