Solubility of Ag3AsO4 in 0.02 M K3AsO4


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> Calculate the solubility of $\ce{Ag3AsO4}$ in $\pu{0.02M}~\ce{K3AsO4}$ *neglecting the activity coefficients*. Find the relative error. $K\_\mathrm{sp}(\ce{Ag3AsO4}) = \pu{6.0e-23}$
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I know how to calculate the relative error but I get a very complicated equation finding the concentration solubility product constant ($K'\_\mathrm{sp}$). There should be a quicker way to solve this since it is a midterm question.


I tried this:


$$\ce{Ag3AsO4 -> 3Ag+ + AsO4^3-}$$


$$K\_\mathrm{sp} = 27x^4$$


$$x = \pu{2.78e-6}$$


$$[\ce{Ag+}] = 3x = \pu{8.34e-6M}$$


$$[\ce{AsO4^3-}] = x = \pu{2.78e-6M}$$


$0.02~\mathrm{M}\ \ce{AsO4^3-}$ comes from $\ce{K2AsO4}$. So there should be an equation like:


$$K\_\mathrm{sp} = (\pu{8.34e-6} - 3x)^3 \times (\pu{2.78e-6} + 0.02 - x) = \pu{6.0e-6}$$


And things get complicated. After finding $x$, I will also have found the final concentrations of silver and $\ce{AsO4^3-}$ ions. Then I will read the activity coefficients of them from the appendix table.


I have two questions:


1. Is my method true or false?
2. What is an easier way of solving this problem?




Your calculations are mathematically correct up to the second step and then I have no clue how you arrived at your value of $x$. Starting from there, $x$ would be:


$$K\_\mathrm{sp} = 27 x^4 \\
\frac{K\_\mathrm{sp}}{27} = x^4 \\
\sqrt[4]{\frac{K\_\mathrm{sp}}{27}} = x \\
x \approx 0.021711\dots$$




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However, to calculate the solubility you should be doing this:


$$K\_\mathrm{sp} = \left[\ce{Ag+}\right]^3 \left[\ce{AsO4^3-}\right]$$


$$c\left (\ce{Ag3AsO4} \right) = x$$


$$K\_\mathrm{sp} = \left(3x \right )^3 \left(x + 0.02~\mathrm{M}\right)\\
K\_\mathrm{sp} = 27x^4 + 0.54~\mathrm{M} \cdot x^3$$


Note that this is the same equation as Uros proposed. I arrived there by saying:


1. $\left [\ce{Ag+}\right] = 3\ c\left(\ce{Ag3AsO4}\right)$ — silver ions all come from dissolved $\ce{Ag3AsO4}$
2. $\left [\ce{AsO4^3-}\right] = c\left(\ce{Ag3AsO4}\right) + 0.02~\mathrm{M}$ — $0.02~\mathrm{M}$ of $\ce{AsO4^3-}$ stem from $\ce{K3AsO4}$, the remaining from dissolved $\ce{Ag3AsO4}$.


Unfortunately, our equation is not (easily) solveable analytically to the best of my knowledge. You need some kind of estimation method. Uros’ estimate of $c\left(\ce{Ag3AsO4}\right) \approx 0.018~\mathrm{M}$ seems pretty accurate.


 2

You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:


Let $S$ be the molar solubility:


$$K\_\mathrm{sp} = 0.02 \times 27 S^3$$


$$S = \pu{0.0223 M}$$


Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:


$$K\_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$


Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.


So the relative error is $23.9~\%$.


 2

The correct solution:


$$K\_\mathrm{sp} = a\_\mathrm{Ag^+}^3 \cdot a\_\mathrm{AsO\_{4}^{3-}} = 6.0 \cdot 10^{-23}$$


$$K\_\mathrm{sp} = K\_\mathrm{sp}^{'} \cdot γ\_\mathrm{Ag^+}^3 \cdot γ\_\mathrm{AsO\_{4}^{3-}}$$


$$K\_\mathrm{sp}^{'} = \frac{K\_\mathrm{sp}}{γ\_\mathrm{Ag^+}^3 \cdot γ\_\mathrm{AsO\_{4}^{3-}}}$$


Ionic strength:


$$μ = \frac{1}{2} \cdot (0,06\cdot1^2 + 0,02\cdot3^2) = 0,12~\mathrm{M}$$


$$0,12~\mathrm{M} ≈ 0,10~\mathrm{M} $$


Activity coefficients at ionic strength $0,12~\mathrm{M}$ are:


$0.75$ for $Ag^+$


$x$ for $AsO\_{4}^{3-}$


Then:


$$K\_\mathrm{sp}^{'} = \frac{K\_\mathrm{sp}}{0.75^3 \cdot x}$$


And the relative error is:
$$\%~error = \frac{|K\_\mathrm{sp}-K\_\mathrm{sp}^{'}|}{K\_\mathrm{sp}} \cdot 100~\% = \frac{|6.0 \cdot 10^{-23}-K\_\mathrm{sp}^{'}|}{6.0 \cdot 10^{-23}} \cdot 100~\%$$


I can't find the activity coefficient for arsenate. If someone can find it, please share it with us.


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