Question

What volume of a 0.22 m silver nitrate solution is required to precipitate all the cl− ion in the solution as agcl?

Answer 1

Answer:

Depends on the data of chloride solution, an example provided

Explanation:

Silver nitrate is used as a common reagent that is used to precipitate the majority of halogen anions, typically chloride.

We may firstly look at the net ionic equation for the process described. Aqueous silver cation combines with aqueous chloride anion to produce silver chloride, the precipitate:

$Ag^+ (aq) + Cl^- (aq)\rightarrow AgCl (s)$

According to the stoichiometry of this equation, 1 mole of silver cations reacts with 1 mole of chloride anions. We don't know the exact amount of chloride in the context of this problem. Let's identify what we need:

• moles of silver cation, they are found by multiplying the molarity of silver cations by the volume of the solution;
• moles of chloride anions, found the same way.

Since the ratio in stoichiometry is 1 : 1, we obtain:

$c_{Ag^+}V_{Ag^+} = c_{Cl^-}V_{Cl^-}$

Solving for the volume of silver nitrate solution:

$V_{Ag^+} = \frac{c_{Cl^-}V_{Cl^-}}{c_{Ag^+}}$

We only know:

$c_{Ag^+} = 0.22 M$

Then the equation becomes:

$V_{Ag^+} = \frac{c_{Cl^-}V_{Cl^-}}{0.22 M}$

Let's take an example. Imagine we had a 0.53 M solution that was 50.0 mL in volume, then:

$V_{Ag^+} = \frac{0.53 M\cdot 50.0 mL}{0.22 M} = 120.5 mL$