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Dilution

Part of the Stoichiometry section of Coddy's Chemistry journey. Lesson 16 of 41.

Adding water dilutes a solution. The moles of solute stay the same and spread through a bigger volume, so the concentration falls:

c₁V₁ = c₂V₂

200 mL200\ \mathrm{mL} of 0.2 M0.2\ \mathrm{M} CaCl₂ plus 800 mL800\ \mathrm{mL} of water makes 1000 mL1000\ \mathrm{mL}: c2=0.2×2001000=0.04 Mc_2 = \frac{0.2 \times 200}{1000} = 0.04\ \mathrm{M}.

The same particles before and after dilution

Same particles, five times the volume

V₂ is the total volume, not the water added. The ions follow: [Ca²⁺] = 0.04 M0.04\ \mathrm{M}, [Cl⁻] = 0.08 M0.08\ \mathrm{M}.

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Challenge

150 mL150\ \mathrm{mL} of 0.4 M0.4\ \mathrm{M} MgCl₂ is poured into a flask, and 450 mL450\ \mathrm{mL} of water is added. MgCl₂ dissolves as MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq). What is [Cl⁻] in the diluted solution?

Try it yourself

Physics diagram: beaker

Solve for cClc_{Cl}

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This lesson includes a short quiz. Start the lesson to answer it and track your progress.

All lessons in Stoichiometry