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Chemistry required practical 4 (AQA)- help with method!!!

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rico7102000
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#1
Report Thread starter 3 years ago
#1
i'm struggling to produce a method to distinguish solutions of 9 ionic compounds labelled A-I by a series of test tube reactions. This method needs to be in a logical sequence to prevent too many test being done unnecessarily. These must be test tube reactions so no flame tests.
These are the ionic compounds
potassium iodide
potassium chloride
potassium bromide
potassium sulfate
potassium hydroxide
potassium carbonate
ammonium chloride
barium chloride
magnesium chloride
thank you
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username3718068
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#2
Report 3 years ago
#2
(Original post by rico7102000)
i'm struggling to produce a method to distinguish solutions of 9 ionic compounds labelled A-I by a series of test tube reactions. This method needs to be in a logical sequence to prevent too many test being done unnecessarily. These must be test tube reactions so no flame tests.
These are the ionic compounds
potassium iodide
potassium chloride
potassium bromide
potassium sulfate
potassium hydroxide
potassium carbonate
ammonium chloride
barium chloride
magnesium chloride
thank you
add acidifed AgNO3(aq):
K2SO4(aq) = no visible reaction
KOH(aq) = no visible rxn
K2CO3(aq) = effervescence
KBr(aq) = cream ppt
KI(aq) = yellow ppt
aqueous KCl, BaCl2,NH4Cl, MgCl2 will all give white ppt
=> further test: add warm aqueous NaOH
KCl(aq) - no visible reaction
BaCl2(aq) - no visible reaction
NH4Cl(aq) - effervescence
MgCl2(aq) - white ppt
K2SO4(aq) = no visible rxn
KOH(aq) = no visible rxn
=> further test: add aq sulphuric acid
KCl(aq) - no visible reaction
BaCl2(aq) - white ppt
K2SO4(aq) = no visible rxn
KOH(aq) = no visible rxn
=> further test: add acidified aq BaCl2
KCl(aq) = no visible rxn
K2SO4(aq) = white ppt
KOH(aq) = no visible rxn
=> further test: add acidified AgNO3(aq)
KCl(aq) = white ppt
KOH(aq) = no visible rxn
4
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rico7102000
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#3
Report Thread starter 3 years ago
#3
(Original post by dip0)
add acidifed agno3(aq):
K2so4(aq) = no visible reaction
koh(aq) = no visible rxn
k2co3(aq) = effervescence
kbr(aq) = cream ppt
ki(aq) = yellow ppt
aqueous kcl, bacl2,nh4cl, mgcl2 will all give white ppt
=> further test:add warm aqueous naoh
kcl(aq) - no visible reaction
bacl2(aq) - no visible reaction
nh4cl(aq) - effervescence
mgcl2(aq) - white ppt
k2so4(aq) = no visible rxn
koh(aq) = no visible rxn
=> further test:add aq sulphuric acid
kcl(aq) - no visible reaction
bacl2(aq) - white ppt
k2so4(aq) = no visible rxn
koh(aq) = no visible rxn
=> further test:add acidified aq bacl2
kcl(aq) = no visible rxn
k2so4(aq) = white ppt
koh(aq) = no visible rxn
=> further test:add acidified agno3(aq)
kcl(aq) = white ppt
koh(aq) = no visible rxn
thank u so much!!!!!!!!!
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Ebby786
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#4
Report 9 months ago
#4
(Original post by username3718068)
add acidifed AgNO3(aq):
K2SO4(aq) = no visible reaction
KOH(aq) = no visible rxn
K2CO3(aq) = effervescence
KBr(aq) = cream ppt
KI(aq) = yellow ppt
aqueous KCl, BaCl2,NH4Cl, MgCl2 will all give white ppt
=> further test: add warm aqueous NaOH
KCl(aq) - no visible reaction
BaCl2(aq) - no visible reaction
NH4Cl(aq) - effervescence
MgCl2(aq) - white ppt
K2SO4(aq) = no visible rxn
KOH(aq) = no visible rxn
=> further test: add aq sulphuric acid
KCl(aq) - no visible reaction
BaCl2(aq) - white ppt
K2SO4(aq) = no visible rxn
KOH(aq) = no visible rxn
=> further test: add acidified aq BaCl2
KCl(aq) = no visible rxn
K2SO4(aq) = white ppt
KOH(aq) = no visible rxn
=> further test: add acidified AgNO3(aq)
KCl(aq) = white ppt
KOH(aq) = no visible rxn
How do you then identify which solutions are anions and which are cations?
Thank you
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Ebby786
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#5
Report 4 months ago
#5
How do you then identify which are anions and which are cations?Thank you
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scimus63
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#6
Report 4 months ago
#6
anions have a negative charge- usually non-metals
cations have a positive charge - usually metals
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