A bit confused by TBR's explanation for question #64 from passage X of their Equilibrium Section (Section III) of their GChem book. If someone can offer an alternative explanation or clear TBR's up that would be great! The question is found on page 212 btw. Thanks!
TBR Gchem Equilibrium Question
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passage states that
INSOLUBLE TO THE EYE IS < 1 mg/mL = < 1 g/L
question is asking about Zn and Ag (Zn = 65 g/mol, Ag = ~100 g/mol)
Simple Conversion
so Ag will be visible if it is < 0.01 M
and Zn will be visible if it is < 0.015 M
question wants you to find the case in which one of the ions is visible and the other isnt so were looking for a Zn/Ag compound where the molar solubility for
one is > the insoluble to the eye M and
the other is < the insoluble to the eye M
Choices
A) 0.1 M Na2S
if you look at the sulfide compounds they are Ag2S 10^-17 and ZnS 10^-11 so do both these molar solubilities fall on both sides of our insoluble to the eye requirement....NO so cross out A
B) 0.1 M NaCl
AgCl 1.2 x 10^-5 = 0.000012
ZnCl2 3.1 x 10^-2 = 0.031
does these both fall on opposite sides of our insoluble to the eye requirement....yes
AgCl molar solubility is less than our insoluble to the eye requirement hence it will precipitate while
ZnCl2 is greater than our requirement meaning that it will dissociate
so choice B is most probably the correct answer but lets continue
C) 0.1 M NaNO3
not sure if my reasoning for this one was right i cancelled it out b/c NaNO3 is slightly soluble and i think both will dissociate into an NO3 soln? better reason is that its not in the passage so no numbers to work off of so cross it out
D) 0.1 M Na2CO3
Ag2CO3 = 1.2 x 10^-4
ZnCO3 = 1.4 x 10^-5
both will be seen as percipitate so NO cross this off
the answer is B 0.1 M NaCl
just as a note i originally missed this question b/c i failed to recognize the bit in the passage where they tell you that insoluble to the eye is < 1 mg/mL.... I'm fairly confident thats what probably trips most people up.
im pretty sure this reasoning is right
INSOLUBLE TO THE EYE IS < 1 mg/mL = < 1 g/L
question is asking about Zn and Ag (Zn = 65 g/mol, Ag = ~100 g/mol)
Simple Conversion
so Ag will be visible if it is < 0.01 M
and Zn will be visible if it is < 0.015 M
question wants you to find the case in which one of the ions is visible and the other isnt so were looking for a Zn/Ag compound where the molar solubility for
one is > the insoluble to the eye M and
the other is < the insoluble to the eye M
Choices
A) 0.1 M Na2S
if you look at the sulfide compounds they are Ag2S 10^-17 and ZnS 10^-11 so do both these molar solubilities fall on both sides of our insoluble to the eye requirement....NO so cross out A
B) 0.1 M NaCl
AgCl 1.2 x 10^-5 = 0.000012
ZnCl2 3.1 x 10^-2 = 0.031
does these both fall on opposite sides of our insoluble to the eye requirement....yes
AgCl molar solubility is less than our insoluble to the eye requirement hence it will precipitate while
ZnCl2 is greater than our requirement meaning that it will dissociate
so choice B is most probably the correct answer but lets continue
C) 0.1 M NaNO3
not sure if my reasoning for this one was right i cancelled it out b/c NaNO3 is slightly soluble and i think both will dissociate into an NO3 soln? better reason is that its not in the passage so no numbers to work off of so cross it out
D) 0.1 M Na2CO3
Ag2CO3 = 1.2 x 10^-4
ZnCO3 = 1.4 x 10^-5
both will be seen as percipitate so NO cross this off
the answer is B 0.1 M NaCl
just as a note i originally missed this question b/c i failed to recognize the bit in the passage where they tell you that insoluble to the eye is < 1 mg/mL.... I'm fairly confident thats what probably trips most people up.
im pretty sure this reasoning is right
