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Calculate the solubility of $\ce{Ag3AsO4}$ in $0.02~\mathrm{M}~\ce{K3AsO4}$ neglecting the activity coefficients. Find the relative error. $K_\mathrm{sp}(\ce{Ag3AsO4}) = 6.0 \cdot 10^{-23}$

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:

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

$$K_\mathrm{sp} = 27x^4$$

$$x = 2,78 \cdot 10^{-6}$$

$$[\ce{Ag+}] = 3x = 8,34 \cdot 10^{-6}~\mathrm{M}$$

$$[\ce{AsO4^3-}] = x = 2,78 \cdot 10^{-6}~\mathrm{M}$$

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

$$K_\mathrm{sp} = (8,34 \cdot 10^{-6} - 3x)^3 \times (2,78 \cdot 10^{-6} + 0,02 - x) = 6.0 \cdot 10^{-6}$$

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?

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    $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved the formatting of your post by adding MathJax markup; for more information on how to do so yourself, check out the help center, this meta-post or this one. As per our homework policy, this is a homework question; but this is okay for you, since you already showed your work. $\endgroup$ – Jan Nov 26 '15 at 19:07
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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:

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


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.

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  • $\begingroup$ At the beginning you didn't account for the amount of arsenate produced by the dissociation of potassium arsenate. Why so? $\endgroup$ – RBW Nov 26 '15 at 20:46
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    $\begingroup$ @Marko At the beginning I was merely correcting OP’s calculations. But I’ll make that clearer. $\endgroup$ – Jan Nov 26 '15 at 20:53
  • $\begingroup$ Thank you. I really wonder how the lecturer will solve this problem in the class. If she shows an easier method, I will share it here. $\endgroup$ – Wellenbrecher Nov 27 '15 at 17:00
  • $\begingroup$ The easier method is to neglect x in 0.02+x: Ksp=(3x)^3(x+0.02 M)=(3x)^3*0.02 M $\endgroup$ – RBW Nov 28 '15 at 11:52
  • $\begingroup$ @Jan check this asnwer: chemistry.stackexchange.com/a/42231/23304 $\endgroup$ – Wellenbrecher Dec 14 '15 at 18:15
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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:

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

$$S = 0.0223~\mathrm{M}$$

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 $S = 0.018~\mathrm{M}$.

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

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    $\begingroup$ Two notes: please write the entire equation with MathJax and please use \ce{...} for chemical formulae such as $\ce{AsO4^3-}$. $\endgroup$ – Jan Nov 26 '15 at 20:21
  • $\begingroup$ @Uros check this asnwer: chemistry.stackexchange.com/a/42231/23304 $\endgroup$ – Wellenbrecher Dec 14 '15 at 18:15
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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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