PbCl2(s) is considerably more soluble in HCl(aq) than in pure water, but its solubility in HNO3(aq) is not much different from what it is in water. Explain this difference in behavior. A lead(II) ion can form a complex ion with chloride ions, but forms no such complex ion with nitrate ions. The formation of this complex decreases the concentration of free lead(II)(aq) and free Cl-(aq). Thus, PBC12 will dissolve in the HCI(aq) up until the value of the solubility product is exceeded. HCl is considered to be a strong acid, whereas HNO3 is a weak acid. Therefore HNO3 does not dissociate appreciably in solution and is not a significant source of nitrate ions for the lead(II) ions to form a complex ion with. The statement is false. A reaction would take place between PbCls(s) and HNO3(aq) to yield HCl(aq) and Pb(NO3)2, which is soluble in water. HCI(g) is more soluble in water than NO2(g), thus the amount of Cl- ions in solution is greater than the concentration of nitrate ions in solution, and thus there is not a significant source of nitrate ions for the lead(II) ions to form a complex ion with. The statement is false. HCI serves as an excellent source of Cl- ions, which would cause the solubility to decrease, not increase, in accord with the common-ion effect.
PbCl2(s) is considerably more soluble in HCl(aq) than in pure water, but its solubility in HNO3(aq) is not much different from what it is in water. Explain this difference in behavior. A lead(II) ion can form a complex ion with chloride ions, but forms no such complex ion with nitrate ions. The formation of this complex decreases the concentration of free lead(II)(aq) and free Cl-(aq). Thus, PBC12 will dissolve in the HCI(aq) up until the value of the solubility product is exceeded. HCl is considered to be a strong acid, whereas HNO3 is a weak acid. Therefore HNO3 does not dissociate appreciably in solution and is not a significant source of nitrate ions for the lead(II) ions to form a complex ion with. The statement is false. A reaction would take place between PbCls(s) and HNO3(aq) to yield HCl(aq) and Pb(NO3)2, which is soluble in water. HCI(g) is more soluble in water than NO2(g), thus the amount of Cl- ions in solution is greater than the concentration of nitrate ions in solution, and thus there is not a significant source of nitrate ions for the lead(II) ions to form a complex ion with. The statement is false. HCI serves as an excellent source of Cl- ions, which would cause the solubility to decrease, not increase, in accord with the common-ion effect.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:12.
PbCl2(s) is considerably more soluble in HCI(aq) than in pure water,
but its solubility in HNO3(aq) is not much different from what it is in
water. Explain this difference in behavior.
A lead(II) ion can form a complex ion with chloride ions, but forms no
such complex ion with nitrate ions. The formation of this complex
decreases the concentration of free lead(II)(aq) and free Cl-(aq). Thus,
PBC12 will dissolve in the HCI(aq) up until the value of the solubility
product is exceeded.
HCl is considered to be a strong acid, whereas HNO3 is a weak acid.
Therefore HNO3 does not dissociate appreciably in solution and is not
a significant source of nitrate ions for the lead(II) ions to form a
complex ion with.
The statement is false. A reaction would take place between PbCls(s)
and HNO3(aq) to yield HCl(aq) and Pb(NO3)2, which is soluble in
water.
HCI(g) is more soluble in water than NO2(g), thus the amount of Cl-
ions in solution is greater than the concentration of nitrate ions in
solution, and thus there is not a significant source of nitrate ions for
the lead(II) ions to form a complex ion with.
The statement is false. HCI serves as an excellent source of Cl- ions,
which would cause the solubility to decrease, not increase, in accord
with the common-ion effect.
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