How many times more soluble. HELP

[aisha]

How many times more soluble... ASAP HELP

Hii, I did a lab and need help with this question it says

How many times more soluble is lead 2 chloride in distilled water than in 0.100 mol/L solution of sodium chloride?

The ksp for lead 2 chloride is = 1.2*10^-5

I don't know what other information is needed to solve this question or how to figure this out.
ksp=[Pb] [Cl]^2

PLEASE HELP ME SOMEONE QUICK

 

[Unco]

There is no additional information required.

The reaction is
PbCl_2 \, \Leftrightarrow Pb^{2+} + 2Cl^-

In distilled water, there is no Pb2+ or Cl- present so [Pb2+] at equilibrium is that formed by the dissociation of PbCl2 only.

Table-wise, if we let the concentration of Pb2+ formed by s:

\begin{array}{ c|ccc|}{}&amp;{PbCl_2 \, \Leftrightarrow}&amp;{Pb^{2+}}&amp;{2Cl^-}\\<br /> \hline{\text{Initial conc.}}&amp;{}&amp;0&amp;0\\<br /> \text{Change in conc.}}&amp;{}&amp;{+s}&amp;{+2s}\\<br /> \text{Conc. at equi.}}&amp;{}&amp;{s}&amp;{2s}\\<br /> \end{array}

\begin{align*}K_{sp} &amp;= [Pb^{2+}][Cl^-]^2 \\<br /> 1.2 \times 10^{-5} &amp;= (s)(2s)^2 \\<br /> \end{align*}

Solve for s, that is [Pb2+], the solubility of PbCl2 in distilled water.

In 0.100 NaCl solution, [Cl-] = 0.100. Do the same as above but this time there is an initial concentration of Cl-. You need to assume that 0.100 + 2s is approximately equal to 0.100. Check the assumption is valid by plugging in the value for s you obtain in the right-hand side of the Ksp equation to see if it equals Ksp.

 

[aisha]

ok i don't get what i have to do do i do dissociation of NaCl next?

NaCl--> Na^+^2 +Cl^-

initial concentration Cl=0.100
initial concentration Na=0
change for Cl=x
change for Na=x
equilibrium for Na =x
equilibrium for Cl=(0.100+x)

where did u get the (0.100+2x) there is no 2 in the equation for dissociation of NaCl.

what do we do after this? I don't get it where does the distilled water come in

 

[Unco]

NaCl fully dissociates: [Cl-] = [NaCl] = 0.100 mol/L.

\begin{array}{ c|ccc|}{}&amp;{PbCl_2 \, \Leftrightarrow}&amp;{Pb^{2+}}&amp;{2Cl^-}\\<br /> \hline{\text{Initial conc.}}&amp;{}&amp;0&amp;0.100\\<br /> \text{Change in conc.}}&amp;{}&amp;{}&amp;{}\\<br /> \text{Conc. at equi.}}&amp;{}&amp;{}&amp;{}\\<br /> \end{array}

 

[aisha]

Ok but i still don't get how to solve the question

 

[Unco]

So you have
1.2 \times 10^{-5} = (s)(0.100 + 2s)^2

We assume that 0.100 + 2s \approx 0.100 because the Ksp value is quite small and so 's' should have a minimal numerical effect on 0.100.

So the equation becomes
1.2 \times 10^{-5} = (s)(0.100)^2

This assumption must be checked by plugging the value for s (ie. [Pb2+]) into the Ksp equation.

 

[aisha]

so 1.2*10^-5= [0.0144] [0.100]^2?

1.2*10^-5= 1.44*10^-4

 

[Unco]

Just an arithmetic thing

1.2 \times 10^{-5} = s(0.100)^2 = 0.01s

s = \frac{1.2 \times 10^{-5}}{0.01} = 0.0012

So [Pb^{2+}][Cl^-]^2 = 0.0012 (0.1 + 2\times0.0012) = 1.25 \times 10^{-5} \approx 1.3 \times 10^{-5}

Which isn't good, is it?!

We have to go back to our unapproximated equation, expand, and solve the cubic equation for s with this site:
http://www.1728.com/cubic.htm
if you don't have access to an alternative.

 

[aisha]

i don't know if we are doing this right its getting complicated i don't think I've done this before

 

[Unco]

The process is fine; the numbers just turned out to bite us. If solving the cubic is too complicated, stick with 0.0012mol/L. The actual value is 0.0011 but this won't make a huge difference when you compare with the solubility in distilled water, which was the goal of the exercise.

You may be expected to just assume the assumption is valid and move on.

 

[aisha]

i don't know how to compare to the solubility of water. Lol I am so confused if you could take me through this step by step it would be appreciated thanks

 

[Unco]

You found the solubility of PbCl2 in distilled water to be 0.0144mol/L (I saw in another thread).

You found the solubility of PbCl2 in 0.100M NaCl solution to be 0.0012mol/L.

So the solubility of PbCl2 in distilled water is 0.0144/0.0012 = 12 times greater than the solubility in the NaCl solution.

This is called the common ion effect.

 

[aisha]

is 0.014 mol/L the correct concentration for the solubility of PbCl2? How do you know that is in distilled water?

and how did you get 0.0012 mol/L NaCl solution?

 

[Unco]

is 0.014 mol/L the correct concentration for the solubility of PbCl2?

Yep.

How do you know that is in distilled water?

Because there wasn't a common ion; scroll up to the top of the page.

and how did you get 0.0012 mol/L NaCl solution?

The solubility of PbCl2 in the NaCl solution (which itself had a concentration of 0.100mol/L) was 0.0012 mol/L.

 

[GCT]

Hii, I did a lab and need help with this question it says
How many times more soluble is lead 2 chloride in distilled water than in 0.100 mol/L solution of sodium chloride?
The ksp for lead 2 chloride is = 1.2*10^-5
I don't know what other information is needed to solve this question or how to figure this out.
ksp=[Pb] [Cl]^2
PLEASE HELP ME SOMEONE QUICK

adding to what unco has said so far

Lead chloride would dissolve less in sodium chloride than in distilled water, due to the common ion effect. The complexity of this problem depends on whether it's for general chemistry or analytical chemistry. For now I'll assume general chemistry

NaCl is assumed to dissociate completely, thus your original concentration of chloride is .100 M, the expression for the dissociation of lead chloride is as follows

ksp=[x][2x+.100M]^2, solve for x.