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13 May, 15:58

What volume of 6.00 m naoh would be required to increase the ph to 4.93?

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  1. 13 May, 17:57
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    According to this formula:

    PH = Pka + ㏒[A^-]/[HA]

    when we have the value of PH=4.93 & Pka (missing in your question) = 4.76 &

    no. of mole of acetate = 40 mmol & no. of mole of acetic acid = 60

    so by substitution:

    4.93 = 4.76 + ㏒ ((40+X) / (60-X))

    0.17 = ㏒ ((40+X) / (60-X))

    ∴X = 19.66 mmol = 0.019 mol

    finally we can get the volume from this formula of the molarity:

    molarity = number of moles / volume

    6 = 0.019 / V

    ∴V = 0.0032 mL = 3.2 L
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