Question

29. The solubility of Lead Fluoride (PbF) is 2.5 x 10-3 mol/L. What is the solubility in 0.10M NaF solution
(1 6.3 * 10-6 M
(2) 1.6 x 10-6 M
(3) 5.0 * 10-2 M
(4) 1.6 * 10-5 M
​

Answer 1

Answer : The correct option is, $6.3\times 10^{-6}M$

Explanation :

The dissociation of lead fluoride is written as:

$PbF_2\rightleftharpoons Pb^{2+}+2F^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Pb^{2+}][F^-]^2$

$K_{sp}=(s)\times (2s)^2$

Solubility, s = $2.5\times 10^{-3}mol/L$

Now put all the given values in this expression, we get:

$K_{sp}=(2.5\times 10^{-3})\times (2\times 2.5\times 10^{-3})^2$

$K_{sp}=6.4\times 10^{-8}$

The dissociation of NaF is written as:

$NaF\rightleftharpoons Na^++F^-$

0.10M   0.10M   0.10M

The new expression for solubility constant will be,

$K_{sp}=(s)\times (2s+0.10)^2$

$6.4\times 10^{-8}=(s)\times (2s+0.10)^2$

By solving the term, we get:

$s=6.3\times 10^{-6}M$

Therefore, the solubility in 0.10 M NaF solution is $6.3\times 10^{-6}M$

Answer 2

Answer:

The dissociation of lead fluoride is written as:

PbF_2\rightleftharpoons Pb^{2+}+2F^-PbF

2

⇌Pb

2+

+2F

−

The expression for solubility constant for this reaction will be,

K_{sp}=[Pb^{2+}][F^-]^2K

sp

=[Pb

2+

][F

−

]

2

K_{sp}=(s)\times (2s)^2K

sp

=(s)×(2s)

2

Solubility, s = 2.5\times 10^{-3}mol/L2.5×10

−3

mol/L

Now put all the given values in this expression, we get:

K_{sp}=(2.5\times 10^{-3})\times (2\times 2.5\times 10^{-3})^2K

sp

=(2.5×10

−3

)×(2×2.5×10

−3

)

2

K_{sp}=6.4\times 10^{-8}K

sp

=6.4×10

−8

The dissociation of NaF is written as:

NaF\rightleftharpoons Na^++F^-NaF⇌Na

+

+F

−

0.10M 0.10M 0.10M

The new expression for solubility constant will be,

K_{sp}=(s)\times (2s+0.10)^2K

sp

=(s)×(2s+0.10)

2

6.4\times 10^{-8}=(s)\times (2s+0.10)^26.4×10

−8

=(s)×(2s+0.10)

2

By solving the term, we get:

s=6.3\times 10^{-6}Ms=6.3×10

−6

M