Question

Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn’t change when the sodium chloride is dissolved in it.

Answer 1

`\large \boxed{\text{ 0.066 mol/L}}`

Step-by-step explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

`\text{Moles of NaCl} = \text{245 mg NaCl} * \frac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_(3)= \text{50. mL AgNO}_(3) * \frac{\text{0.0180 mmol AgNO}_(3)}{\text{1 mL AgNO}_(3)} = \text{0.900 mmol AgNO}_(3)`

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

`\text{Moles of AgCl} = \text{4.192 mmol NaCl} * \frac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}`

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

`\text{Moles of AgCl} = \text{0.900 mmol AgNO}_(3) * \frac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_(3)} = \text{0.900 mmol AgCl}`

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

`\text{[Cl$^(-)$] } = \frac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}`