Answer:
7.11 × 10^-4M
Explanation:
PbCl2 will dissociate as follows
PbCl2 ⇌ Pb2+(x) + 2Cl- (2x)
HCl will dissociate as follows:
HCl ⇌H+ (0.15M) + Cl- (0.15M)
Ksp = {Pb2+} {Cl-}^2
Ksp = {x} {2x+ 0.15}^2
Due to common ion effect i.e. 2x<0.15, 2x will be neglected making the total concentration of chloride ion in the solution to be 0.15M
Ksp = {x} {0.15}^2
1.6 x 10^-5 = 0.0225x
x = 1.6 x 10^-5 ÷ 2.25 × 10^-2
x = 1.6/2.25 × 10^(-5+2)
x = 0.7111 × 10^-3
x = 7.11 × 10^-4
Therefore, the solubility is 7.11 × 10^-4M
The solubility (in M) for the given solution of HCl would be:
[tex]7.11[/tex] × [tex]10^-4[/tex] M
Given that,
[tex]Pb2+(x) + 2Cl- (2x)[/tex]
For [tex]HCl[/tex],
[tex]HCl[/tex] ⇌ [tex]H+ (0.15M) + Cl- (0.15M)[/tex]
The reaction of [tex]PbCl2[/tex] in dissociation.
[tex]PbCl2[/tex] ⇌ [tex]Pb2+(x) + 2Cl- (2x)[/tex]
We have,
Ksp of [tex]PbCl2[/tex] = 1.6 × [tex]10^-5[/tex]
Solution of [tex]HCl[/tex] = 0.15M
For Ksp,
[tex]Ksp = {Pb2+} {Cl-}^2[/tex]
⇒ [tex]Ksp = {x} {2x+ 0.15}^2[/tex]
Using this, we can determine the solubility:
Ksp = [tex]{x} {0.15}^2[/tex]
⇒ [tex]1.6[/tex] x [tex]10^-5[/tex] = [tex]0.0225x[/tex]
⇒ [tex]x = 1.6[/tex] x [tex]10^-5[/tex] ÷ [tex]2.25[/tex] × [tex]10^-2[/tex]
by solving the next steps, we get,
⇒ x [tex]= 0.7111[/tex] × [tex]10^-3[/tex]
⇒ x = [tex]7.11[/tex] × [tex]10^-4[/tex] M
∵ [tex]7.11[/tex] × [tex]10^-4[/tex] M
Thus, the above answer is correct.
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