When 0.150 L of 0.10 M lead (II) nitrate and 0.100 L of 0.20 M sodium chloride are mixed? For PbCl 2 , K sp = 1.2 x10 - 5

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Explanation: 

Precipitation reaction :

Pb(NO3)2(aq)+ 2NaCl(aq) ----------> PbCl2(s)+ 2NaNO3(aq)

Solubility reaction :

PbCl2(s) -----> Pb2+  +  2Cl-

Ksp is given and in order to find out if a precipitate form we need to find out Q

Q = [Pb2+][Cl-]2

1:1 mole ratios so initial [Pb2+] = [Pb(NO3)2] = 0.10 M;initial[Cl-] = [NaCl] = 0.20 M
after we combine the solution the total volume becomes .250 L , so the concentration of Pb2+ and Cl- ions is :
[Pb2+] =.10 M Pb2+ x .150 / .250 L = 0.060 M Pb2+
[Cl-] = 0.20 M Cl- x . 1 L / .250 l = 0.080 M Cl-
 
Q = (0.060)(0.080)2= 3.8 x 10-4
Q>Ksp, so PbCl2 does precipitate.Eq shifts to the left .
 
3.8
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Answer: 
Q>Ksp, so PbCl2 does precipitate.Eq shifts to the left .
 
3.8
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