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3 July, 08:17

An initial 25.00 mL solution of nitric acid (HNO3) was diluted to 250.00 mL, and15.00 mL of this diluted solution was titrated with a 0.2244 M solution of sodium hydroxide. If 21.33 mL of the sodium hydroxide solution were required to neutralize the acid, what is the nitric acid concentration of the initial 25.00 mL solution?

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  1. 3 July, 08:31
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    HNO3 concentration in the initial 25.00 mL solution is 3.19 M

    Explanation:

    We are working backward:

    First we will determine the [HNO3] in 15 mL that reacted with Sodium hydroxide (NaOH).

    mole of nitric acid = mole of NaOH

    [HNO3] * 15 mL = 0.2244 M * 21.33 mL

    [HNO3] = 0.319 M

    The above is the [HNO3] in 250 mL solution that was transferred as 15 mL to react with NaOH.

    Second, we can determine the [HNO3] in 25 mL because the 250 mL solution was made from the 25 mL.

    [HNO3] * 25 mL = 0.319 M * 250 mL

    [HNO3] = 3.19 M which is the concentration in the initial 25 mL solution.
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