Answer to Question #94417 in General Chemistry for Miracle

Question #94417
What weight of H3PO4 is needed to make 600 ml of a 5 N solution?
1
Expert's answer
2019-09-13T05:11:59-0400

From the definition of normality,

N(H3PO4)=ng.eq.(H3PO4)V(H3PO4)N(H_3PO_4)=\frac{n_{g.eq.}(H_3PO_4)}{V(H_3PO_4)}

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

ng.eq.(H3PO4)=m(H3PO4)Meq(H3PO4)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 Meq(H3PO4) is equivalent mass of acid.

Meq(H3PO4)=M(H3PO4)n(H+)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(H3PO4)=ng.eq.(H3PO4)Meq(H3PO4)=N(H3PO4)V(H3PO4)M(H3PO4)n(H+)=5eqL0.6L98gmol3=98gm(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


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