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Dilutions. Solve problems involving the dilution of solutions. Include: dilution of stock solutions, mixing common solutions with different volumes and.

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Presentation on theme: "Dilutions. Solve problems involving the dilution of solutions. Include: dilution of stock solutions, mixing common solutions with different volumes and."— Presentation transcript:

1 Dilutions

2 Solve problems involving the dilution of solutions. Include: dilution of stock solutions, mixing common solutions with different volumes and concentrations Additional KEY Terms

3 Labs buy highly concentrated solutions to save storage space - stock solutions. To dilute a solution means to add more solvent to small volume of stock without adding more solute. Moles of solute remains unchanged. moles s = moles d

4 moles d moles s

5 M = moles Where: C 1 is the initial concentration (stock) V 1 is the initial volume of solution (stock) C 2 is the concentration after dilution V 2 is the volume after dilution moles s = moles d C 1 V 1 = C 2 V 2 L

6 If 500.0 mL of water is added to 300.0 mL of a 0.100 mol/L solution of NaCl, what is the new concentration? V 2 = 300.0 + 500.0 = 800.0 mL = C1C1 V1V1 C2C2 V2V2 = (0.100)C2C2 (0.800) (0.300)0.0375 M

7 What is the concentration of the stock solution if 50.0 mL of the stock is diluted to make 220.0 mL of a 0.400 M solution? = C1C1 V1V1 C2C2 V2V2 = (0.400) C1C1 (0.0500) (0.220) 1.76 mol/L

8 How much water must be added to 25.0 mL of a 1.00 M stock solution of NaOH to make a 0.100 M solution? = C1C1 V1V1 C2C2 V2V2 = (1.00) V2V2 (0.100) (0.0250) 0.250 L 0.250 L – 0.0250 L = 0.225 L

9 To prepare a diluted solution: 1.Determine the volume of stock solution needed. 2.Add half of the volume of water to the beaker. 3.Add the stock solution. 4.Add remaining water to the total volume and mix.

10 Mixing two or more solutions having the same solute, but different concentrations: = (C 1 V 1 + C 2 V 2 + …) C Final (V 1 + V 2 + …) = C Final Total volume Total moles

11 450.0 mL of a 0.150 M NaCl solution is mixed with 125 mL of a 0.220 M NaCl solution. = 0.165 M = (C 1 V 1 + C 2 V 2 ) C Final (V 1 + V 2 ) = (0.150)(0.450) + (0.220)(0.125) CFCF (0.450 + 0.125)

12  A diluted solution has the same number of moles of solute as the given stock, but a new volume.  Since the moles of solute remains the same, the dilution equation is C 1 V 1 = C 2 V 2  Mix two or more solutions containing the same solute - use the equation

13 CAN YOU / HAVE YOU? Solve problems involving the dilution of solutions. Include: dilution of stock solutions, mixing common solutions with different volumes and concentrations Additional KEY Terms


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