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 Fe3+ in trial 2 is
____ M.
I can't make heads or tails of the post. We have a tough time spacing on these boards. In general, however, calculate (Fe^+3) from mols in the solution (molarity x L) and divide that by total volume of the diluted solution.
To determine the concentration of Fe3+ in trial 2, we need to consider the initial concentrations of Fe(NO3)3 and KSCN reagents, as well as the total volume of the solution.
According to the table provided, the initial concentration of Fe(NO3)3 in each trial is 0.020 M, and the initial concentration of KSCN is also 0.020 M. In trial 2, we mixed 5 mL of Fe(NO3)3 with 3 mL of KSCN.
To find the total volume of the solution in trial 2, we sum the volumes of Fe(NO3)3 and KSCN:
Total volume = volume of Fe(NO3)3 + volume of KSCN = 5 mL + 3 mL = 8 mL.
Now, we can calculate the concentration of Fe3+ in trial 2 using the equation:
C1 * V1 = C2 * V2
where C1 is the initial concentration of Fe(NO3)3, V1 is the volume of Fe(NO3)3, C2 is the concentration of Fe3+ (what we want to find), and V2 is the total volume of the solution.
Plugging in the values we know:
0.020 M * 5 mL = C2 * 8 mL
Now, we can solve for C2 (the concentration of Fe3+):
(0.020 M * 5 mL) / 8 mL = C2
0.0125 M = C2
Therefore, the concentration of Fe3+ in trial 2, initially, is 0.0125 M.