you have75.0ml of 2.5 M Solution of Na2CrO4(aq). You also have 125ml of a 1.74M solution of AgNO3(aq)Calculate the concentration of NO3 when the two solutions are added together
Nitrate never enters into the reaction.
Na2CrO4 + 2AgNO3 ==> Ag2CrO4 + 2NaNO3
(AgNO3) = 1.74M x 125 mL = 217.5 mmoles.
(NO3^-) at the end is 217.5 mmoles/total mL volume. That is 125 mL AgNO3 + 75 mL from the Na2CrO4. So nitrate = mols/L = or 217.5 mmoles/200 mL ??.
No se
To calculate the concentration of NO3 when the two solutions are added together, we need to determine the number of moles of Na2CrO4 and AgNO3 present in each solution.
Step 1: Calculate the number of moles of Na2CrO4:
We have 75.0 ml of a 2.5 M solution of Na2CrO4.
Molarity (M) is defined as moles of solute per liter of solution.
Therefore, we can calculate the number of moles by using the formula:
moles of solute = molarity × volume (in liters)
Converting 75.0 ml to liters:
75.0 ml × (1 L / 1000 ml) = 0.075 L
Now, we can calculate the number of moles of Na2CrO4:
moles of Na2CrO4 = 2.5 M × 0.075 L = 0.1875 moles
Step 2: Calculate the number of moles of AgNO3:
We have 125 ml of a 1.74 M solution of AgNO3.
Converting 125 ml to liters:
125 ml × (1 L / 1000 ml) = 0.125 L
Now, we can calculate the number of moles of AgNO3:
moles of AgNO3 = 1.74 M × 0.125 L = 0.2175 moles
Step 3: Calculate the total moles of NO3:
Since one mole of AgNO3 produces one mole of NO3, the total moles of NO3 is equal to the moles of AgNO3.
Total moles of NO3 = 0.2175 moles
Step 4: Calculate the total volume of the solution:
The total volume of the solution is the sum of the individual volumes of Na2CrO4 and AgNO3.
Total volume = 75.0 ml + 125 ml = 200 ml
Finally, we can calculate the concentration of NO3 in the combined solution:
Concentration of NO3 = Total moles of NO3 / Total volume (in liters)
Concentration of NO3 = 0.2175 moles / (200 ml × (1 L / 1000 ml)) = 0.2175 moles / 0.2 L = 1.0875 M
Therefore, the concentration of NO3 when the Na2CrO4 and AgNO3 solutions are combined is 1.0875 M.