What mass of li3po4 is needed to prepare 500 ml of a solution having a lithium ion concentration of 0.175 m?

M (Li₃PO₄) = 115.8 g/mol

c (Li⁺) = 0.175 mol/L

v=500 mL = 0/5 L

n (Li⁺) = 3n (Li₃PO₄) = 3m (Li₃PO₄) / M (Li₃PO₄) = c (Li⁺) v

m (Li₃PO₄) = c (Li⁺) vM (Li₃PO₄) / 3

m (Li₃PO₄) = 0.175*0.5*115.8/3=3.3775 * g

*Solubility lithium phosphate in water about 0,34 g/L. Litium phosphate can be dissolved in solution of a phosphoric acid. For example:

2Li₃PO₄ (s) + H₃PO₄ (aq) = 3Li₂HPO₄ (aq)