Why does a #23*g# mass of calcium hydroxide dissolved in a #400*mL# volume of water, have #[HO^-]-=1.55*mol*L^-1#?

1 Answer
anor277
Oct 16, 2016

It seems that you have grasped the concept of stoichiometry:

#Ca(OH)_2(s) rarr Ca^(2+) + 2HO^-#

Explanation:

#[HO^-]# #=# #"Moles of hydroxide ion"/"Volume of solution"#

#"Moles of calcium hydroxide"=(23*g)/(74.1*g*mol^-1)# #=# #0.310*mol#. #"Moles of calcium"=0.155*mol#.

#[HO^-]# #=# #(2xx0.310*mol)/(0.400*L)# #=# #1.55*mol*L^-1#, because each formula unit of #Ca(OH)_2# delivers one #Ca^(2+)#, but #2xxHO^-#.

Each mole of calcium hydroxide clearly gives 2 moles of hydroxide anion upon dissolution, simply because of the formulation of calcium hydroxide. Is this clear?

To put it another way, the given solution is #1.55*mol*L^-1# with respect to #HO^-#, BUT half this concentration with respect to #Ca(OH)_2(aq)# and #Ca^(2+)#.

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