Answer:
Explanation has been given below.
Explanation:
Let's consider solubility equilibrium of an ionic insoluble compound e.g. ![BaSO_{4}](https://tex.z-dn.net/?f=BaSO_%7B4%7D)
![BaSO_{4}\rightleftharpoons Ba^{2+}+SO_{4}^{2-}](https://tex.z-dn.net/?f=BaSO_%7B4%7D%5Crightleftharpoons%20Ba%5E%7B2%2B%7D%2BSO_%7B4%7D%5E%7B2-%7D)
Equilibrium constant of this solubility equilibrium is represented in terms of solubility product (
) which is expressed as-
![K_{sp}=[Ba^{2+}][SO_{4}^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BSO_%7B4%7D%5E%7B2-%7D%5D)
Now, if we add an ionic salt e.g.
with a common ion
then concentration of
increases.
But, at a constant temperature,
is constant.
Therefore, to keep
constant, excess amount of
will combine with free
ion in solution and produce
.
Hence, as a whole, solubility of
decreases.
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Answer:
sry I don't know much about chemistry
A 25.00 ml sample of hydrochloric acid solution, HCl, is titrated with 0.0512 m NaOH solution. the volume of NaOH solution required is 21.68 ml then the molarity of the HCl solution is 0.044 M .
Calculation ,
Formula used :
...( i )
Where M is the molarity or concentration and V is the volume in ml .
concentration of hydrochloric acid solution (
) = ?
concentration of NaOH (
) = 0.0512 M
volume of hydrochloric acid solution (
) = 25.00 ml
volume of NaOH (
) = 21.68 ml
Putting the value of concentration , volume of both in equation ( i ) we get .
× 25.00 ml = 0.0512 × 21.68 ml
= 0.0512 × 21.68 ml / 25.00 ml= 0.044 M
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