Respuesta :
Answer: The concentration of chloride ion in the resulting solution is 12.04 M
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .......(1)
- For iron (III) chloride:
Molarity of solution = 5.00 M
Volume of solution = 75.00 mL
Putting values in equation 1, we get:
[tex]5.00M=\frac{\text{Moles of iron (III) chloride}\times 1000}{75}\\\\\text{Moles of iron (III) chloride}=\frac{5.00\times 75}{1000}=0.375moles[/tex]
1 mole of iron (III) chloride produces 1 mole of [tex]Fe^{3+}[/tex] ions and 3 moles of [tex]Cl^-[/tex] ions
Moles of chloride ions in iron (III) chloride solution = (3 × 0.375) = 1.125 moles
- For calcium chloride:
Molarity of solution = 4.66 M
Volume of solution = 81 mL
Putting values in equation 1, we get:
[tex]4.66M=\frac{\text{Moles of calcium chloride}\times 1000}{81}\\\\\text{Moles of calcium chloride}=\frac{4.66\times 81}{1000}=0.377moles[/tex]
1 mole of calcium chloride produces 1 mole of [tex]Ca^{2+}[/tex] ions and 2 moles of [tex]Cl^-[/tex] ions
Moles of chloride ions in calcium chloride solution = (2 × 0.377) = 0.754 moles
Now, calculating the chloride ions in the solution by using equation 1, we get:
Total moles of chloride ions = [1.125 + 0.754] = 1.879 moles
Total volume of base solution = [75 + 81] = 156 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of }Cl^-\text{ ions}=\frac{1.879mol\times 1000}{156}\\\\\text{Molarity of }Cl^-\text{ ions}=12.04M[/tex]
Hence, the concentration of chloride ion in the resulting solution is 12.04 M
