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.

You might be interested in

Who can help me with this? Asap

aivan3 [116]

Answer:

A. = 143

B. = 8

C. =32

Explanation:

Number of neutron = mass number - atomic number

3 0

3 years ago

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points

labwork [276]

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

$Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

$NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

$H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

$NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$