Question

How many milliliters of 10.0 M HNO3 solution must be added to a 1.0 L buffer that is .01 M acetic acid ( HC2H3O2) and .2M sodium acetate (NaC2H3O2) to bring the pH to 5.6? the pKa of acetic acid is 4.75

Answer 1

pH of the solution after addition of HNO3 = 5.6 Let x be the number moles of HNO3 to be added to the buffer to change the pH.

Number of moles of CH3COOH = M*V = 0.01 M * 1.0 L = 0.01 mol Number of moles of CH3COO-= M*V = 0.2 M * 1.0 L = 0.2 mol CH3COO + H = CH3COOH Initial 0.2 0 0.01 Add Change -x -x + x Equilibrium 0.2-X 0.01 + x pH = pKa+log [CH,COOH] 55.6= 4.75+log 0.2-X 0.01+x log 0.2-X – 0.85 108 0.01+x 0.2 – X = 100.85 = 7.079458 0.01+x 0.2-x=7.079458x(17.55 + x) 0.2- x=0.070795 +7.079458x 0.2 -0.070795 = 7.079458x+x 8.079458x = 0.129205 x=0.015992 mol number of moles of HNO, = 0.015992 mol Volume of HNO, = moles _ 0.015992 mola -= 0.001599 L = 1.60 mL Molarity 10 M Volume of HNO, to be added = 1.60 mL