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26 August, 23:23

How many mL of 0.150M NaOH (base) solution are required to neutralize 35.0mL of 0.220M HCl (acid) solution?

Answers (1)
  1. 27 August, 00:38
    The neutralization reaction between an acid and a base is given through the equation,

    M₁V₁ = M₂V₂

    Substituting the known values from the given,

    (0.150 M) (V₁) = (35 mL) (0.220M)

    The value of V₁ from the equation above is 51.33 mL.

    Thus, 51.33 mL of NaOH is needed for the process.
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