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What volume in milliliters of 0.100 M HCIO3 is required to neutralize
40.0 mL of 0.165 M KOH?

Respuesta :

Answer:

66.0 mL HClO3

Explanation:

M1V1 = M2V2

M1 = 0.100 M HClO3

V1 = ?

M2 = 0.165 M KOH

V2 = 40.0 mL KOH

Solve for V1 --> V1 = M2V2/M1

V1 = (0.165 M)(40.0 mL) / (0.100 M) = 66.0 mL HClO3

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