Question

How many mL of 0.500 M HCI would have to be added to the buffer prepared by dissolving 12.0 g of tris (FM 121.14) plus 7.0 g of tris hydrochloride (FM 157.60) in 250 ml water to adjust the pH of the solution to 8.21? Enter your answer in mL NH3 NH HOCH2 + H+ CH OH HOCH,C HOCH, BH+ PK, = 8.07 HOCH2CH OH HOCH, B = "tris" Tris(hydroxymethyl)aminomethane B+H +BH

Answer 1

pH of the solution after addition of HCI = 9.00 Let x be the number moles of HCl to be added to the buffer to change the pH. Number of moles of Tris = mass/mol.wt. = 12.0 g/(121.14 g/mol) = 0.09906 mol Number of moles of TrisH+ = mass/mol.wt. = 7.0 g/(157.60 g/mol) = 0.04442 mol mol Tris + H = Trisht I Initial 0.09906 0 0.04442 Add Change -x + x Equilibrium 0.09906 -x O 0.04442 + x log [Tris] pH = pKa+log [TrisH+1 0.09906 - x >>8.21=8.07 +log - 0.04442 + x 0.09906- X = 0.14 0.04442+x 0.09906-X 120.14 21 280384 0.04442+ x 0.09906- x =1.380384x(0.04442+x) 0.09906- x=0.061311+1.380384x 0.09906 -0.061311=1.380384x+x 2.380384x=0.03775 x=0.015858 mol number of moles of HC1 = 0.015858 mol Molarity of HCl = 0.500 M moles 0.015858 mol Volume of HCl required=- -= 0.031715 L=31.72 mL molarity 0.500M Volume of HCl to be added = 31.72 mL