A lab technician adds 3.50 mL of a 0.00450 M stock solution of barium chloride to a test tube, then adds other solutions so that the total volume is 10.00 mL. In the absence of reactions, the molarity of barium ions is M in the solution and the molarity of chloride ions is M in solution.
IMPORTANT: Report each answer to 3 significant figures, I made this as a free response fill-in question and the question will be graded credit/no credit. For example, 0.0000123 or 1.23. Do NOT include units as they are already provided for you.

[Ba²⁺] = 1.58 × 10⁻³ M

[Cl⁻] = 3.16 × 10⁻³ M.

Explanation:

Step 1: Calculate the molar concentration of BaCl₂

We have 3.50 mL (V₁) of a 0.00450 M (C₁) BaCl₂ solution and we dilute it to a final volume of 10.00 mL (V₂). We can calculate the final concentration of BaCl₂ by using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.00450 M × 3.50 mL / 10.00 mL = 1.58 × 10⁻³ M

Step 2: Calculate the molar concentrations of the Ba²⁺ and Cl⁻ ions

Let's consider the dissociation of BaCl₂.

BaCl₂(aq) ⇒ Ba²⁺(aq) + 2 Cl⁻(aq)

The molar ratio of BaCl₂ to Ba²⁺ is 1:1. The molar concentration of Ba²⁺ is 1/1 × 1.58 × 10⁻³ M = 1.58 × 10⁻³ M.

The molar ratio of BaCl₂ to Cl⁻ is 1:2. The molar concentration of Cl⁻ is 2/1 × 1.58 × 10⁻³ M = 3.16 × 10⁻³ M.

3 1/4 %

explanation:

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