Question

What volume of 0.082 M sodium carbonate solution is required to precipitate 99% of the Mg from 1.00 L of 0.100 M magnesium nitrate solution?

Answer 1

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let Required volume of sodium Carbonate = alt Total volenne = litt all = (1+x) et co? added = alt x 0.092 M. = 0.082 mooke & mole Mg 2 remaining - Droge montant = 0.1 x let x o.o = 10-3 mod Iron precipate Mg2+ precipitated - 0. 1 1 - 103 = 0.099 moe cor remaing = (0.082+ - 0.099) male Ksp = [Mg ²+ ] [102] 0.0821 - 0.099 103 Itx Itx 6.824106 = 103 x (0.082 x - 0.099) (+22 0.0822-0.099 = 6.82 x 103 [x = 1.92 it (ta)2