Otdfbt

I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.

40. Du tillsätter $pu100 ml$ $pu0,01 M$ $ceNa2SO4$-lösning i $pu100 ml$ $pu0,02 M$ $ceCaCl2$-lösning. Det bildas en $ceCaSO4$-fällning. Hur mycket av $pu0,01 M$ $ceNa2SO4$-lösning måste ännu tillsättas för att den bildade $ceCaSO4$-fällningen just och just ska upplösas helt? $K_mathrms(ceCaSO4) = pu2,5e-5 M$. $(t = pu25 °C)$ (2p)

a. $pu0,43 l$

b. $pu0,48 l$

c. $pu0,53 l$

d. $pu0,58 l$

Translation: We add 100 ml of 0.01 M $ceNa2SO4$-solution to 100 ml of 0.02 M $ceCaCl2$-solution. A precipitate of $ceCaSO4$ is formed. What is the (minimal) volume of 0.01 M $ceNa2SO4$-solution that needs to be added to the mix for the $ceCaSO4$ precipitate to be just dissolved completely?

The problem then lists the solubility constant, $K_mathrms$, for $ceCaSO4$ as $2.5cdot 10^-5$.

The only way I can think of solving this myself is by looking at the added $ceNa2SO4$-solution as just water and using the $K_mathrms$ value for $ceCaSO4$ to see what concentration of $ceCaSO4$ is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:

$$K_mathrms(ceCaSO4) = pu2.5e-5 M^2 = [ceCa^2+][ceSO4^2-]$$

$$therefore [ceCa^2+] = [ceSO4^2-] = sqrtpu2.5e-5 M^2 = pu5e-3 M$$

$$C = fracnV to n = CV$$

Amount of precipitate formed:

$$n(ceCaSO4) = pu0.02 M cdot pu0.1 dm^3 = pu0.002 mol$$

Required total volume of water to dissolve $pu0.002 mol$ of $ceCaSO4$:

$$V = fracnC = fracpu0.002 molpu5e-3 M = pu0.4 dm^3$$

This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well $ceCaSO4$ is dissolved in a $ceNa2SO4$-solution as opposed to just water.

As you've already noted, there are 3 substances of interest here: $ceCa^2+, ceSO4^2-, ceH2O$.

Let $V$ be the additional volume of sodium sulfate solution. Then:

The total volume is $V + 0.2 mathrmL$.

The total amount of sulfate is $(V + 0.1 mathrmL) (0.01 mathrmM)$.

The total amount of calcium is $(0.1 mathrmL) (0.02 mathrmM)$.

We have then the following relationship:

$$ce[Ca^2+][SO4^2-]=frac(0.1 mathrmL) (0.02 mathrmM)V + 0.2 mathrmLfrac(V + 0.1 mathrmL) (0.01 mathrmM)V + 0.2 mathrmLleq K_mathrmsp = 2.5times 10^-5$$

Now, it's just algebra to solve for the threshold value of $V$, keeping in mind that there are two solutions for $V$ because this is a quadratic equation. In this case, the correct solution is the positive root:

$$V approx 0.48 mathrmL$$

The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.

When all of the $ceCaSO4$ is dissolved completely, all the calcium ions originate from the $pu100 mL ; ceCaCl2$ solution, so it is easy to determine the amount:

$$n_ceCa^2+ = pu100 mL times 0.02 mol/L = 2 mmol$$

To get the total sulfate amount, we have to consider what is present initially and what we add when we add a volume of $V_textadd$ of $ceNa2SO4$.

$$n_ceSO4^2- = left(pu100 mL + V_textaddright) times pu0.01 mol/L$$

From this, we can figure out the concentration:

$$c_ceCa^2+ = fracpu2 mmol V_textadd +pu200 mL $$

$$c_ceSO4^2- = fracpu(100 mL + V_textadd) times pu0.01 mol/L V_textadd +pu200 mL $$

The product has to be equal to $K_s$:

$$K_s = fracpu2 mmol V_textadd +pu200 mL times fracpu(100 mL + V_textadd) times pu0.01 mol/L V_textadd +pu200 mL$$

This gives a quadratic equation for $V_textadd$, giving you the numeric result upon solving it. The answer will be approximate because it assumes that volumes are additive and that activities are independent of spectator ion concentrations.

Goggle translate from Swedish yields a slightly different translation.

You add $pu100 ml$ of $pu0.01 M ; ceNa2SO4$ solution in $pu100 ml$ of $pu0.02 M ; ceCaCl2$ solution. A $ceCaSO4$ precipitate forms. How much of $pu0.01 M; ceNa2SO4$ solution has yet to be added to completely dissolve the $ceCaSO4$ precipitate formed? $K_mathrms ce(CaSO4) = pu2.5e−5 M. (t =pu25 ^circC)$

The key point being that the problem is looking for the "extra" amount of the $ceNa2SO4$ solution to be added, not including the initial $pu100 ml$.

To start let's convert the volumes to liters. So $pu0.100 L$ of $pu0.01 M ; ceNa2SO4$ solution is added to $pu0.100 L$ of $pu0.02 M ; ceCaCl2$ solution.

Let's let $V_e$ be the extra volume of the $pu0.01 M ; ceNa2SO4$ solution, which needs to be added.

We know that for the product of the final concentrations, $ce[Ca^2+]_f$ and $ce[SO4^2-]_f$, is equal to the $K_mathrmsp$.

$K_mathrmsp = ce[Ca^2+]_f cdot [SO4^2-]_f $

Now the amount of $ceCa^2+$ is $pu0.1 mol/L times pu0.02 L = pu0.002 mol$. There is $pu0.2 L$ of solution initially to which $V_e ; puL$ of additional $pu0.01 M ; ceNa2SO4$ solution is added. So the final volume is $left(0.2 + V_eright)$ liters. Thus,

$ce[Ca^2+]_f = dfrac0.0020.2+V_e$

The amount of $ceSO4^2-$ is $(0.1+V_e)times0.01$ moles. There is $0.2$ liters of solution initially to which $V_e$ liters of additional $pu0.01 M ; ceNa2SO4$ solution is added. So the final volume is again $left(0.2 + V_eright)$ liters. Thus:

$ce[SO4^2-]_f = dfrac(0.1+V_e)times0.010.2+V_e$

Now making the substitutions:

$K_mathrmsp = dfrac0.0020.2+V_e times dfrac(0.1+V_e)times0.010.2+V_e$

Solving the quadratic equation yields $V_e approx pu0.483 L$