Calculate the value of the ion product of PbCl2 (Ksp = 1.6 ×10-5) when 255 mL of 0.016 M KCl (aq) are added to 245 mL of 0.175M Pb(NO3)2.
Value of the ion product ;
=1.05×10−5mol/l=1.05×10^{-5}mol/l=1.05×10−5mol/l
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