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15 October, 00:26

If 22.4 mL of 0.0100 M HCI are required to completely react with 15.0 mL NaOH, what was the initial molarity of the NaOH solution?

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  1. 15 October, 00:28
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    NaOH → 0.015 M

    Explanation:

    We propose the neutralization formula to solve the problem:

    N acid. Volume acid = N base. Volume base

    N means normality. A sort of concentration that is defined as:

    M / val where the val means the number of H⁺ and OH⁻ from the acid or the base.

    In this case, we have NaOH and HCl, so the M = N

    Let's replace dа ta:

    0.01 M. 22.4 mL = 15 mL. M NaOH

    0.01 M. 22.4 mL / 15 mL = M NaOH → 0.015 M
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