What mass of #"magnesium hydroxide"# is required to give a #1L# volume of #[Mg(OH)_2]# whose concentration is #0.01mol*L^-1# with respect to the hydroxide?

Question

2451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-12T04:03:38

For a #1*L# volume of #0.01*mol*L^-1# #Mg(OH)_2#, #0.28*g# salt are required.

#"Molarity"# #=# #"Moles"/"Volume"# #=# #(0.01*mol)/(1.0*L)#.

And thus we need to dissolve #0.01*molxx58.32*g*mol^-1xx1/2# #=# #0.28*g# in a #1*L# volume. Why did I include the #1/2#?

However, #K_"sp",Mg(OH)_2=5.61×10^(−12)# at #298K#, which gives a solubility of #6.4xx10^-3*g*L^-1#. The question was thus not well-proposed.

Magnesium hydroxide is thus too insoluble to provide such a concentration.

What mass of #"magnesium hydroxide"# is required to give a #1*L# volume of #[Mg(OH)_2]# whose concentration is #0.01*mol*L^-1# with respect to the hydroxide?