Question

A 25 mL sample of H3PO4 requires 31.15 mL of 0.242 N KOH for its titration. What is the normality of this H3PO4 (aq) if it is always to be used in reaction?

Answer 1

$C_n(H_3PO_4) *V(H_3PO_4) =$

$= C_n(KOH)*V(KOH)$

$C_n(H_3PO_4) = C_n(KOH)*\frac{V(KOH)}{V(H_3PO_4)} =$

$= 0.242\ N * \frac{31.15\ ml}{25\ ml} =$

$= 0.301532\ N$

Answer: $C_n(H_3PO_4) = 0.301532\ N$