How many times more soluble. HELP

  • Thread starter aisha
  • Start date
The 2x term comes from the stoichiometric coefficient in the balanced equation, and the .100M is the original concentration of chloride. If you wanted to get really technical you could subtract 2x from .100M, but the simplification unco used is more than sufficient for this problem. The rest is just plugging numbers in, and subtracting the solubilities to get your answer.I hope this helps you.In summary, the solubility of lead 2 chloride is 12 times greater in distilled water than in a 0.100 mol/L solution of sodium chloride, due to the common ion effect. The Ksp for lead 2 chloride is 1.2*10^-
  • #1
aisha
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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
 
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  • #2
There is no additional information required.

The reaction is
[tex]PbCl_2 \, \Leftrightarrow Pb^{2+} + 2Cl^-[/tex]

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:

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

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

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.
 
Last edited:
  • #3
ok i don't get what i have to do do i do dissociation of NaCl next?

NaCl--> [tex] Na^+^2 +Cl^- [/tex]

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
 
  • #4
NaCl fully dissociates: [Cl-] = [NaCl] = 0.100 mol/L.

[tex] \begin{array}{ c|ccc|}{}&{PbCl_2 \, \Leftrightarrow}&{Pb^{2+}}&{2Cl^-}\\
\hline{\text{Initial conc.}}&{}&0&0.100\\
\text{Change in conc.}}&{}&{}&{}\\
\text{Conc. at equi.}}&{}&{}&{}\\
\end{array}[/tex]
 
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  • #5
Ok but i still don't get how to solve the question
 
  • #6
So you have
[tex]1.2 \times 10^{-5} = (s)(0.100 + 2s)^2[/tex]

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

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

This assumption must be checked by plugging the value for s (ie. [Pb2+]) into the Ksp equation.
 
  • #7
so 1.2*10^-5= [0.0144] [0.100]^2?

1.2*10^-5= 1.44*10^-4
 
  • #8
Just an arithmetic thing

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

[tex]s = \frac{1.2 \times 10^{-5}}{0.01} = 0.0012[/tex]

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

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.
 
  • #9
i don't know if we are doing this right its getting complicated i don't think I've done this before
 
  • #10
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.
 
  • #11
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
 
  • #12
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.
 
  • #13
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?
 
  • #14
aisha said:
is 0.014 mol/L the correct concentration for the solubility of PbCl2?
Yep.
aisha said:
How do you know that is in distilled water?
Because there wasn't a common ion; scroll up to the top of the page.
aisha said:
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.
 
  • #15
aisha said:
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.
 

1. How do you determine how many times more soluble a substance is?

The solubility of a substance can be determined by conducting a solubility test, where a known amount of the substance is added to a known volume of solvent and the amount that dissolves is measured. This measurement can then be compared to the solubility of another substance to determine how many times more soluble it is.

2. What factors affect the solubility of a substance?

The solubility of a substance is affected by a variety of factors, including temperature, pressure, polarity, and the presence of other solutes. For example, increasing the temperature and pressure can often increase the solubility of a substance, while the presence of other solutes can decrease its solubility.

3. How is solubility different from concentration?

Solubility refers to the maximum amount of a substance that can dissolve in a given amount of solvent, while concentration refers to the amount of a substance that is actually dissolved in a given amount of solvent. Solubility is a physical property of a substance, while concentration can vary depending on the amount of solvent and solute used.

4. Can a substance be both soluble and insoluble?

Yes, a substance can be soluble in one solvent and insoluble in another. This is because the solubility of a substance is dependent on the interactions between the solvent and solute, and different solvents can have different interactions with a substance.

5. How can solubility be expressed quantitatively?

Solubility can be expressed quantitatively in several ways, including in grams per liter, moles per liter, or as a percentage. These measurements can be used to compare the solubility of different substances or to track changes in solubility under different conditions.

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