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Question
At 25oC, will a 1×10−4M solution of Mg(NO3)2 form a precipitate of Mg(OH)2 if pH of the solution is adjusted to 9.0.
Ksp(Mg(OH))2=8.9×10−12M3. At which minimum pH will the precipitation start?
√8.9=2.98
log 2.98=0.474
At 25oC, will a 1×10−4M solution of Mg(NO3)2 form a precipitate of Mg(OH)2 if pH of the solution is adjusted to 9.0.
Ksp(Mg(OH))2=8.9×10−12M3. At which minimum pH will the precipitation start?
√8.9=2.98
log 2.98=0.474
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Solution
The correct option is C 10.46
For, pH=9.0,[H+]=10−9 M, and
[OH−]=Kw[H+]=10−1410−9=10−5 M
The ionic product of Mg(OH)2 in the solution will be
[Mg2+][OH−]2=(10−4)(10−5)2=10−14 M3
Since, the value of ionic product is smaller than Ksp(8.9×10−12), no precipitate of Mg(OH)2 will be formed.
The minimum concentration of OH− needed to precipitate Mg2+from the solution is
[OH−]=√Ksp[Mg2+]=√8.9×10−1210−4[OH−]=√8.9×10−8[OH−]=2.98×10−4M
pOH=−log 2.98×10−4
pOH=4−log 2.98
∴ Corresponding pOH=3.54
and minimum pH=14−3.54=10.46
For, pH=9.0,[H+]=10−9 M, and
[OH−]=Kw[H+]=10−1410−9=10−5 M
The ionic product of Mg(OH)2 in the solution will be
[Mg2+][OH−]2=(10−4)(10−5)2=10−14 M3
Since, the value of ionic product is smaller than Ksp(8.9×10−12), no precipitate of Mg(OH)2 will be formed.
The minimum concentration of OH− needed to precipitate Mg2+from the solution is
[OH−]=√Ksp[Mg2+]=√8.9×10−1210−4[OH−]=√8.9×10−8[OH−]=2.98×10−4M
pOH=−log 2.98×10−4
pOH=4−log 2.98
∴ Corresponding pOH=3.54
and minimum pH=14−3.54=10.46
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