It took 14.50 mL of 0.455M NaOH to fully neutralize 12.0mL of HCl. What is the concentration of the HCl? HCl + NaOH \rightarrow→ NaCl + H2O

The appropriate answer is "0.549 M".

Explanation:

The given values are:

N₁ = 14.50 mL

V₁ = 0.455 M

N₂ = 12 mL

Let

V₂ = C = ?

As we know,

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On substituting the values, we get

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