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Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.

(a) TlCl( s ) in 1.250 M HCl

(b) PbI 2 ( s ) in 0.0355 M CaI 2

(c) Ag 2 CrO 4 ( s ) in 0.225 L of a solution containing 0.856 g of K 2 CrO 4

(d) Cd(OH) 2 ( s ) in a solution buffered at a pH of 10.995

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Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.

(a) TlCl( s ) in 0.025 M TlNO 3

(b) BaF 2 ( s ) in 0.0313 M KF

(c) MgC 2 O 4 in 2.250 L of a solution containing 8.156 g of Mg(NO 3 ) 2

(d) Ca(OH) 2 ( s ) in an unbuffered solution initially with a pH of 12.700

(a) [Cl ] = 7.6 × 10 −3 M
Check: 7.6 × 10 3 0.025 × 100 % = 30 %
This value is too large to drop x . Therefore solve by using the quadratic equation:
[Ti + ] = 3.1 × 10 –2 M
[Cl ] = 6.1 × 10 –3
(b) [Ba 2+ ] = 7.7 × 10 –4 M
Check: 7.7 × 10 4 0.0313 × 100 % = 2.4 %
Therefore, the condition is satisfied.
[Ba 2+ ] = 7.7 × 10 –4 M
[F ] = 0.0321 M ;
(c) Mg(NO 3 ) 2 = 0.02444 M
[ C 2 O 4 2− ] = 2.9 × 10 −5
Check: 2.9 × 10 −5 0.02444 × 100 % = 0.12 %
The condition is satisfied; the above value is less than 5%.
[ C 2 O 4 2− ] = 2.9 × 10 −5 M
[Mg 2+ ] = 0.0244 M
(d) [OH ] = 0.0501 M
[Ca 2+ ] = 3.15 × 10 –3
Check: 3.15 × 10 −3 0.050 × 100 % = 6.28 %
This value is greater than 5%, so a more exact method, such as successive approximations, must be used.
[Ca 2+ ] = 2.8 × 10 –3 M
[OH ] = 0.053 × 10 –2 M

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Explain why the changes in concentrations of the common ions in [link] can be neglected.

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Explain why the changes in concentrations of the common ions in [link] cannot be neglected.

The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.

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Calculate the solubility of aluminum hydroxide, Al(OH) 3 , in a solution buffered at pH 11.00.

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Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.

CaSO 4 ∙2H 2 O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L.

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Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract ( [link] ). This use of BaSO 4 is possible because of its low solubility. Calculate the molar solubility of BaSO 4 and the mass of barium present in 1.00 L of water saturated with BaSO 4 .

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Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10 –3 M ) of SO 4 2− because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO 4 (“gyp” water) as a result or passing through soil containing gypsum, CaSO 4 ·2H 2 O, meet these standards? What is SO 4 2− in such water?

4.8 × 10 –3 M = [ SO 4 2− ] = [Ca 2+ ]; Since this concentration is higher than 2.60 × 10 –3 M , “gyp” water does not meet the standards.

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Practice Key Terms 4

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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