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5 February, 13:17

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

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  1. 5 February, 13:28
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    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
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