The concentration and volume of reagents combined in each trial are listed in the table below:
Trial 00020 M 00020 M H2O(mL)
Fe(NO3)3 (mL) KSCN (mL)
1 5 2 3
2 5 3 2
3 5 4 1
4 5 5 0
Initially (right after combining and before any reaction occurs) the concentration of SCNÐ in trial 1 is? ____M
See my comment at your post above.
To find the initial concentration of SCN- in trial 1, you need to consider the balanced chemical equation for the reaction between Fe(NO3)3 and KSCN:
Fe(NO3)3 + 3KSCN → Fe(SCN)3 + 3KNO3
From the table, you can see that in trial 1, 5 mL of Fe(NO3)3 and 2 mL of KSCN are combined. The total volume of the solution is 5 mL + 2 mL + 3 mL (H2O) = 10 mL.
Since the concentrations of both Fe(NO3)3 and KSCN are given as 0.0020 M, you can assume that they are the same in the final solution.
First, convert the volumes to liters:
5 mL = 0.005 L
2 mL = 0.002 L
Since concentration (C) = moles (n) / volume (V), you can use the equation:
C1V1 = C2V2
C1 × 0.005 L = 0.0020 M × 0.005 L + SCN- concentration × 0.002 L
Rearranging the equation and substituting the known values:
C1 × 0.005 L = 0.0020 M × 0.005 L
C1 = (0.0020 M × 0.005 L) / 0.005 L
C1 = 0.0020 M
Therefore, the initial concentration of SCN- in trial 1 is 0.0020 M.