Answer to Question #94417 in General Chemistry for Miracle

Question

What weight of H3PO4 is needed to make 600 ml of a 5 N solution?

Answer 1

From the definition of normality,

$N(H_3PO_4)=\frac{n_{g.eq.}(H_3PO_4)}{V(H_3PO_4)}$

where n_g.eq.(H₃PO₄) is the number of gram equivalents of solute, and V(H₃PO₄) is the volume of solution.

$n_{g.eq.}(H_3PO_4)=\frac{m(H_3PO_4)}{M_{eq}(H_3PO_4)}$

where m(H₃PO₄) - is the mass of phosphoric acid, and M_eq(H₃PO₄) is equivalent mass of acid.

$M_{eq}(H_3PO_4)=\frac{M(H_3PO_4)}{n(H^+)}$

where M(H₃PO₄) is molar mass, and n(H⁺) is the number of replaceable H⁺ in H₃PO₄.

Thus, combining these equations we can find the mass of H₃PO₄:

$m(H_3PO_4)=n_{g.eq.}(H_3PO_4) \cdot M_{eq}(H_3PO_4) = N(H_3PO_4) \cdot V(H_3PO_4) \cdot \frac{M(H_3PO_4)}{n(H^+)}=5\frac{eq}{L} \cdot 0.6 L \cdot \frac{98\frac{g}{mol}}{3}=98 g$