Question

A solution is prepared by combining 5.00 mL of 4.8x10-4 M NaSCN solution, 2.00 mL of 0.21 M Fe(NO3)3 solution and 13.00 mL of 0.3 M HNO3.
Calculate the analytical concentrations of SCN- and Fe3+ in the resulting solution.

Answer 1

Answer:

[SCN-] = 0.00012 M

[Fe3+] = 0.021 M

Explanation:

Step 1: Data given

Volume of a 4.8 *10^-4 M NaSCN solution = 5.00 mL = 0.005 L

Volume of a 0.21 M Fe(NO3)3 = 2.00 mL = 0.002 L

Volume of a 0.3 M HNO3 = 13.00 mL =0.013 L

Step 2: Calculate moles NaSCN

Moles NaSCN = molarity * volume

Moles NaSCN = 4.8 *10^-4 M * 0.005 L

Moles NaSCN = 0.0000024 moles

Step 3: Calculate moles SCN-

For 1 mol NaSCN we have 1 mol SCN-

For 0.0000024 moles NaSCN e have 0.0000024 moles SCN-

Step 4: Calculate moles of Fe(NO3)3

Moles Fe(NO3)3 = 0.21 M * 0.002 L

Moles Fe(NO3)3 = 0.00042 moles

Step 5: Calculate moles Fe3+

For 1 mol Fe(NO3)3 we have 1 mol Fe3+

For 0.00042 mol Fe(NO3)3 we have 0.00042 mol Fe3+

Step 6: Calculate the analytical concentrations

Total volume = 5.00 mL + 2.00 mL + 13.00 mL = 20.00 mL = 0.02 L

[SCN-] = 0.0000024 moles / 0.02 L

[SCN-] = 0.00012 M

[Fe3+] = 0.00042 moles / 0.02 L

[Fe3+] = 0.021 M

Answer 2

Answer:

The analytical concentrations of thiocyanate ions:

$[SCN^-]=0.00012 mol/L$

The analytical concentrations of ferric ions:

$[Fe^{3+}]=0.063 mol/L$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

1) Moles of sodium thiocyanate  = n

Volume of sodium thiocyanate solution = 5.00 mL = 0.005 L

(1 mL = 0.001L)

Molarity of the sodium thiocyanate = $4.8\times 10^{-4} M$

$n=4.8\times 10^{-4} M\times 0.005 L=2.4\times 10^{-6}mol$

1 mole of sodium thiocyanate has 1 mol of thiocyante ions.

So, moles of thioscyanate ions in $2.4\times 10^{-6}mol$  of NaSCN.

$=1\times 2.4\times 10^{-6}mol=2.4\times 10^{-6}mol$

2) Moles of ferric nitrate = n'

Volume of ferric nitrate solution = 2.00 mL = 0.002 L

Molarity of the ferric nitrate = 0.21 M

$n'=0.002 M\times 0.21 L=0.00042 mol$

1 mole of ferric nitrate has 3 moles of ferric ions.

So number of moles of ferric ions in 0.00042 moles of ferric nitrate is :

$3\times 0.00042 mol=0.00126 mol$

Volume of nitric acid = 13.00 mL

Total volume by adding all three volumes of solutions = V

V = 5.00 mL + 2.00 mL + 13.00 mL = 20.00 mL = 0.020 L

The analytical concentrations of thiocyanate ions:

$[SCN^-]=\frac{2.4\times 10^{-6}mol}{0.020 L}=0.00012 mol/L$

The analytical concentrations of ferric ions:

$[Fe^{3+}]=\frac{0.00126 mol}{0.020 L}=0.063 mol/L$