Question

The titration reaction can be represented as MgCl2(aq) 2 NaOH(aq) → Mg(OH)2(s) 2 NaCl(aq) The solubility equilibrium is written as Mg(OH)2(s) ⇌ Mg2 (aq) 2 OH-(aq) Kemmi is trying to teach Noke how to calculate the concentration of the Mg2 cation at the titration endpoint. She explains that the initial MgCl2 concentration must be multiplied by the volume of MgCl2 pipetted and divided by the total volume at the endpoint. (Note: the Calculation page in the Lab Manual has a similar equation for the lead cation.) Help Noke do her calculation. For the first titration in the set the initial concentration of MgCl2(aq) was 0.0050 M, the volume measured was 100.00 mL, and the total volume of solution at the endpoint was 104.12 mL. Calculate the concentration of Mg2 at the endpoint. (Answer with 2 significant digits and units of M)

Answer 1

The concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

Step-by-step explanation:

The titration reaction involved: MgCl₂(aq) + 2 NaOH(aq) → Mg(OH)₂(s) + 2 NaCl(aq)

and the solubility equilibrium for the product: Mg(OH)₂(s) ⇌ Mg²⁺ (aq) + 2 OH⁻(aq)

Given: Concentration of MgCl₂: M₁ = 0.005 M, volume of MgCl₂: V₁ = 100 mL, Total volume of solution at end point = 104.12 mL

Concentration of Mg²⁺ cation at end point: M₂ = ?M

Now, to calculate the concentration of Mg²⁺ cation at end point, we use the equation: M₁ × V₁ = M₂ × V₂

⇒ M₂ = M₁ × V₁ ÷ V₂

⇒ M₂ = (0.005 M × 100 mL) ÷ 104.12 mL

⇒ M₂ = 0.0048 M (2 significant digits)

Therefore, the concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M