Question

What is the concentration of Br- (aq) in a solution prepared by mixing 75.0 mL of 0.62 M iron (III) bromide with 75.0 mL of water? Assume volumes are additive. the answer is 0.93 but I got 1.86 and 3.1 I'm not even sure anymore!

Answer 1

Answer on Question#55191 – Chemistry – General Chemistry

Question:

What is the concentration of Br⁻ (aq) in a solution prepared by mixing 75.0 mL of 0.62 M iron (III) bromide with 75.0 mL of water? Assume volumes are additive. The answer is 0.93 but I got 1.86 and 3.1 I'm not even sure anymore!

Solution:

v – The number of moles (mol);

V – The volume (L);

C – The molar concentration (mol×L⁻¹);

$$C = \frac{v}{V};$$

V1 = 0.075 L; C(FeBr₃) = 0.62 \, \text{M};

v(FeBr₃) = C(FeBr₃) × V1;

v(FeBr₃) = 0.62 × 0.075 = 0.0465 \, \text{mol};

The molecule of FeBr₃ contains 3 Br⁻ ions. According to this statement: v(Br⁻) = v(FeBr₃) × 3;

v(Br⁻) = 3 × 0.0465 = 0.1395 mol;

V2 = V1 + ΔV; ΔV = 0.075 L; V2 = 0.075 + 0.075 = 0.15 L;

$$C2(\text{Br}) = \frac{v(\text{Br})}{V2};$$ $$C2(\text{Br}^{-}) = \frac{0.1395}{0.15} = 0.93 \, \text{mol} \times \text{L}^{-1};$$ $$C2(\text{Br}^{-}) = 0.93 \, \text{mol} \times \text{L}^{-1};$$

Answer: 0.93 M;

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