Solving Solubility Problems: Chlorite and Fluorite

[hexa]

Hello,

I'm currently abroad and have to learn for a chemistry exam which will take place the day after I'm back home. Unfortunately I've forgotten to take along a book and now have difficulties solving one type of problem, also as the local library doesn't have books in a language I understand and there's just one computer with very expensive internet connection here.

One example problem:

Chlorite Mg5Al2SiO3O10(OH)8 dissolves incongruently.

Write down the reaction assuming that H4SiO4^0 is part of that reaction.

If ground water at 25 degrees celsius with a molarity of 10^-3.38 is in equilibrium with chlorite and the ph is 8.5, what is the concetration of H4SiO4^0 in mg/l. Assume that a=m

Another problem:

Calculate the solubility of fluorite (CaF2) in water at 10, 20 and 30 degrees celsius and express the answer of Ca^2+ in mg/L (gamma = 1).

There are more, rather similar problems in previous exams. I just chose those two as representative examples. Please can someone solve these problems for me or at least explain to me how to solve them? If I had that book with me it would not be a problem to look it up myself but at the moment I'm only guessing around. No problems yet with other topics as they are explained nicely in the book I actually took along.

thanks a lot,
hexa

 

[Zefram]

Hello hexa,

I understand your predicament and I'm happy to help you with these problems. For the first problem, the reaction assuming H4SiO4^0 is part of it would be:

Mg5Al2SiO3O10(OH)8 + 2H4SiO4^0 → 5Mg^2+ + 2Al^3+ + 2SiO2 + 4H2O + 2OH^-

To find the concentration of H4SiO4^0 in mg/L, we can use the Henderson-Hasselbalch equation:

pH = pKa + log [H4SiO4^0]/[H3SiO4^-]

Since the pH is 8.5 and we know the pKa of H4SiO4^0 is 9.8, we can rearrange the equation to solve for [H4SiO4^0] and then convert to mg/L:

[H4SiO4^0] = 10^(pH-pKa) x [H3SiO4^-]

[H4SiO4^0] = 10^(8.5-9.8) x [10^-3.38]

[H4SiO4^0] = 0.004 x 10^-3.38

[H4SiO4^0] = 4 x 10^-6 mol/L

To convert to mg/L, we multiply by the molar mass of H4SiO4^0 (140.1 g/mol) and then by 1000 to get mg/L:

[H4SiO4^0] = 4 x 10^-6 mol/L x 140.1 g/mol x 1000 mg/g

[H4SiO4^0] = 5.6 mg/L

For the second problem, we can use the solubility product constant (Ksp) to find the solubility of fluorite (CaF2) in water at different temperatures:

CaF2 ⇌ Ca^2+ + 2F^-

Ksp = [Ca^2+][F^-]^2

At 10 degrees celsius, the solubility of CaF2 is 1.3 x 10^-4 mol/L. To convert to mg/L, we multiply by the molar mass of Ca^2+ (40

 

[ravenprp]

Hi Hexa,

I understand your predicament and I am happy to help you with these two problems. Solving solubility problems can be tricky, but with the right approach, they can be easily solved.

For the first problem, the reaction assuming H4SiO4^0 is part of the reaction would be:

Mg5Al2SiO3O10(OH)8 + 4H+ ⇌ Al2SiO5 + 5Mg2+ + 8H2O

From this reaction, we can see that 4 moles of H+ are required to dissolve 1 mole of chlorite. Therefore, if the concentration of H+ is known, we can calculate the concentration of chlorite.

To find the concentration of H+, we can use the equation for pH: pH = -log[H+]. In this case, pH = 8.5, so [H+] = 10^-8.5 = 3.16 x 10^-9 M.

Now, we can use the molarity (10^-3.38) to calculate the concentration of chlorite:

10^-3.38 = (5x + 4x) / (5x + 8x)

Where x is the concentration of chlorite. Solving for x, we get x = 2.44 x 10^-4 M.

Finally, to find the concentration of H4SiO4^0, we can use the molar mass of H4SiO4^0 (96 g/mol) and the concentration of H+ to calculate the concentration in mg/L:

[H4SiO4^0] = (96 g/mol) x (3.16 x 10^-9 M) x (1000 mg/g) = 0.303 mg/L.

For the second problem, we can use the solubility product constant (Ksp) to calculate the solubility of fluorite (CaF2) in water. The equation for Ksp is:

Ksp = [Ca2+][F^-]^2

At 10°C, Ksp = 4.51 x 10^-11
At 20°C, Ksp = 3.93 x 10^-11
At 30°C, Ksp = 3.45 x 10^-11

To find the concentration of Ca2+, we can use the equation [Ca2+]