Explanation:
search-icon-header
Search for questions & chapters
search-icon-image
Class 11
>>Chemistry
>>Equilibrium
>>Solubility Equilibria
>>The solubility product of Ag2CrO4 is 1.9
Question
Bookmark
The solubility product of Ag
2
CrO
4
is 1.9X10
−12
.The volume of water in mL that can dissolve 4mg of Ag
2
CrO
4
is about:
Hard
Solution
verified
Verified by Toppr
Correct option is A)
Ag
2
CrO
4
⇌2Ag
+
+CrO
4
2−
Let S mol/L be the solubility of Ag
2
CrO
4
[Ag
2
CrO
4
]=[CrO
4
2−
]=S
[Ag
+
]=2S
K
sp
=[Ag
+
]
2
[CrO
4
2−
]
1.9×10
−12
=(2S)
2
×S=4S
3
S=7.8×10
−5
M
The molar mass of Ag
2
CrO
4
=331.7 g/mol
Convert solubility of Ag
2
CrO
4
from mol/L to g/L.
S=7.8×10
−5
mol/L×331.7 g/mol
S=0.0259 g/L
Convert solubility of Ag
2
CrO
4
from g/L to mg/mL.
S=0.0259 g/L×
1 g
1000 mg
×
1000 mL
1 L
S=0.0259 mg/mL
To dissolve 4 mg of Ag
2
CrO
4
, volume of water required will be 4 mg×
0.0259 mg
1 mL
=154 mL≃150 mL.
Author:
jaumeztam
Rate an answer:
5