# Solubility of Ag3AsO4 in 0.02 M K3AsO4

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

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?

• 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.
– Jan
Nov 26, 2015 at 19:07

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.

• At the beginning you didn't account for the amount of arsenate produced by the dissociation of potassium arsenate. Why so?
– EJC
Nov 26, 2015 at 20:46
• @Marko At the beginning I was merely correcting OP’s calculations. But I’ll make that clearer.
– Jan
Nov 26, 2015 at 20:53
• 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. Nov 27, 2015 at 17:00
• The easier method is to neglect x in 0.02+x: Ksp=(3x)^3(x+0.02 M)=(3x)^3*0.02 M
– EJC
Nov 28, 2015 at 11:52
• @Jan check this asnwer: chemistry.stackexchange.com/a/42231/23304 Dec 14, 2015 at 18:15

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 $$23.9~\%$$.

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.