A 2.0 mL aliquot of 0.001 M NaSCN is diluted to 20.0 mL with 0.2 M Fe(NO3)3 and 0.1 M HNO3. How many moles of SCN– are present?
- hcbiochemLv 74 months agoBest answer
0.0020 L X 0.001 mol/L SCN- = 2X10^-6 moles SCN- were added.
In this reaction, because [Fe3+] is so high, all of the SCN- will be in the form of the Fe(SCN)2+ complex. Its molarity will be 1X10^-6 mol / 0.0200 L = 5.0X10^-5 M Fe(SCN)2+
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