Answer:
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Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as
The equation for the ionization of the is given as:
By stoichiometry of the reaction:
1 mole of
gives 1 mole of
and 2 mole of
When the solubility of
is S moles/liter, then the solubility of
will be S moles\liter and solubility of
will be 2S moles/liter.
![6.5\times 10^{-6}=[S][2S]^2](https://tex.z-dn.net/?f=6.5%5Ctimes%2010%5E%7B-6%7D%3D%5BS%5D%5B2S%5D%5E2)
Thus solubility of this compound in pure water is 0.012 M
Answer: The concentration of hydrogen ion in the given solution is
.
Explanation:
pOH of a solution is the negative logarithm of concentration of hydroxide ions.
Hence, if
is
then its pOH value is calculated as follows.
![pOH = -log [OH^{-}]\\= -log (3.5 \times 10^{-5})\\= 4.45](https://tex.z-dn.net/?f=pOH%20%3D%20-log%20%5BOH%5E%7B-%7D%5D%5C%5C%3D%20-log%20%283.5%20%5Ctimes%2010%5E%7B-5%7D%29%5C%5C%3D%204.45)
Now, the relation between pH and pOH is as follows.
pH + pOH = 14
pH + 4.45 = 14
pH = 14 - 4.45
= 9.55
And, pH is the negative logarithm of concentration of hydrogen ions. Hence,
is calculated as follows.
![pH = -log [H^{+}]\\9.55 = -log [H^{+}]\\H^{+} = 2.81 \times 10^{-10}](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C9.55%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5CH%5E%7B%2B%7D%20%3D%202.81%20%5Ctimes%2010%5E%7B-10%7D)
Thus, we can conclude that the concentration of hydrogen ion in the given solution is
.