# 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

Aman Bansal
592 Points
11 years ago

Dear Nancy,

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

q = mass x specific heat capacity x change in temperature
q = [mPb(NO3)2(aq) + mKI(aq)] x cg x (Tf - Ti
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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Thanks

Aman Bansal