How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?

V KOH = 41 mL

Explanation:

for neutralization:

(V*C) acid = (V*C) base

∴ C H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ C KOH = 0.0050 N = 0.0050 eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ C KOH = (0.0050 eq-g/L) * (mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = (V*C) acid / C KOH

⇒ V KOH = (0.041 L) (0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L