Lead(II) chloride has limited solubility in water (Ksp for PbCl2 is 1.7 x 10-5).

Question

Lead(II) chloride has limited solubility in water (Ksp for PbCl2 is 1.7 x 10-5).

PbCl2(s) Pb2+(aq) + 2 Cl-(aq)

Consider combining 2.5 x 10-2 moles of NaCl and 2.50 x 10-2 moles of Pb(NO3)2 in enough water to make 1.00 L of solution. What is the value of the ion product constant, Qsp and will a precipitate of PbCl2 form?

a) Qsp = 1.6 x 10-5 and a precipitate of PbCl2 forms.

b) Qsp = 1.6 x 10-5 and no precipitate of PbCl2 forms.

c) Qsp = 6.2 x 10-4 and a precipitate of PbCl2 forms.

d) Qsp = 6.2 x 10-4 and no precipitate of PbCl2 forms.

e) Qsp = 6.2 x 10-4 and the solubility limit of PbCl2 in water has just been met (a saturated solution is present).

***THE ANSWER IS B, PLEASE EXPLAIN WHY***

Explanation / Answer

Answer - Given, moles of NaCl = 2.5 x 10-2 moles , moles of Pb(NO3)2 = 2.50 x 10-2 moles, volume is 1.0 L, so [NaCl] = 2.50 x 10-2 M , [Pb(NO3)2] = 2.50 x 10-2 M

[Cl-] = 2.50 x 10-2 M , [Pb2+] = 2.50 x 10-2 M

Ksp for PbCl2 is 1.7 x 10-5

We need to calculate the Qsp for the PbCl2

We know the Qsp is the product of the initial concentration

So, Qsp expression is

Qsp = [Pb2+] [Cl-]2

=   (2.50 x 10-2 M) (2.50 x 10-2 M)2

= 1.6*10-5

So we the Qsp < Ksp , so there is no precipitate will be formed

So answer is b) Qsp = 1.6 x 10-5 and no precipitate of PbCl2 forms.

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