This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).
However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.
Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:
Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:
Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.
Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:
This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).
However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.
Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:
Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:
Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.
Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:
I guess I should explain more clearly. For the given chemical reaction in equation 1, the activity of the solid is defined as 1. It doesnt matter at equilibrium if there is 1 microgram of FeS(s) or 1 kg of FeS(s) in equilibrium with the aqueous solution, the amount of FeS dissolved in the solution will be the same.More
I think this is the correct way to solve this. Published solubility constants vary; the OP was supposed to just use the given one (provided in reciprocal form) and demonstrate the larger concepts (e.g. dilution effects) with it.More
The $\mathrm{K}_\mathrm{sp}$ was provided in the problem statement, just in an unusual way. $$\mathrm{K}_\mathrm{sp} = \dfrac{1}{1.6\cdot10^{17}} =6.25\cdot10^{-18}$$More
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When you mix the solutions, the iron that was in solution A is now diluted over both A and B. Same is true for the sulfide coming from B. With the A and B components being equal in volume you then have to divide your concentrations in half, thus
When you mix the solutions, the iron that was in solution A is now diluted over both A and B. Same is true for the sulfide coming from B. With the A and B components being equal in volume you then have to divide your concentrations in half, thus
Consider this reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}\tag{1}$$
The equilibrium constant for the reaction is $1.6 \cdot 10^{17}$ at $\pu{298K}$.
This is nothing but a weird statement of the $\mathrm{K}_\mathrm{sp}$ for FeS. Instead of concentrations let's use activities, $a$, for the species which is formally correct.
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.2 M}$ of $\ce{S^{-2}}$ are mixed, the concentration of $\ce{Fe^{+2}}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.
Since the volume of the mixed solution is twice the volume of either of the solutions of the individual ions, the concentrations will be cut in half. So nominally the final solution would have $\pu{0.03 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.1 M}$ of $\ce{S^{-2}}$.
We can assume that the FeS precipitates quantitatively. Obviously the $\ce{S^{-2}}$ is $\pu{0.07 M}$ in excess, so we can assume that is the final concentration of the $\ce{S^{-2}}$. Thus
Consider this reaction: $$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}\tag{1}$$ The equilibrium constant for the reaction is $1.6 \cdot 10^{17}$ at $\pu{298K}$.
This is nothing but a weird statement of the $\mathrm{K}_\mathrm{sp}$ for FeS. Instead of concentrations let's use activities, $a$, for the species which is formally correct.
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.2 M}$ of $\ce{S^{-2}}$ are mixed, the concentration of $\ce{Fe^{+2}}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.
Since the volume of the mixed solution is twice the volume of either of the solutions of the individual ions, the concentrations will be cut in half. So nominally the final solution would have $\pu{0.03 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.1 M}$ of $\ce{S^{-2}}$.
We can assume that the FeS precipitates quantitatively. Obviously the $\ce{S^{-2}}$ is $\pu{0.07 M}$ in excess, so we can assume that is the final concentration of the $\ce{S^{-2}}$. Thus
I hate these kinds of stupid problems that use significant figures incorrectly. If the answer is supposed to be given to three significant figures then the $\mathrm{K}_\mathrm{sp}$, $\ce{[Fe^{+2}]}$ and $\ce{[S^{-2}]}$ should be given to three significant figures too. More
This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)} \tag{1}$$
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).
However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.
Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:
$$\ce{FeS (s) <=> F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$
Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:
$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$
$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07} = 7.0 $$
Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.
Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:
$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$
$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$
Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).
This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)} \tag{1}$$
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).
However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.
Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:
$$\ce{FeS (s) <=> F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$
Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:
$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$
$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07} = 7.0 $$
Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.
Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:
$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$
$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$
Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).
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When you mix the solutions, the iron that was in solution A is now diluted over both A and B. Same is true for the sulfide coming from B. With the A and B components being equal in volume you then have to divide your concentrations in half, thus
$K=\dfrac{1}{(0.03-x)(0.1-x)}=\dfrac{1}{(y)(0.07+y)}$
where $y=0.03-x$ is the iron concentration you want. Conclude that
$y(y+0.07)=1/K$
and simplify by noting that $y$ will be much less than $0.07$.
When you mix the solutions, the iron that was in solution A is now diluted over both A and B. Same is true for the sulfide coming from B. With the A and B components being equal in volume you then have to divide your concentrations in half, thus
$K=\dfrac{1}{(0.03-x)(0.1-x)}=\dfrac{1}{(y)(0.07+y)}$
where $y=0.03-x$ is the iron concentration you want. Conclude that
$y(y+0.07)=1/K$
and simplify by noting that $y$ will be much less than $0.07$.
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This is nothing but a weird statement of the $\mathrm{K}_\mathrm{sp}$ for FeS. Instead of concentrations let's use activities, $a$, for the species which is formally correct.
$$\mathrm{K}_\mathrm{eq} = \dfrac{a_\mathrm{FeS}}{a_\ce{Fe^{+2}}a_\ce{S^{-2}}}\tag{2}$$
Since FeS is a solid, its activity, in an aqueous solution, is 1 by definition. But $\mathrm{K}_\mathrm{sp} = a_\ce{Fe^{+2}}a_\ce{S^{-2}}$ so:
$$\mathrm{K}_\mathrm{eq} = \dfrac{1}{\mathrm{K}_\mathrm{sp}}\tag{3a}$$ $$\mathrm{-- or --}$$ $$\mathrm{K}_\mathrm{sp} = \dfrac{1}{\mathrm{K}_\mathrm{eq}} = 6.25\cdot10^{-18}\tag{3b}$$
Now, let's make the typical substitutions:
$$a_\ce{Fe^{+2}} \approx [\ce{Fe^{+2}}]\tag{4}$$ $$a_\ce{S^{-2}} \approx [\ce{S^{-2}}]\tag{5}$$
Therefore:
$$\mathrm{K}_\mathrm{sp} =\ce{[Fe^{+2}][S^{-2}]}\tag{6}$$
Since the volume of the mixed solution is twice the volume of either of the solutions of the individual ions, the concentrations will be cut in half. So nominally the final solution would have $\pu{0.03 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.1 M}$ of $\ce{S^{-2}}$.
We can assume that the FeS precipitates quantitatively. Obviously the $\ce{S^{-2}}$ is $\pu{0.07 M}$ in excess, so we can assume that is the final concentration of the $\ce{S^{-2}}$. Thus
$$\ce{[Fe^{+2}]_\mathrm{final}} = \dfrac{\mathrm{K}_\mathrm{sp}}{[S^{-2}]_\mathrm{final}}= \dfrac{6.25\cdot10^{-18}}{0.07}= 8.9\cdot10^{-17}$$
This is nothing but a weird statement of the $\mathrm{K}_\mathrm{sp}$ for FeS. Instead of concentrations let's use activities, $a$, for the species which is formally correct.
$$\mathrm{K}_\mathrm{eq} = \dfrac{a_\mathrm{FeS}}{a_\ce{Fe^{+2}}a_\ce{S^{-2}}}\tag{2}$$
Since FeS is a solid, its activity, in an aqueous solution, is 1 by definition. But $\mathrm{K}_\mathrm{sp} = a_\ce{Fe^{+2}}a_\ce{S^{-2}}$ so:
$$\mathrm{K}_\mathrm{eq} = \dfrac{1}{\mathrm{K}_\mathrm{sp}}\tag{3a}$$$$\mathrm{-- or --}$$$$\mathrm{K}_\mathrm{sp} = \dfrac{1}{\mathrm{K}_\mathrm{eq}} = 6.25\cdot10^{-18}\tag{3b}$$
Now, let's make the typical substitutions:
$$a_\ce{Fe^{+2}} \approx [\ce{Fe^{+2}}]\tag{4}$$$$a_\ce{S^{-2}} \approx [\ce{S^{-2}}]\tag{5}$$
Therefore:
$$\mathrm{K}_\mathrm{sp} =\ce{[Fe^{+2}][S^{-2}]}\tag{6}$$
Since the volume of the mixed solution is twice the volume of either of the solutions of the individual ions, the concentrations will be cut in half. So nominally the final solution would have $\pu{0.03 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.1 M}$ of $\ce{S^{-2}}$.
We can assume that the FeS precipitates quantitatively. Obviously the $\ce{S^{-2}}$ is $\pu{0.07 M}$ in excess, so we can assume that is the final concentration of the $\ce{S^{-2}}$. Thus
$$\ce{[Fe^{+2}]_\mathrm{final}} = \dfrac{\mathrm{K}_\mathrm{sp}}{[S^{-2}]_\mathrm{final}}= \dfrac{6.25\cdot10^{-18}}{0.07}= 8.9\cdot10^{-17}$$
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