How Do You Calculate the New Concentration of IO3- After Dilution?

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SUMMARY

The concentration of IO3- after dilution can be calculated using the dilution formula C1V1 = C2V2. In this scenario, the initial concentration (C1) is 0.20 mol/L, the initial volume (V1) is 15 mL, and the final volume (V2) is 20 mL after adding 5 mL of water. By rearranging the formula, the final concentration (C2) is determined to be 0.15 mol/L. This calculation demonstrates the direct relationship between concentration and volume in dilution processes.

PREREQUISITES
  • Understanding of molarity and concentration calculations
  • Familiarity with the dilution formula C1V1 = C2V2
  • Basic knowledge of unit conversions (mL to L)
  • Ability to perform algebraic rearrangements
NEXT STEPS
  • Study advanced dilution techniques in chemistry
  • Learn about molarity calculations in different solutions
  • Explore the implications of dilution on reaction rates
  • Investigate the effects of temperature on solubility and concentration
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Chemistry students, laboratory technicians, and educators looking to deepen their understanding of solution concentration and dilution calculations.

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How could i (mathematically) show what the concentration of IO3- in the following solution?

15mL of 0.20mol/L IO3- solution
5mL of water
 
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[tex](15mL)\times\frac{1L}{1000mL}\times\frac{0.20mol~IO_{3}^{-}}{L}=\mbox{moles of the ion in the solution}[/tex]

You know the new volume. Can you find the concentration?
 
Last edited:


To mathematically show the concentration of IO3- in this solution, we can use the formula for dilution: C1V1 = C2V2. In this case, C1 is the initial concentration of the IO3- solution (0.20 mol/L), V1 is the initial volume (15 mL), C2 is the final concentration (unknown), and V2 is the final volume (20 mL, since 15 mL of the initial solution is mixed with 5 mL of water). Rearranging the formula, we get C2 = (C1V1)/V2 = (0.20 mol/L * 15 mL) / 20 mL = 0.15 mol/L. Therefore, the concentration of IO3- in this solution is 0.15 mol/L.
 

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