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Question

The solubility product constant Ksp of $$Mg{ (OH) }_{ 2 }$$ is $$9.0\times { 10 }^{ -12 }$$. If a solution is 0.010 M with respect to $${ Mg }^{ 2+ }$$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of  $$Mg{ (OH) }_{ 2 }$$? 


A
1.5×107M
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B
3.0×107M
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C
1.5×105M
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D
3.0×105M
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Solution

The correct option is B $$3.0\times { 10 }^{ -5 }M$$
$${ K }_{ SP }=\left[ { Mg }^{ 2+ } \right] { \left[ { OH }^{ - } \right]  }^{ 2 }\\ 9\times { 10 }^{ -12 }=0.01\times { \left[ { OH }^{ - } \right]  }^{ 2 }\\ \cfrac { { \left[ { OH }^{ - } \right]  }^{ 2 } }{ { 10 }^{ -2 } } ={ \left[ { OH }^{ - } \right]  }^{ 2 }\\ 9\times { 10 }^{ -18 }={ \left[ { OH }^{ - } \right]  }^{ 2 }\\ { \left[ { OH }^{ - } \right]  }=\sqrt { 9\times { 10 }^{ -10 } } =3\times { 10 }^{ -5 }$$

Chemistry

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