Precipitating silver(I) salts from the solution of barium(II) cyanate and iodideHow to determine which salt will precipitate from a solution containing multiple ions?Is solubility in Qsp affected by coefficient?Why should I acidify twice in the procedure for qualitative analysis of chloride anions?Adding powdered Pb and Fe to a solutionFinding x and y in Pt(NH3)xClyHow to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?Precipitation of AgCl from the tap water solution of the group 2 chlorideMixing cobalt(III) hydroxide and thallium(III) hydroxide
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Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide
How to determine which salt will precipitate from a solution containing multiple ions?Is solubility in Qsp affected by coefficient?Why should I acidify twice in the procedure for qualitative analysis of chloride anions?Adding powdered Pb and Fe to a solutionFinding x and y in Pt(NH3)xClyHow to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?Precipitation of AgCl from the tap water solution of the group 2 chlorideMixing cobalt(III) hydroxide and thallium(III) hydroxide
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
The answer was only $ceAgCN$ will precipitate, but I don't understand why $ceAgI$ wouldn't precipitate as well since there is more than enough excess $ceAgNO3$ available to precipitate with both $ceI-$ and $ceCN-$?
inorganic-chemistry aqueous-solution solubility
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
$endgroup$
add a comment
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$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
The answer was only $ceAgCN$ will precipitate, but I don't understand why $ceAgI$ wouldn't precipitate as well since there is more than enough excess $ceAgNO3$ available to precipitate with both $ceI-$ and $ceCN-$?
inorganic-chemistry aqueous-solution solubility
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
$endgroup$
2
$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
$begingroup$
I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
$endgroup$
– user77021
Apr 15 at 16:16
4
$begingroup$
Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
$endgroup$
– Zhe
Apr 15 at 16:28
add a comment
|
$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
The answer was only $ceAgCN$ will precipitate, but I don't understand why $ceAgI$ wouldn't precipitate as well since there is more than enough excess $ceAgNO3$ available to precipitate with both $ceI-$ and $ceCN-$?
inorganic-chemistry aqueous-solution solubility
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
$endgroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
The answer was only $ceAgCN$ will precipitate, but I don't understand why $ceAgI$ wouldn't precipitate as well since there is more than enough excess $ceAgNO3$ available to precipitate with both $ceI-$ and $ceCN-$?
inorganic-chemistry aqueous-solution solubility
inorganic-chemistry aqueous-solution solubility
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
edited Apr 15 at 16:30
andselisk♦
22.2k8 gold badges78 silver badges149 bronze badges
22.2k8 gold badges78 silver badges149 bronze badges
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
asked Apr 15 at 16:04
user77021user77021
211 bronze badge
211 bronze badge
2
$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
$begingroup$
I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
$endgroup$
– user77021
Apr 15 at 16:16
4
$begingroup$
Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
$endgroup$
– Zhe
Apr 15 at 16:28
add a comment
|
2
$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
$begingroup$
I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
$endgroup$
– user77021
Apr 15 at 16:16
4
$begingroup$
Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
$endgroup$
– Zhe
Apr 15 at 16:28
2
2
$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
$begingroup$
I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
$endgroup$
– user77021
Apr 15 at 16:16
$begingroup$
I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
$endgroup$
– user77021
Apr 15 at 16:16
4
4
$begingroup$
Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
$endgroup$
– Zhe
Apr 15 at 16:28
$begingroup$
Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
$endgroup$
– Zhe
Apr 15 at 16:28
add a comment
|
2 Answers
2
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oldest
votes
$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.
$ce[CN-]_i = [I-]_i = 2cdot10^-10$ molar
Neglecting any volume change of solution the initial concentration of $ceAg+$ will be
$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$
Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.
$ce[CN-]_i + [I-]_i = 4cdot10^-10$ molar
$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$
So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $ceCN-$ is
$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$
The the final concentration of $ceI-$ is
$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$
Conclusion:
Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.
Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
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$begingroup$
Alternative method to MaxW method:
Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:
$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$
Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$
For precipitation of $ceAgCN(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgCN(s)$ will precipitate.
For precipitation of $ceAgI(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgI(s)$ will not precipitate in this condition.
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.
$ce[CN-]_i = [I-]_i = 2cdot10^-10$ molar
Neglecting any volume change of solution the initial concentration of $ceAg+$ will be
$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$
Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.
$ce[CN-]_i + [I-]_i = 4cdot10^-10$ molar
$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$
So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $ceCN-$ is
$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$
The the final concentration of $ceI-$ is
$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$
Conclusion:
Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.
Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
$endgroup$
add a comment
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$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.
$ce[CN-]_i = [I-]_i = 2cdot10^-10$ molar
Neglecting any volume change of solution the initial concentration of $ceAg+$ will be
$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$
Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.
$ce[CN-]_i + [I-]_i = 4cdot10^-10$ molar
$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$
So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $ceCN-$ is
$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$
The the final concentration of $ceI-$ is
$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$
Conclusion:
Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.
Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
$endgroup$
add a comment
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$begingroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.
$ce[CN-]_i = [I-]_i = 2cdot10^-10$ molar
Neglecting any volume change of solution the initial concentration of $ceAg+$ will be
$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$
Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.
$ce[CN-]_i + [I-]_i = 4cdot10^-10$ molar
$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$
So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $ceCN-$ is
$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$
The the final concentration of $ceI-$ is
$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$
Conclusion:
Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.
Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
$endgroup$
Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.
Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.
$ce[CN-]_i = [I-]_i = 2cdot10^-10$ molar
Neglecting any volume change of solution the initial concentration of $ceAg+$ will be
$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$
Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.
$ce[CN-]_i + [I-]_i = 4cdot10^-10$ molar
$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$
So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $ceCN-$ is
$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$
The the final concentration of $ceI-$ is
$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$
Conclusion:
Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.
Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
edited Apr 15 at 17:28
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
answered Apr 15 at 17:07
MaxWMaxW
16.7k2 gold badges23 silver badges64 bronze badges
16.7k2 gold badges23 silver badges64 bronze badges
add a comment
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add a comment
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$begingroup$
Alternative method to MaxW method:
Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:
$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$
Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$
For precipitation of $ceAgCN(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgCN(s)$ will precipitate.
For precipitation of $ceAgI(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgI(s)$ will not precipitate in this condition.
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
$endgroup$
add a comment
|
$begingroup$
Alternative method to MaxW method:
Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:
$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$
Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$
For precipitation of $ceAgCN(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgCN(s)$ will precipitate.
For precipitation of $ceAgI(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgI(s)$ will not precipitate in this condition.
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
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Alternative method to MaxW method:
Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:
$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$
Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$
For precipitation of $ceAgCN(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgCN(s)$ will precipitate.
For precipitation of $ceAgI(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgI(s)$ will not precipitate in this condition.
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
$endgroup$
Alternative method to MaxW method:
Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:
$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$
Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$
For precipitation of $ceAgCN(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgCN(s)$ will precipitate.
For precipitation of $ceAgI(s)$:
$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$
Therefore, $ceAgI(s)$ will not precipitate in this condition.
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
edited Apr 15 at 19:38
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
answered Apr 15 at 19:33
Mathew MahindaratneMathew Mahindaratne
11.3k2 gold badges14 silver badges39 bronze badges
11.3k2 gold badges14 silver badges39 bronze badges
add a comment
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add a comment
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$begingroup$
"There is more than enough." How did you determine that with out any quantitative calculations?
$endgroup$
– Zhe
Apr 15 at 16:08
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I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-
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– user77021
Apr 15 at 16:16
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Only if the concentrations are such that the solubility product exceeds $K_mathrmsp$.
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– Zhe
Apr 15 at 16:28