# 1. a) Calculate the volume of

1. a) Calculate the volume of 2.00 X 10^-3 M SCN^- required toprepare 50 mL of 2.00 X 10^-4 M SCN^-.

b) Calculate [FeSCN^2+] for each standard solution:

Sample ID | 2.00 X 10^-4 M SCN^- (mL) (limiting reagent) | H2O (mL) | 0.200 M Fe(NO3)3 (mL) (excess reagent) |

1 (blank) | 0 | 9.00 | 1.00 |

2 | 1.00 | 8.00 | 1.00 |

3 | 3.00 | 6.00 | 1.00 |

4 | 5.00 | 4.00 | 1.00 |

5 | 7.00 | 2.00 | 1.00 |

6 | 9.00 | 0 | 1.00 |

Assume that all SCN^- ions react (mol SCN^- = mol FeSCN^2+ ).Thus the calculation of [FeSCN^2+] is: mol FeSCN^2+ / L of totalsolution.

Answer:

**a)**

Concentration of available solution of SCN^{–} =

Let the volume of the available solution needed to make therequired solution be

Target concentration of the required solution =

Target volume of the required solution =

Now, we know that the number of moles of SCN^{–} will besame before and after dilution.

Hence, the product of concentration and volume will also be samebefore and after dilution. Hence, we can write

**Hence, 5.0 mL of the given solution is required to make50 mL of the required solution.**

**b)**

Note that SCN^{–} is the limiting reactant andFe(NO_{3})_{3} is the excess reagent.

The reaction for the complex formation can be written as

Hence, the number of moles of SCN^{–} present in thesolution will be equal to the number of moles of complexFe(SCN)^{2+} formed.

Given that the concentration of SCN- solution used =

Number of moles of SCN- can be calculated by multiplying theconcentration of SCN- solution used and the volume used in litres.**Hence, we can create the following table to calculate thefinal concentrations of complex formed.**

Sample ID |
Volume of SCN^{–}(L) |
Moles of SCN^{–} =Moles of Fe(SCN)^{2+} |
Total Volume(L) |
,M |

1 | 0 | |||

2 | 0.001 | |||

3 | 0.003 | |||

4 | 0.005 | |||

5 | 0.007 | |||

6 | 0.009 |

*Note: the answers are in red.*