Three solutions 40mL of 0.2M KNO3; 66mL of 0.3 M Ca(NO3)2; and 39mL 0.185M KCl were poured together to make one new solution. what is the concentration of Cl- and NO3-
figure out how many moles of each ion. For instance, the K ion
moleK+=.185*.039+.2*.040 (sum of both solutions K).
Then, to find the total concentration of each ion(you only have to do two of them), concentrationion= moleion/totalvolume.
To find the concentration of Cl- and NO3- in the new solution, we need to use a method called dilution. Dilution involves combining solutions of different concentrations to create a new solution with a different concentration.
First, we need to determine the moles of each ion in each solution:
For KNO3:
Concentration (C1) = 0.2 M
Volume (V1) = 40 mL = 0.04 L
Moles (n1) = C1 * V1 = 0.2 M * 0.04 L = 0.008 mol
For Ca(NO3)2:
Concentration (C2) = 0.3 M
Volume (V2) = 66 mL = 0.066 L
Moles (n2) = C2 * V2 = 0.3 M * 0.066 L = 0.0198 mol
For KCl:
Concentration (C3) = 0.185 M
Volume (V3) = 39 mL = 0.039 L
Moles (n3) = C3 * V3 = 0.185 M * 0.039 L = 0.007215 mol
Next, we will calculate the total moles of each ion in the new solution by summing up the moles from each solution:
Total moles of Cl- = n1 + n2 + n3
Total moles of NO3- = 2 * (n1 + n2)
Using the calculated moles, we can find the concentration of each ion in the new solution:
Total volume (V_total) = V1 + V2 + V3 = 0.04 L + 0.066 L + 0.039 L = 0.145 L
Concentration of Cl- in the new solution:
C_Cl- = (Total moles of Cl-) / (V_total)
Concentration of NO3- in the new solution:
C_NO3- = (Total moles of NO3-) / (V_total)
Now, substituting the calculated values:
C_Cl- = (0.007215 mol) / (0.145 L) = 0.0498 M
C_NO3- = 2 * (0.008 mol + 0.0198 mol) / (0.145 L) = 0.2193 M
Therefore, the concentration of Cl- in the new solution is 0.0498 M, and the concentration of NO3- is 0.2193 M.