Question

1.write the balanced equation for the reaction of aqueous Pb (ClO3)^2 with aqueous Nal?
2. 0.250L of 0.450 M H2SO4 is mixed with 0.200L of 0.240M KOH. What concentration of Sulfuric acid remains after neutralization?

Answer 1

(1) The balanced reaction is : Pb(ClO₃)₂(aq) + NaI(aq) --------> PbI + Na(ClO₃)₂

(2) The number of moles of H2SO4 is , n = Molarity x volume in L

                                                            = 0.450 M x 0.250 L

                                                            = 0.1125 moles

The number of moles of KOH, n' = Molarity x volume in L

                                               = 0.240 M x 0.200 L

                                               = 0.048 mol

The balanced reaction is : H2SO4 + 2KOH ----> K2SO4 + 2 H2O

From the balanced reaction,

1 mole of H2SO4 reacts with 2 moles of KOH

M moles of H2SO4 reacts with 0.048 moles of KOH

M = ( 0.048x1)/2

   = 0.024 moles

So number of moles of H2SO4 left = 0.1125-0.024 = 0.0885 moles

Total volume of the reaction mixture = 0.250 + 0.200 L = 0.450 L

So concentration of H2SO4 remains = number of moles / volume in L

                                                     = 0.0885 mol / 0.450 L

                                                     = 0.197 M