Answer :
Milliliters of 0.500 m NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.40 in a final volume of 500 ml is 31.8 mL.
The expression for the pH of buffer solution is given as :
pH = pKa + log ( base / acid)
moles of Tris HCl = mass / molar mass
= 10 / 157.59
= 0.0635 moles
pH = pKa + log ( base / acid)
7.40 = 8.072 + log ( [A⁻] / [HA] )
[A⁻] / [HA] = 0.335
but total moles of [A⁻] + [HA] = 0.0635 mol
therefore , 0.335 = y/ (0.0635 - y)
y = no. of moles of NaOH
mow, y = 0.015
volume of NaOH = moles / molarity
= 0.015 / 0.500
= 0.0318 L = 31.8 mL
Thus, Milliliters of 0.500 m NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.40 in a final volume of 500 ml is 31.8 mL.
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