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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on January 12, 2021, 11:00:14 AM

[1] Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
a. TlCl_{3}(s) in 0.025 M TlNO_{3}.
b. BaF_{2}(s) in 0.0313 M KF.
c. MgC_{2}O_{4} in 2.250 L of a solution containing 8.156 g of Mg(NO_{3})_{2}
d.Ca(OH)_{2}(s) in an unbuffered solution initially with a pH of 12.700
[2] Explain why the changes in concentrations of the common ions in question [1] can be neglected.
[3] Explain why the changes in concentrations of the common ions in Question [2] cannot be neglected
What are the answers to all these three question?
I am working on them. Any member may guide in this regard if he/she knows the answers.

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[1] ... Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
[2] Explain why the changes in concentrations of the common ions in question [1] can be neglected.
[3] Explain why the changes in concentrations of the common ions in Question [2] cannot be neglected
These three questions are inconsistent with each other. Can you post the exact wording of the original questions?
Also, it looks like you need solubility product values to answer the questions. Are you given those values?

Answer to [1] a.
TlCl(s) in Tl(NO_{3})
K_{sp}=1.7×10^{4}=[Tl^{+}][Cl^{}]; Letx=[Cl^{}]:
1.7×10^{4}=(x+0.025)x
Solving quadratic equation x^{2}+ 0.025x 1.7×10^{4}=0; we get x=0.006 M
[Tl^{+}]=0.025 + 0.006=3.1× 10^{2} M; [Cl^{}]=6.0×10^{3} M
Answer to [1] b.
BaF_{2}(s) in 0.0313 M KF
K_{sp}=2.4× 10^{5}=[Ba^{2+}][F^{}]^{2}, Let x=[Ba^{2+}]
2.4×10^{5}=x(x+0.0313)^{2}
Solving cubic equation x^{3}+0.0626x^{2} + 0.000969x 2.4× 10^{5}=0; we get x=0.01251 M; [Ba^{2+}]=0.01251 M; [F^{}]=0.01251 +0.0313 =0.04381 M
Answer to [1] c.
MgC_{2}O_{4} in 2.250 L of solution containing 8.156 g of Mg(NO_{3})_{2}
Molar mass of Mg(NO_{3})_{2}=148.3148 g/mol So,[itex]\frac{8.156 g}{148.3148 g/mol}=0.05499 [/itex] mol are number of moles of Mg(NO_{3})_{2}
Molarity of Mg(NO_{3}]_{2}=[itex]\frac{0.05499 mol}{2.250 L}=0.02444 M[/itex]
Let x=[C_{2}O_{4}^{2}]
K_{sp}=7× 10^{7}= [Mg^{2+}][C_{2}O_{4}^{2}]=x(x+0.02444 )
Solving the quadratic equation x^{2} +0.02444x 7× 10^{7}=0; we get x=0.000028608 M
Thus, [C_{2}O_{4}^{2}]=2.8608× 10^{5} M; [Mg^{2+}]=0.000028608 + 0.02444 = 0.0245 M
Answer to [1] d.
Ca(OH)_{2} (s) in an unbuffered solution initially with a pH of 12.700
pH=12.700; pOH=1.300, [OH^{}]=0.0501 M; Let x= [Ca^{2+}]
K_{sp}= 1.3× 10^{6}=[Ca^{2+}][OH^{}]^{2}=x(x+0.050)^{2}
Solving cubic equation x^{3} +0.1x^{2} +0.0025x  1.3× 10^{6}=0; we get x=0.0005095 M
[Ca^{2+}]=5.09560×10^{4} M, [OH^{}]=0.00050956 +0.050=0.05050 M
Now how to answer [2] and [3] ?

Check the dissolution products of your compounds. Only the Ksp TlCl value matches my table data. Of course, sometimes the authors of various computational tasks modify the numbers to facilitate the calculation.
However, the value of Ksp Ca(OH)_{2} is clearly wrong, and also your calculation in this case.
Besides, there is a calculation error related to BaF_{2}.

Check the dissolution products of your compounds. Only the Ksp TlCl value matches my table data. Of course, sometimes the authors of various computational tasks modify the numbers to facilitate the calculation.
However, the value of Ksp Ca(OH)_{2} is clearly wrong, and also your calculation in this case.
Besides, there is a calculation error related to BaF_{2}.
Hello,
Please check the K_{sp} value for Ca(OH)_{2} in below table.

In the case of Ca(OH)_{2}, the calculation method is different, because OH comes only from the dissociation of calcium hydroxide, and then the concentration of Ca^{2+} must be about half the concentration of hydroxide ions. The pH of the calcium hydroxide complies with Ksp 5.5·10^{6}. In the case of BaF_{2}, you assumed stoichiometry for BaF_{1} in your calculations.
As mjc123 has already noted, questions 2 and 3 in the form as you presented them contradict each other. So it is difficult to answer them.

In the case of Ca(OH)_{2}, the calculation method is different, because OH comes only from the dissociation of calcium hydroxide, and then the concentration of Ca^{2+} must be about half the concentration of hydroxide ions. The pH of the calcium hydroxide complies with Ksp 5.5·10^{6}. In the case of BaF_{2}, you assumed stoichiometry for BaF_{1} in your calculations.
As mjc123 has already noted, questions 2 and 3 in the form as you presented them contradict each other. So it is difficult to answer them.
Hello,
In the case of [1] b. , I have rectified my computational errors. But in the case of [1] d. [OH^{}] ions also come from an unbuffered solution having pH 12.700.

"Unbuffered solution" means that OH^{} ions only come from the dissociation of Ca(OH)_{2} (those from the dissociation of water are negligible). Then K_{sp} = 4x^{3}, where x is the concentration of calcium ions and [OH^{}] = 2x.
In fact, there is one more equilibrium Ca(OH)_{2} ::equil:: CaOH^{+} + OH^{} in the calcium hydroxide solution, which is not taken into account by the solubility product. As a result, the concentration of calcium ions calculated from the actual pH value of the solution is slightly lower (a few%) than half of the concentration of hydroxide ions. With your K_{sp} value, nonsense comes out of your incorrect calculation.

"Unbuffered solution" means that OH^{} ions only come from the dissociation of Ca(OH)_{2} (those from the dissociation of water are negligible). Then K_{sp} = 4x^{3}, where x is the concentration of calcium ions and [OH^{}] = 2x.
In fact, there is one more equilibrium Ca(OH)_{2} ::equil:: CaOH^{+} + OH^{} in the calcium hydroxide solution, which is not taken into account by the solubility product. As a result, the concentration of calcium ions calculated from the actual pH value of the solution is slightly lower (a few%) than half of the concentration of hydroxide ions. With your K_{sp} value, nonsense comes out of your incorrect calculation.
Hello,
I may be wrong for computing the answer for [1] d. But what is your opinion about the belowgiven answer which i found on other chemistry educational website. Is this answer also nonsense?

This is a miscalculation for two reasons: the value of the solubility product is overstated; the calculation is for the presence of a common ion and the substitution of the numbers into the formula contains a fatal error. If OH^{} ions are derived only from calcium hydroxide the expression should be K_{sp} = 4x^{3},
if we have additional OH^{} ions with concentration A, the expression K_{sp} = x(2x+A)^{2}.

Answer to [1] a.
TlCl(s) in Tl(NO3)
Your original post said TlCl_{3}. Was that a mistake?

Answer to [1] a.
TlCl(s) in Tl(NO3)
Your original post said TlCl_{3}. Was that a mistake?
Yes, That was a mistake.

This is a miscalculation for two reasons: the value of the solubility product is overstated; the calculation is for the presence of a common ion and the substitution of the numbers into the formula contains a fatal error. If OH^{} ions are derived only from calcium hydroxide the expression should be K_{sp} = 4x^{3},
if we have additional OH^{} ions with concentration A, the expression K_{sp} = x(2x+A)^{2}.
Hello,
Thanks for your guidance and explaining the definition of "Unbuffered solution" in your reply #8.