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Solubility and moles check

Okay, i don't know the solubility formula to start off with, so if anyone just provides this in the method i'd really appreciate it :smile:

This Q has 3 parts, 2 of the parts i did so if i ahve any mistakes plz show me.

Q: In an experiment to measure the solubility of barium hydroxide, 10.0cm3 of a saturated solution of Ba(OH)2 were exactly neutralised by 24.0cm3 of a 0.2M HCL acid solution.

i) How many moles of Ba(OH)2 were neutralized in this reaction?

Ba(OH)2 + 2HCL -------> BaCl2 + 2H2O

Ba(OH)2 : HCL is 1:2,

24/1000 x 0.2 = 0.0048mol of HCL

Moles of Ba(OH)2 = 0.0048/2
= 0.0024 mol

ii) Conc of Ba(OH)2 solution?

Moles/Volume

Conc of Ba(OH)2 = 0.0024/(10/1000)
= 0.24M

ii) What is the solubility of Ba(OH)2 in g.dm^-3

I cannot do this as i dont knw the equation and stuff, pls help me out on this one :smile: +rep for sure :biggrin:
The first lot looks good to me.

Because it's 0.24mol/dm^3, we can simply work out the mass per dm using the molar value.

mass = number of moles x Ar
=0.24 x (137.3 + 2(16 + 1))
=0.24 X 171.3
=41.1g/dm^3

I think thats right, makes sense to me.

Hope that helps :smile:
Reply 2
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winnie the poo
The first lot looks good to me.

Because it's 0.24mol/dm^3, we can simply work out the mass per dm using the molar value.

mass = number of moles x Ar
=0.24 x (137.3 + 2(16 + 1))
=0.24 X 171.3
=41.1g/dm^3

I think thats right, makes sense to me.

Hope that helps :smile:


but the moles of barium hydroxide is 0.0024mol. 0.24M is the concentration right?
It's 0.0024mol in 10.0cm3, so in 1 dm there are 0.24mol.
Reply 4
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ohhhhh get it now, so there is nos et equation for solubility, wld rep you if it wld let me lol :biggrin:
Yeah you can only rep once a day :smile:

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