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Sergio039 [100]
3 years ago
10

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 fina

l volume of 250 mL? pk, for the tris = 8.072​
Chemistry
1 answer:
liubo4ka [24]3 years ago
3 0

Answer:

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]

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

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

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 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

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, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

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

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

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