What is the solubility of PbCl2 in a 0.075M solution of CaCl2?
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Answer: 7.6x10^-4 M
Explanation: You need to know the Ksp of PbCl2, which should be a pretty small number since chlorides of lead aren't very soluble at all. From this site:
http://www.ktf-split.hr/periodni/en/abc/…
I find that the Ksp of PbCls is 1.70x10^-5.
The equation for the dissolution of PbCls is:
PbCl2(s) <=> Pb(2+)(aq) + 2Cl(-)(aq)
So the Ksp expression is:
Ksp = [Pb(2+)][Cl(-)]²
We wish to find the solubility of PbCl2 in a solution that already contains 0.075 M CaCl2. That's going to further reduce the solubility of PbCl2 due to the Cl(-) ions already present in solution. Because each mole of CaCl2 liberates two moles of Cl(-) ions when it dissolves, the molar concentration of Cl(-) ions is 2 x 0.075 M, or 0.15 M. Let's set up an ICE box to see what happens when the PbCl2 is added.
Species______[Pb(2+)]_____[Cl(-)]
Initial__________0 M_______0.15 M
Change________+ x_______+ 2x
Equilibrium______x_______0.15 + 2x
So the Ksp expression becomes:
1.70x10^-5 = (x)(0.15 + 2x)²
Now let's pause for a moment. If we try to solve this as it is, we're going to come up with a very ugly third-order equation. We can make a simplifying assumption that will speed the process along. Because Ksp is so small, we can assume that x is also going to be very small. In fact, x is so small that 0.15 + 2x is really not much larger than 0.15. We can ignore the + 2x, thereby avoiding a third-order equation.
1.70x10^-5 = (x)(0.15)²
1.70x10^-5 = (x)(0.0225)
x = 7.6x10^-4 M
Our assumption was valid; the value of x is so small that 0.15 + 2x would be no greater than 0.15, once it was rounded back down to two decimal places.
Checking: We can check our math by plugging x into the modified Ksp expression:
1.70x10^-5 = (7.6x10^-4)(0.15 + 2(7.6x10^-4))²
1.70x10^-5 = (7.6x10^-4)(0.15 + 0.00152)²
1.70x10^-5 = (7.6x10^-4)(0.15152)²
1.70x10^-5 = (7.6x10^-4)(0.022958)
1.70x10^-5 = 1.74x10^-5
Which is close enough for government work! We can assume that the minor difference is due to round-off error and our simplification. I hope that helps. Good luck!
Explanation: You need to know the Ksp of PbCl2, which should be a pretty small number since chlorides of lead aren't very soluble at all. From this site:
http://www.ktf-split.hr/periodni/en/abc/…
I find that the Ksp of PbCls is 1.70x10^-5.
The equation for the dissolution of PbCls is:
PbCl2(s) <=> Pb(2+)(aq) + 2Cl(-)(aq)
So the Ksp expression is:
Ksp = [Pb(2+)][Cl(-)]²
We wish to find the solubility of PbCl2 in a solution that already contains 0.075 M CaCl2. That's going to further reduce the solubility of PbCl2 due to the Cl(-) ions already present in solution. Because each mole of CaCl2 liberates two moles of Cl(-) ions when it dissolves, the molar concentration of Cl(-) ions is 2 x 0.075 M, or 0.15 M. Let's set up an ICE box to see what happens when the PbCl2 is added.
Species______[Pb(2+)]_____[Cl(-)]
Initial__________0 M_______0.15 M
Change________+ x_______+ 2x
Equilibrium______x_______0.15 + 2x
So the Ksp expression becomes:
1.70x10^-5 = (x)(0.15 + 2x)²
Now let's pause for a moment. If we try to solve this as it is, we're going to come up with a very ugly third-order equation. We can make a simplifying assumption that will speed the process along. Because Ksp is so small, we can assume that x is also going to be very small. In fact, x is so small that 0.15 + 2x is really not much larger than 0.15. We can ignore the + 2x, thereby avoiding a third-order equation.
1.70x10^-5 = (x)(0.15)²
1.70x10^-5 = (x)(0.0225)
x = 7.6x10^-4 M
Our assumption was valid; the value of x is so small that 0.15 + 2x would be no greater than 0.15, once it was rounded back down to two decimal places.
Checking: We can check our math by plugging x into the modified Ksp expression:
1.70x10^-5 = (7.6x10^-4)(0.15 + 2(7.6x10^-4))²
1.70x10^-5 = (7.6x10^-4)(0.15 + 0.00152)²
1.70x10^-5 = (7.6x10^-4)(0.15152)²
1.70x10^-5 = (7.6x10^-4)(0.022958)
1.70x10^-5 = 1.74x10^-5
Which is close enough for government work! We can assume that the minor difference is due to round-off error and our simplification. I hope that helps. Good luck!
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pretty soluble