Question
Which of the following will have lowest value of $${K_{sp}}$$ at room temperature?
A.
$$Be{\left( {OH} \right)_2}$$
B.
$$Mg{\left( {OH} \right)_2}$$
C.
$$Ca{\left( {OH} \right)_2}$$
D.
$$Ba{\left( {OH} \right)_2}$$
Answer :
$$Be{\left( {OH} \right)_2}$$
Solution :
$$Be{\left( {OH} \right)_2}$$ is least soluble in water hence, it will have lowest value of $${K_{sp}}.$$
$$Be{\left( {OH} \right)_2} \rightleftharpoons B{e^{2 + }} + 2O{H^ - };$$ $${K_{sp}} = \left[ {B{e^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2}$$