Page 89 - Basic physical chemistry for the atmospheric sciences

Solution chemistry and aqueous equilibria

is the molarity of cuprous chloride in water, C moles of cuprous
chloride will dissolve in 1 L of water to produce C moles of the
chloride ion (Cl - ) and C moles of the cuprous ion (Cu + ) . Hence, K,r
2
= 3 . 2 x 1 0 - 7 = C or = 0 .00057 M. Therefore, only . 7 x 0 - 4 moles
1
C
5
of cuprous chloride dissolve in 1 L of water. (Note: We could
have predicted from Table 4. 1 that the solubility of CuCl in water
.
is low )
Since Ksp in Eq. (4 9 ) i s the product o f i o n concentrations, i t i s
.
called the ion product constant r solubility product (hence the sub­
o
script "sp"). U s e of the solubility product is generally (and most
accurately) used for substances with small solubilities. The solubility
products for a number of salts and ions are listed in Appendix IV(c).
Exercise 4.3. Calculate the solubility of barium hydroxide,
Ba(OHh(s), in water at 25°C given that its solubility product is
5 x 1 0 - 3 at 25°C.
Solution. The reaction for the dissociation of barium hydroxide in
i
water s
2
Ba(O h (s) µ Ba + (aq) + 2(0H - ( aq)) (4. 10)
H
The solubility product is
2 2 3
[Ba + (aq)] [OH-(aq)] = Ksp = 5 X 1 0 - (4. 1 1 )
From Reaction (4. 1 0) we see that two moles of OH -(aq) are produced
for every one mole of Ba2+ (aq). Therefore,
2
[OH - (aq)] = 2[Ba + ( aq)] (4. 1 2 )
2
2
2
3
From Eqs. (4. 1 1 ) and (4. 1 2 , [Ba + (aq)J{2[Ba + (aq)]} = 5 x 10 - , or
)
3
2
2
4 [Ba + ( aq)] = 5 x 1 0 - 3 • Therefore, [Ba + ( aq)] = O. l M . Since, from
Reaction (4. 1 0), I mole of barium hydroxide dissolves for every 1
2
mole of Ba + (aq) that forms, the solubility of barium hydroxide in
water is 0. 1 M .
Even very small solubility products can be measured electrically,
and these values are listed in chemical tables. As the above exercise
s
illustrate , the solubility of a substance can be derived from its solubil­
ity product . Solubility products are generally listed only for slightly or
sparingly soluble substances. If K,P is very small, the substance is
often termed insoluble (in water). In the case of moderately and highly
soluble substances ( s uch as NaCl or NaOH), the use of solubility
products is not very useful. This is because instead of defining an
equilibrium constant in terms of concentrations we would have to