Question

A 12.00g sample of Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) chloride precipitate. What was the molar concentration of chloride ions in the solution (in mol/L)?

Answer 1

Answer: The concentration of chloride ions in the solution is 1.056 mol/L

Step-by-step explanation:

The chemical equation for the reaction of lead (II) nitrate and magnesium chloride follows:

`Pb(NO_3)_2+MgCl_2\rightarrow PbCl_2+Mg(NO_3)_2`

All the chloride ions are getting converted to lead (II) chloride

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of lead (II) chloride = 7.35 g

Molar mass of lead (II) chloride = 278.1 g/mol

Putting values in above equation, we get:

`\text{Moles of }PbCl_2=(7.35g)/(278.1g/mol)=0.0264mol`

1 mole of lead (II) chloride produces 1 mole of lead ions and 2 moles of chloride ions.

So, moles of chloride ions = (2 × 0.0264) = 0.0528 moles

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

`\text{Molarity of chloride ions in solution}=\frac{\text{Moles of chloride ions}* 1000}{\text{Volume of solution (in mL)}}`

Moles of chloride ions = 0.0528 moles

Volume of solution = 50.0 mL

Putting values in above equation, we get:

`\text{Concentration of chloride ions in solution}=(0.0528* 1000)/(50.0)\\\\\text{Concentration of chloride ions in solution}=1.056mol/L`

Hence, the concentration of chloride ions in the solution is 1.056 mol/L