Respuesta :
Answer:
The analytical concentrations of thiocyanate ions:
[tex][SCN^-]=0.00012 mol/L[/tex]
The analytical concentrations of ferric ions:
[tex][Fe^{3+}]=0.063 mol/L[/tex]
Explanation:
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
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 = [tex]4.8\times 10^{-4} M[/tex]
[tex]n=4.8\times 10^{-4} M\times 0.005 L=2.4\times 10^{-6}mol[/tex]
1 mole of sodium thiocyanate has 1 mol of thiocyante ions.
So, moles of thioscyanate ions in [tex]2.4\times 10^{-6}mol[/tex] of NaSCN.
[tex]=1\times 2.4\times 10^{-6}mol=2.4\times 10^{-6}mol[/tex]
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
[tex]n'=0.002 M\times 0.21 L=0.00042 mol[/tex]
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 :
[tex]3\times 0.00042 mol=0.00126 mol[/tex]
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:
[tex][SCN^-]=\frac{2.4\times 10^{-6}mol}{0.020 L}=0.00012 mol/L[/tex]
The analytical concentrations of ferric ions:
[tex][Fe^{3+}]=\frac{0.00126 mol}{0.020 L}=0.063 mol/L[/tex]
Based on the data provided, the analytical concentrations of SC N⁻ and Fe³⁺ in the solution is 1.2 * 10⁻⁴ M and 2.4 * 10⁻² M.
What is the analytical concentration of SC N⁻ and Fe³⁺ in the solution?
The concentration of a solution is calculated using the formula below:
- Concentration = moles/volume in L
- moles = molarity * volume in L
The moles of the ions are first determined:
moles of NaSC N⁻ in 5.00 mL in 4.8 * 10⁻⁴ is calculated below:
moles of NaSC N⁻ = 0.005 * 4.8 * 10⁻⁴
moles of NaSC N⁻ = 2.4 * 10⁻⁶ moles
1 mole NaSC N ⁻ produces 1 mole SC N⁻
moles of SC N⁻ = 2.4 * 10⁻⁶ moles
moles of Fe(NO₃)₃ in 2.00 mL in 0.21 M solution is calculated below;
moles of Fe(NO₃)₃ = 0.002 * 0.21
moles of Fe(NO₃)₃ = 4.2 * 10⁻⁴ moles
1 mole Fe(NO₃)₃ produces 1 mole Fe³⁺
moles of Fe³⁺ = 4.2 * 10⁻⁴ moles
Total volume of solution = 0.002 + 0.005 + 0.013
Total volume of solution = 0.020 L
Concentration of SC N⁻ = 2.4 * 10⁻⁶/0.020
- Concentration of SC N⁻ = 1.2 * 10⁻⁴ m
Concentration of Fe³⁺ = 4.2 * 10⁻⁴/0.02
- Concentration of Fe³⁺ = 2.4 * 10⁻² M
Therefore, the analytical concentrations of SC N⁻ and Fe³⁺ in the solution is 1.2 * 10⁻⁴ M and 2.4 * 10⁻² M.
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