Volume needed to precipitate something

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SUMMARY

To precipitate 25.0 mL of 0.0832 M nickel(II) sulfate using 0.100 M lead nitrate, first calculate the moles of sulfate ions present in the nickel sulfate solution. This requires determining the moles of SO42- in the solution, which equals 0.00208 moles. Since lead(II) ions react with sulfate ions on a 1:1 mole basis, 0.00208 moles of lead(II) nitrate are needed. Finally, calculate the volume of 0.100 M lead nitrate solution required to provide this amount of lead ions.

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  • Understanding of molarity and molar calculations
  • Knowledge of precipitation reactions and stoichiometry
  • Familiarity with lead(II) nitrate and nickel(II) sulfate chemical properties
  • Basic skills in performing volume and concentration conversions
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  • Calculate the moles of ions in various precipitation reactions
  • Study the solubility product constant (Ksp) for lead(II) sulfate
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starsun
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what volume of 0.100 M lead nitrate is required to precipitate completely 25.0 mL of 0.0832 M nickel(II) sulfate?

what are the steps to completing this problem? thanks
 
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Determine how many moles of SO4(-2) there are in the NiSO4 solution. Since Pb(+2) reacts with sulfate on a mole for mole basis, that's the number of moles of Pb(+2) you need. Then determine the volume of Pb(NO3)2 solution which contains that many moles. By the way, since PbSO4 is not COMPLETELY insoluble (nothing is), you'll never precipitate out every last sulfate ion.
 

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