25.0 mL of 0.100 M H2SO4 requires what volume of 0.200 M NaOH for complete neutralization?

25.0 mL (n{H^+} = n{OH^-}, with H2SO4 providing 2 H^+) Moles H^+ = 2(0.100)(0.0250) = 0.00500. Moles NaOH needed = 0.00500. V = 0.00500/0.200 = 0.0250 L = 25.0 mL.

25.0 mL (n{H^+} = n{OH^-}, with H2SO4 providing 2 H^+)

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Acemy

Introduction to Titration

AP Chemistry· difficulty 4/5

_2

A
1. mL
B
$50.0 mL$
C
$12.5 mL$
D
$n_{H^+} = n_{OH^-}$
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Explanation

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