Please help me answer the empty boxes I understand you need to use M1V1=M2V2 but
ID: 890007 • Letter: P
Question
Please help me answer the empty boxes I understand you need to use M1V1=M2V2 but I don't know how to apply it. The table below contains the volume of reagents combined in a series of trials for the following reaction. ng concentrations after all reagent solutions have been combined. Trial 4 Trial mL of o.0020 M Fe(NOs)s 5.0 mL of 0.20 M Fe(NO3)3 Trial 2 5.0 0 3.0 2.0 Trial 3 5.0 0 4.0 1.0 Trial 1 9.0 1.0 0 2.0 3.0 mL of H2O [Fe3] ions in the reaction mixture Number Number Number Number | M |||| | M |||| (just aftermixing.before anyja reaction has taken place) Number SCN 1 ions in the reaction Number mixture Number Gust after mixing, before any reaction has taken place) * 10Explanation / Answer
Solution
Trial 1
Fe(NO3)3 = 0.0020 M = 5.0 ml
KSCN=0.0020 M = 2.0 ml
Water = 3.0 ml
Total volume = 5.0+2.0+3.0 = 10 ml
New molarity of the Fe^3+ = 0.0020 M * 5.0 ml / 10 ml = 0.001 M
New molarity of SCN^- = 0.0020 M * 2 ml / 10 ml = 0.0004 M
Trial 2
Fe(NO3)3 = 0.0020 M = 5.0 ml
KSCN=0.0020 M = 3.0 ml
Water = 2.0 ml
Total volume = 5.0+3.0+2.0 = 10 ml
New molarity of the Fe^3+ = 0.0020 M * 5.0 ml / 10 ml = 0.001 M
New molarity of SCN^- = 0.0020 M * 3 ml / 10 ml = 0.0006 M
Trial 3
Fe(NO3)3 = 0.0020 M = 5.0 ml
KSCN=0.0020 M = 4.0 ml
Water = 1.0 ml
Total volume = 5.0+4.0+1.0 = 10 ml
New molarity of the Fe^3+ = 0.0020 M * 5.0 ml / 10 ml = 0.001 M
New molarity of SCN^- = 0.0020 M * 4 ml / 10 ml = 0.0008 M
Trial 4
Fe(NO3)3 = 0.20 M = 9.0 ml
KSCN=0.0020 M = 1.0 ml
Total volume = 9.0+1.0 = 10 ml
New molarity of the Fe^3+ = 0.20 M * 9.0 ml / 10 ml = 0.18 M
New molarity of SCN^- = 0.0020 M * 1.0 ml / 10 ml = 0.0002 M
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