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5 May, 19:30

A chemist must prepare of potassium hydroxide solution with a pH of at. He will do this in three steps: Fill a volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to significant digits

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  1. 5 May, 19:54
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    The question is incomplete; the complete question is:

    A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

    Answer:

    4.5g (to 2 significant digits)

    Explanation:

    Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

    To find the pH we can use the equation pH + pOH = 14.

    Firstly to find the pOH:

    13.00 + pOH = 14

    pOH = 1.00

    To find the [OH-]

    Since

    pOH = - log[OH^-]

    [OH^-] = antilog (-pOH)

    [OH^-] = antilog (-1)

    [OH^-] = 0.1 molL-1

    Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

    Also, we know that concentration = number of moles/volume

    we have the concentration and the volume now so we can calculate the number of number of moles as follows:

    The 800mL volume is the same as 0.8L

    0.1 molL-1 = number of moles/0.8L

    0.08 moles = number of moles

    now we can calculate the amount of solid KOH required

    the molar mass of KOH = 39 + 16 + 1 = 56 gmol-1

    56 x 0.08 moles = 4.48g

    So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.
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