Volume needed to precipitate something

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To determine the volume of 0.100 M lead nitrate needed to completely precipitate 25.0 mL of 0.0832 M nickel(II) sulfate, start by calculating the moles of sulfate ions (SO4(-2)) in the nickel sulfate solution. The reaction between lead ions (Pb(+2)) and sulfate occurs on a one-to-one mole basis, meaning the moles of Pb(+2) required will equal the moles of SO4(-2) present. After finding the required moles of lead, calculate the corresponding volume of the lead nitrate solution needed to provide that amount. It's important to note that lead sulfate (PbSO4) is not completely insoluble, indicating that not all sulfate ions will precipitate out.
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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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