# Chemical Forums

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

Title: Precipitation and dissolution question
Post 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. TlCl3(s) in 0.025 M TlNO3.

b. BaF2(s) in 0.0313 M KF.

c. MgC2O4 in 2.250 L of a solution containing 8.156 g of Mg(NO3)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.
Title: Re: Precipitation and dissolution question
Post by: billnotgatez on January 12, 2021, 12:06:56 PM
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http://www.chemicalforums.com/index.php?topic=65859.0
Title: Re: Precipitation and dissolution question
Post by: mjc123 on January 13, 2021, 05:31:06 AM
Quote
[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?
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 09:38:06 AM

TlCl(s) in Tl(NO3)
Ksp=1.7×10-4=[Tl+][Cl-]; Letx=[Cl-]:
1.7×10-4=(x+0.025)x
Solving quadratic equation x2+ 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

BaF2(s) in 0.0313 M KF
Ksp=2.4× 10-5=[Ba2+][F-]2, Let x=[Ba2+]
2.4×10-5=x(x+0.0313)2
Solving cubic equation x3+0.0626x2 + 0.000969x -2.4× 10-5=0; we get x=0.01251 M; [Ba2+]=0.01251 M; [F-]=0.01251 +0.0313 =0.04381 M

MgC2O4 in 2.250 L of solution containing 8.156 g of Mg(NO3)2

Molar mass of Mg(NO3)2=148.3148 g/mol So,$\frac{8.156 g}{148.3148 g/mol}=0.05499$ mol are number of moles of Mg(NO3)2

Molarity of Mg(NO3]2=$\frac{0.05499 mol}{2.250 L}=0.02444 M$
Let x=[C2O42-]
Ksp=7× 10-7= [Mg2+][C2O42-]=x(x+0.02444 )
Solving the quadratic equation x2 +0.02444x -7× 10-7=0; we get x=0.000028608 M

Thus, [C2O42-]=2.8608× 10-5 M; [Mg2+]=0.000028608 + 0.02444 = 0.0245 M

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= [Ca2+]
Ksp= 1.3× 10-6=[Ca2+][OH-]2=x(x+0.050)2

Solving cubic equation x3 +0.1x2 +0.0025x - 1.3× 10-6=0; we get x=0.0005095 M
[Ca2+]=5.09560×10-4 M, [OH-]=0.00050956 +0.050=0.05050 M

Now how to answer [2] and [3] ?

Title: Re: Precipitation and dissolution question
Post by: AWK on January 13, 2021, 10:34:48 AM
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 BaF2.
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 11:05:19 AM
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 BaF2.

Hello,
Please check the Ksp value for Ca(OH)2 in below table.
Title: Re: Precipitation and dissolution question
Post by: AWK on January 13, 2021, 11:25:14 AM
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 Ca2+ 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 BaF2, you assumed stoichiometry for BaF1 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.
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 12:05:29 PM
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 Ca2+ 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 BaF2, you assumed stoichiometry for BaF1 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.
Title: Re: Precipitation and dissolution question
Post by: AWK on January 13, 2021, 12:26:17 PM
"Unbuffered solution" means that OH- ions only come from the dissociation of Ca(OH)2 (those from the dissociation of water are negligible). Then Ksp = 4x3, 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 Ksp value, nonsense comes out of your incorrect calculation.
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 02:54:02 PM
"Unbuffered solution" means that OH- ions only come from the dissociation of Ca(OH)2 (those from the dissociation of water are negligible). Then Ksp = 4x3, 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 Ksp 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 below-given answer which i found on other chemistry educational website. Is this answer also nonsense?

Title: Re: Precipitation and dissolution question
Post by: AWK on January 13, 2021, 03:40:29 PM
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 Ksp = 4x3,
if we have additional OH- ions with concentration A, the expression Ksp = x(2x+A)2.
Title: Re: Precipitation and dissolution question
Post by: mjc123 on January 13, 2021, 05:07:55 PM
Quote

TlCl(s) in Tl(NO3)
Your original post said TlCl3. Was that a mistake?
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 10:55:54 PM
Quote

TlCl(s) in Tl(NO3)
Your original post said TlCl3. Was that a mistake?

Yes, That was a mistake.
Title: Re: Precipitation and dissolution question
Post by: Win,odd Dhamnekar on January 13, 2021, 11:18:04 PM
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 Ksp = 4x3,
if we have additional OH- ions with concentration A, the expression Ksp = x(2x+A)2.

Hello,