What volume of 0.265 M KOH will be neutralized completely by 34.56 mL of 0.183 M HNO3?

A. 2.39 x 10 to the negative second power
B. 1.40 L
C. 713 x 10 to the negative 1 power
D. 8.55 x 10 to the negative 2nd power

Answer:
Moles HNO3 = 34.56 x 0.183 /1000 =0.00532 = moles KOH
M = moles / V
V = moles / M = 0.00532 / 0.265 = 0.0239 L = 2.39 x 10^-2
The correct answer is A
M1 x V1 = M2 x V2

volume of KOH = (34.56 x 0.183)/ 0.265
= 23.9 mL
= 2.39 . 10^-1

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