In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid into enough

Question

In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid into enough distilled water to make 500.0 mL of solution. 30.0 mL of the base was measured, and in the titration, 16.74 mL of the acid titrant was needed to neutralize. Determine the concentration of the Sodium hydroxide.

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Thành Công 6 months 2021-07-17T08:25:02+00:00 1 Answers 18 views 0

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  1. thienthanh
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    2021-07-17T08:26:08+00:00
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    Answer:

    1.395M NaOH

    Explanation:

    Sodium hydroxide, NaOH, reacts with nitric acid, HNO3, as follows:

    NaOH + HNO3 → NaNO3 + H2O

    Where 1mol of NaOH reacts with 1mol of HNO3

    To solve this question we must find the concentration of the titrant. With the concentration and the needed acid we can find the moles of HNO3 added = moles NaOH in the solution. With the moles of NaOH and its volume we can find its concentration as follows:

    HNO3 concentration:

    10.00mol/L HNO3 * (125.0mL/500.0mL) = 2.500M HNO3

    Moles HNO3 = Moles NaOH:

    16.74mL = 0.01674L*(2.500mol/L) = 0.04185 moles HNO3 = Moles NaOH

    Concentration NaOH:

    0.04185 moles / 0.0300L =

    1.395M NaOH

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