what mass of Pb^2+ ion is left in solution when 50 mL of 0.20 M Pb(NO3)2 is added to 50 mL of 1.5 NaCl? given k sp of PbCl2 is 1.7 x 10^-4

Dear Nancy,

1. Calculate the heat released, in J, by the precipitation reaction:

   q = mass x specific heat capacity x change in temperature 
   q = [m_Pb(NO3)2(aq) + m_KI(aq)] x c_g x (T_f - T_i) 
   q = [50 + 30] x 4.184 x (22.2 - 19.6) = 870.27 J

    

2. Calculate the moles of reactant specified, n[Pb(NO₃)_2(aq)]:

   n = M x V 
   n[Pb(NO₃)_2(aq)] = 0.20 x 50 x 10^-3 = 0.010 mol

    

3. Calculate the heat of precipitation, H, in kJ/mol of Pb(NO₃)_2(aq):

   H = -q/1000 ÷ n 
   H = -0.870/1000 ÷ 0.010 = -87 kJ mol^-1 
   H is negative because the reaction is exothermic.

    

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