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6 November, 18:30

A 0.388 g sample of a monoprotic acid is dissolved in water and titrated with 0.280 M NaOH. What is the molar mass of the acid if 10.5 mL of the NaOH solution is required to neutralize the sample? A flask with a solution sits on the base of a ring stand. A buret filled with liquid is suspended above the flask by the ring stand.

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  1. 6 November, 18:42
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    The molar mass of the monoprotic acid is 131.97 g/mol

    Explanation:

    Step 1: Data given

    Mass of the sample monoprotic acid = 0.388 grams

    Molarity of NaOH = 0.280 M

    Volume of NaOH = 10.5 mL = 0.0105 L

    Step 2: The balanced equation

    HX + NaOH → NaX + H2O

    Step 3: Calculate moles NaOH

    Moles NaOH = molarity NaOH * volume NaOH

    Moles NaOH = 0.280 M * 0.0105 L

    Moles NaOH = 0.00294 moles

    Step 3: Calculate moles HX

    For 1 mol NaOH we need 1 mol HX to react

    For 0.00294 moles NaOH we need 0.00294 moles HX to react

    Step 4: Calculate molar mass of the acid

    Molar mass = mass / moles

    Molar mass = 0.388 grams / 0.00294 moles

    Molar mass = 131.97 g/mol

    The molar mass of the monoprotic acid is 131.97 g/mol
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