The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?
Answers
Explanation:
When ionic product is greater than solubility product, the precipitation will occur.
Total volume after mixing is 10+5=15 mL.
[S 2−]= 15
1.0×10 −19 ×10=6.67×10 −20M[M 2+ ]= 155 ×0.04=0.0133M
Ionic product will be [M 2+][S 2− ]=0.0133×6.67×10−20=8.87×10 −22
In case of ZnS and CdS, the ionic product is greater than the solubility product.
Hence, ZnCl 2 and CdCl 2
solution will precipitate.
Let s M be the solubility of silver benzoate in water.
[Ag
+
]=[PhCOO
−
]=sM
K
sp
=[PhCOO
−
][Ag
+
]=s×s=s
2
=2.5×10
−13
s=5.0×10
−7
M. This is solubility of silver benzoate in pure water.
pH=−log[H
+
]=3.19
[H
+
]=6.456×10
−4
Benzoate ions combine with protons to form benzoic acid but hydrogen ion concentration is constant due to buffer solution.
PhCOOH⇌PhCOO
−
+H
+
K
a
=
[PhCOOH]
[PhCOO
−
][H
+
]
[PhCOO
−
]
[PhCOOH]
=
K
a
[H
+
]
=
6.46×10
−5
6.456×10
−4
=10
Let y M be the soubiity in the buffer solution.
y=[Ag
+
]=[PhCOO
−
]+[PhCOOH]=[PhCOO
−
]+10[PhCOO
−
]=11[PhCOO
−
]
[PhCOO
−
]=
11
y
K
sp
=[PhCOO
−
][Ag
+
]
2.5×10
−13
=
11
y
×y
y=1.66×10
−6
s
y
=
5.0×10
−7
1.66×10
−6
=3.32
Thus, silver benzoate is 3.32 times more soluble in buffer.
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