5 February, 13:17
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
5 February, 13:28
V KOH = 41 mL
(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
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