Determine the molarity of the individual ions in the following: 2M Al(OH)3 4M Na2 SO4 1/2M Ca3 (PO4)2 0.5M Al2(SO4)3 3M K2 SO4 if you could explain how to work any one of these then I would appreciate it a lot thanks.
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I'm going to quote in Tank Girl now so she can move your thread to the right place if it's needed.
Determine the molarity of the individual ions in the following: 2M Al(OH)3 4M Na2 SO4 1/2M Ca3 (PO4)2 0.5M Al2(SO4)3 3M K2 SO4 if you could explain how to work any one of these then I would appreciate it a lot thanks.
Ionic compounds dissociate 100% in solution, so if there are two ions in the formula unit then the concentration of ions in solution will be double that of the compound itself.
eg:
CuCl2 -- (aq) --> Cu2+(aq) + 2Cl-(aq)
If the copper(II) chloride concentration is 1M then the chloride concentration is 2M ...
Ionic compounds dissociate 100% in solution, so if there are two ions in the formula unit then the concentration of ions in solution will be double that of the compound itself.
eg:
CuCl2 -- (aq) --> Cu2+(aq) + 2Cl-(aq)
If the copper(II) chloride concentration is 1M then the chloride concentration is 2M ...