Question

Suppose that 75 mL of 5.00 M iron (III) chloride is combined with 81 mL of 4.66 M calcium chloride. What is the concentration of Cl- ion in the resulting solution?

Answer 1

Answer: The concentration of chloride ion in the resulting solution is 12.04 M

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}` .......(1)

• For iron (III) chloride:

Molarity of solution = 5.00 M

Volume of solution = 75.00 mL

Putting values in equation 1, we get:

`5.00M=\frac{\text{Moles of iron (III) chloride}* 1000}{75}\\\\\text{Moles of iron (III) chloride}=(5.00* 75)/(1000)=0.375moles`

1 mole of iron (III) chloride produces 1 mole of
`Fe^(3+)` ions and 3 moles of
`Cl^-` 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:

`4.66M=\frac{\text{Moles of calcium chloride}* 1000}{81}\\\\\text{Moles of calcium chloride}=(4.66* 81)/(1000)=0.377moles`

1 mole of calcium chloride produces 1 mole of
`Ca^(2+)` ions and 2 moles of
`Cl^-` 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:

`\text{Molarity of }Cl^-\text{ ions}=(1.879mol* 1000)/(156)\\\\\text{Molarity of }Cl^-\text{ ions}=12.04M`

Hence, the concentration of chloride ion in the resulting solution is 12.04 M