Chemistry
10th Edition
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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Question

Transcribed Image Text:Group I Cations
A
1. Explain why litmus paper is used in step 4. Why is blue litmus paper used rather than red
litmus paper?
litmus Paper is used in Step y because
2. Several steps involve the use of reactive compounds. Based on your laboratory
experience, identify these steps and indicate any appropriate precautions you need
to take.
![Qualitative Analysis
Step 2. The solubility of lead is sufficiently high that at low lead concentrations, all of the
in addition, the solubility of lead increases about threefold in hot water compared to room tem
perature water; thus, a significant amount of lead can be solubilized from the mixture of AgCl,
H92CI2, and P6CI2 by treating the mixture with hot water. Therefore, centrifuging the resulting
solution leaves only a solution of lead (if lead is present).
Step 3. The presence of lead can be confirmed or eliminated by the addition of chromate to
form the bright yellow solid PbCrO. (Equation 8). You may recognize this color because lead
chromate is often the pigment in the yellow paint used to paint yellow lines on highways.
Equation (8)
Pb2*(aq) + Cro?-(aq) PbCrO4(s) Ksp = 1.8 × 10-14
Step 4. The addition of aqueous ammonia results in the formation of the complex ion
[Ag(NH3)21*, solubilizing the silver, but also in the disproportionation of the Hg* (i.e., Hg2") into
Hg° metal and Hg2*, which forms the solid H9NH,CI under these conditions (Equations 9, 10,
and 11).
Equation (9)
Hg2*(aq) :
= Hg°(s) + Hg2*(aq)
Equation (10)
Hg2*(aq) + CI(aq) + 2 NH3(aq) = H9NH2CI(s) + NHẬ(aq)
Equation (11)
AgCl(s) + 2 NH3(aq) =
2[Ag(NH3)2]*(aq) + CI¯(aq)
Finally, to confirm the presence of silver, the addition of acid results in the protonation of the
ammonia to the ammonium cation, resulting in the reprecipitation of silver chloride (Equation 12).
Equation (12)
[Ag(NH3)2]*(aq) + 2 H*(aq) + CI"(aq)=AgCI(s) + 2[NH4J*(aq)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ce781c6-ca6f-48b8-a006-1cccfb769dcf%2F3aca54e7-6156-4f9b-8688-adf875196d86%2F6av4xz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Qualitative Analysis
Step 2. The solubility of lead is sufficiently high that at low lead concentrations, all of the
in addition, the solubility of lead increases about threefold in hot water compared to room tem
perature water; thus, a significant amount of lead can be solubilized from the mixture of AgCl,
H92CI2, and P6CI2 by treating the mixture with hot water. Therefore, centrifuging the resulting
solution leaves only a solution of lead (if lead is present).
Step 3. The presence of lead can be confirmed or eliminated by the addition of chromate to
form the bright yellow solid PbCrO. (Equation 8). You may recognize this color because lead
chromate is often the pigment in the yellow paint used to paint yellow lines on highways.
Equation (8)
Pb2*(aq) + Cro?-(aq) PbCrO4(s) Ksp = 1.8 × 10-14
Step 4. The addition of aqueous ammonia results in the formation of the complex ion
[Ag(NH3)21*, solubilizing the silver, but also in the disproportionation of the Hg* (i.e., Hg2") into
Hg° metal and Hg2*, which forms the solid H9NH,CI under these conditions (Equations 9, 10,
and 11).
Equation (9)
Hg2*(aq) :
= Hg°(s) + Hg2*(aq)
Equation (10)
Hg2*(aq) + CI(aq) + 2 NH3(aq) = H9NH2CI(s) + NHẬ(aq)
Equation (11)
AgCl(s) + 2 NH3(aq) =
2[Ag(NH3)2]*(aq) + CI¯(aq)
Finally, to confirm the presence of silver, the addition of acid results in the protonation of the
ammonia to the ammonium cation, resulting in the reprecipitation of silver chloride (Equation 12).
Equation (12)
[Ag(NH3)2]*(aq) + 2 H*(aq) + CI"(aq)=AgCI(s) + 2[NH4J*(aq)
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