Question

Consider the titration of 30.00 mL of 0.1000 M Pb(NO3)2 with 0.3000 M Na2CO3. Pb2 (aq) COs2 (aq) PbCO3 (s) Ksp of PbC03 (s)-7.4x10-14 Calculate [Pb2+] in the solution after a total of 5.00 mL of Na2CO3 is added. a. 1.72x10-12 M b. 2.47x10-13 M C. 4.29x10-2 M d. 8.57x102 M

i have already calculated this part and answer is 'b' but i don't know how to do the next part.

Answer 1

2)

Answer

b) 2.220×10^-12

Explanation

Consider the complete reaction between Pb²⁺ and CO3^2-

Pb²⁺(aq) + CO3^2-(aq) -------> PbCO3(s)

given moles of Pb²⁺ = (0.1000mol/1000ml) ×30ml = 0.0030 moles

given moles of CO3^- = (0.3000mol/1000ml)×15ml = 0.0045mol

0.0030 moles of Pb²⁺ react with 0.0030moles of CO3^2-

remaining moles of CO3^2- = 0.0045mol - 0.0030mol = 0.0015mol

remaining [CO3^2-] = (0.0015mol/45ml)×1000ml = 0.0333M

Now consider the solubility equillibrium of PbCO3

PbCO3(s) <-------> Pb²⁺(aq) + CO3^2-(aq)

Ksp = [Pb²⁺] [CO3^2-] = 7.4×10^-14

Initial concentration

[Pb²⁺] = 0

[CO3^2-] = 0.0333

Change in concentration

[Pb²⁺] = + x

[CO3^2-] = +x

Equillibrium concentration

[Pb²⁺] = x

[CO3^2-] = 0.0333 + x

So,

x ( 0.0333 +x) = 7.4×10^-14

x is small quantity , so, we can assume 0.0333 + x = 0.0333

0.0333x = 7.4×10^-14

x = 2.222× 10^-12

Therefore ,

at equillibrium , [Pb²⁺] = 2.222×10^-12M