Molarity and Dilution: Concentration Calculations
Calculate solution concentrations, perform dilution calculations, and determine volumes needed for solution preparation using M₁V₁ = M₂V₂.
Problem Scenario
A lab needs to prepare three solutions from a stock solution of 12.0 M HCl. (a) What volume of stock is needed to make 500.0 mL of 2.00 M HCl? (b) If 25.0 mL of stock solution is diluted to 1.50 L, what is the final concentration? (c) How many grams of NaOH (molar mass 40.00 g/mol) are needed to prepare 250.0 mL of 0.500 M NaOH solution?
Given Data
Requirements
- Use M₁V₁ = M₂V₂ to find the stock volume for part (a)
- Use M₁V₁ = M₂V₂ to find the final concentration for part (b)
- Calculate mass of NaOH from molarity and volume for part (c)
Solution
Step 1:
Part (a): Apply M₁V₁ = M₂V₂. (12.0 M)(V₁) = (2.00 M)(500.0 mL). V₁ = (2.00 × 500.0)/12.0 = 83.3 mL of stock HCl needed.
Step 2:
Part (b): Apply M₁V₁ = M₂V₂. (12.0 M)(25.0 mL) = (M₂)(1500 mL). Note: convert 1.50 L to 1500 mL. M₂ = (12.0 × 25.0)/1500 = 0.200 M HCl.
Step 3:
Part (c): This is NOT a dilution — we are dissolving a solid. First find moles needed: n = M × V = 0.500 mol/L × 0.2500 L = 0.125 mol NaOH.
Step 4:
Convert moles to grams: mass = n × molar mass = 0.125 mol × 40.00 g/mol = 5.00 g NaOH. Dissolve 5.00 g of NaOH in water and dilute to a final volume of 250.0 mL.
Final Answer
(a) 83.3 mL of 12.0 M HCl stock needed. (b) Final concentration is 0.200 M HCl. (c) 5.00 g of NaOH must be dissolved and diluted to 250.0 mL.
Key Takeaways
- ✓M₁V₁ = M₂V₂ works for dilutions because moles of solute are conserved
- ✓Always keep units consistent — convert L to mL or vice versa before calculating
- ✓For solid solutes, use n = M × V to find moles, then convert to grams
- ✓Molarity = moles of solute / liters of solution (not solvent)
Common Errors to Avoid
- ✗Mixing up M₁V₁ = M₂V₂ with the wrong variable — always identify stock vs dilute
- ✗Forgetting to convert volume units (mL vs L) before plugging into equations
- ✗Applying M₁V₁ = M₂V₂ to solid dissolution problems where it doesn't apply
- ✗Confusing volume of solution with volume of solvent — molarity uses total solution volume
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Common questions about this problem type
Yes, as long as the solute doesn't change chemically during dilution. It works because dilution only adds solvent — the moles of solute stay the same. It doesn't work for mixing two different solutes or for reactions.
Because volumes are not always additive when mixing. Intermolecular forces between solute and solvent can cause the total volume to be slightly more or less than the sum of the parts. Diluting to a final volume in a volumetric flask ensures the exact concentration.