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]
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 =
