The relation between stability and solubility

Chemistry

1 answer:

Alex17521 [72]3 years ago

3 0

Answer:

Nobody is able to foresee the solubility of a product. There are some experimental rules, but they all have exceptions, that nobody is able to explain.

Just have a look on the Calcium salts made with the halogens (F, Cl, Br, I). There is a nice analogy among Cl, Br and I, but not F. Look ! The Calcium chloride CaCl2, bromide CaBr2 and iodide CaI2 are all extremely soluble in water. They are all soluble in less than their weight of water. But calcium fluoride CaF2 is among the least soluble product on Earth. The principal mineral for Fluoride is CaF2, and it can be found everywhere at the surface of the Earth. If this mineral would have been at least a little soluble in water, the rains would have washed away this mineral in the geological times. Why is there such a huge difference between calcium fluoride and the other halogenides ? Nobody knows !

Another example. Potassium perchlorate is the only potassium salt which is very weakly soluble in water. By comparison, Sodium perchlorate is extremely soluble in water. Why? Another example: the number of Silver compounds which are soluble in water is limited. In organic solvents, it is even worse. But the Handbook of Chemistry and Physics says that Silver perchlorate is soluble in toluene. Why? Nobody knows.

From time to time there are articles published in journals like the Journal of Chemical Education. The author is proud of displaying a theory filled with new parameters, for explaining the solubility of quite a lot of chemicals. But there are always exceptions, that he regrets not to be able to explain.

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Answer:

The relevant equation is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Explanation:

1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.

The formed CO₂ is the reason why you noticed bubbles as the reaction took place

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A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$