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28 December, 16:08

How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a final volume of 250 mL? pk, for the tris = 8.072

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  1. 28 December, 16:36
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    41.64mL of NaOH 0.500M must be added to obtain the desire pH

    Explanation:

    It is possible to find pH of a buffer by using H-H equation, thus:

    pH = pka + log [A⁻] / [HA]

    Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.

    Replacing in H-H equation:

    7.60 = 8.072 + log [A⁻] / [HA]

    0.3373 = [A⁻] / [HA] (1)

    10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

    10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

    [A⁻] + [HA] = 0.08255 (2)

    Replacing (1) in (2):

    0.3373 = 0.08255 - [HA] / [HA]

    0.3373[HA] = 0.08255 - [HA]

    1.3373[HA] = 0.08255

    [HA] = 0.06173 moles

    Thus:

    [A⁻] = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

    The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

    HA + NaOH → A⁻ + H₂O + K⁺

    Thus, you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. As NaOH solution is 0.500M:

    0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

    41.64mL of NaOH 0.500M must be added to obtain the desire pH
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