Asked Oct 11, 2019

How many liters of pure water will be required to disslove 0.885g of solid lead iodate, if the water already intially contains 0.135 M Pb(NO3)2


Expert Answer

Step 1

First the moles of Pb(IO3)2 present in given grams is calculated as below,

Water already contains Pb(N...


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0.885 g Mass Pb(IOMoles 461.01 g/mol Molar mass 1.91x 103 moles Pb(IO) Pb2 210 2 [Pb Io;T Here s solubility and K, 2+ sp Kgs+0.135]x (2s) 2.5x 1013 4s2x К. X 4s2 x0.135 s 6.80 x 10 M Here, K value of Pb(IO), is 2.5x1013


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