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Question

At $$25^0$$C, the solubility product of $$Mg(OH)_2$$ is 1.0 x $$10^{-11}$$. At which pH, will $$Mg^{2+}$$ ions start precipiating in the form of $$Mg(OH)_2$$ from a solution of 0.001 M $$Mg^{2+}$$ ions?


A
8
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B
9
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C
10
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D
11
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Solution

The correct option is C 10
$${ K }_{ sp }$$ of $${ Mg\left( OH \right)  }_{ 2 }=\left[ { Mg }^{ 2+ } \right] { \left[ { OH }^{ - } \right]  }^{ 2 }$$
$${ Mg\left( OH \right)  }_{ 2 }\rightleftharpoons { Mg }^{ 2+ }+2{ OH }^{ - }$$
$$\left[ { OH }^{ - } \right] =\sqrt { \dfrac { { K }_{ sp } }{ \left[ { Mg }^{ 2+ } \right]  }  } $$              given, $${ K }_{ sp }=1.0\times { 10 }^{ -4 }$$
             $$=\sqrt { \dfrac { { 10 }^{ -11 } }{ { 10 }^{ -3 } }  } $$                        $$\left[ { Mg }^{ 2+ } \right] ={ 10 }^{ -3 }$$
$$\left[ { OH }^{ - } \right] ={ 10 }^{ -4 }$$
$$pOH=4$$
$$pH=14-4=10$$
at $$pH=10$$, $${ Mg }^{ 2+ }$$ ions starts precipitating.

Chemistry

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