Question

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

Answer 1

Answer:

$3.37~g~Li_3PO_4$

Explanation:

The first step is to find the moles of  $Li^+$, using the equation:

$M=\frac{mol}{L}$

Where, $500 mL=0.5 L$

Plug in the values into the equation:

$0.175~M=\frac{mol}{0.5~L}$

Solving for mole:

$mol=0.175~M*0.5~L$

$mol=0.0875~mol~Li^+$

Now, the have to write the chemical equation to find the molar ratio between $Li_3PO_4$ and $Li^+$, so:

$Li_3PO_4~->~3Li^+~+~(PO_4)^-3$

The molar ratio then is 1:3, using this molar ratio we can convert from moles of $Li^+$ to moles of $Li_3PO_4$, so:

$mol=0.0875~mol~Li^+~\frac{1~mol~Li_3PO_4}{3~mol~Li^+} =0.0291~mol~Li_3PO_4$

Finally, we have to conver from moles of $Li_3PO_4$ to grams of  $Li_3PO_4$ using the molar mass of $Li_3PO_4$ (115.79 g/mol), so:

$0.0291~mol~Li_3PO_4\frac{115.79~gLi_3PO_4}{1~mol~Li_3PO_4}=3.37~g~Li_3PO_4$

Answer 2

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)