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What volume of 0.3 M HI is needed to exacly neutralize 50.0 ml of 0.3 M KOH?
Wiki User
∙ 13y ago
V1.c1 = V2.c2 = number of equivalent moles (analyte = titrant)
in which:
subscript 1 = titrant
subscript 2 = analyte
V is volume in mL or L (use same units on both sides)
c is concentration of reactant in equiv.mole/L, corrected for stoiechiometric factor (from balanced equation)
50.0 * 0.3 [H+] = V2 * 0.3 [OH-]
V2 = 50.0 mL
Wiki User
∙ 13y ago
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