A solution of 0.00016 M lead(II) nitrate, Pb(NO3)2, was poured into 456 mL of 0.00023 M sodium

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A solution of 0.00016 M lead(II) nitrate, Pb(NO3)2, was poured into 456 mL of 0.00023 M sodium sulfate, Na2SO4. Would a precipitate of lead(II) sulfate, PbSO4, be expected to form if 255 mL of the lead nitrate solution were added?
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General Chemistry

ISBN: 978-1439043998

9th edition

Authors: Darrell Ebbing, Steven D. Gammon

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