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

Question

Andreas Pennington

Chemistry

29 November, 22:57

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

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Answers (1)

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Fletcher Ruiz · Answer 1

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.