B2.(a)Aspirin,a common pain reliever,is a monoprotic acid with the molecular formula HCHO.K for this weak acid is 3.27x 104 Calculate the pH of a solution made by dissolving 0.325 g of Aspirin in 200.0 mL of water. 0.325g =0.00181mol 180gmol-1 0.00181mol =0.0090M 0.200L Initial 0.0090 0 0 Change +X Equilibrium 0.0090.X Thus at equilibrium, x(x) =3.27×10 0.0090-× Note that although the acid is weak.we can not assume that x<<00090.since there is little acid dissolved in this solution.Thus we must use the quadratic formula to solve this one x2=3.27x10(0.0090)-3.27x104x x2+3.27x104x-2.94x106=0 ×=-b±B2-490-3.27x10-±327×10y-40-2.94×10可 2a 2(① -0.000327±0.00344 =0.001560r-0.00188 thus,×=0.00156 Thus,[H3O'(]=0.00156 M pH=-log1o[H30a=-log10(0.00156)=2.81 B2. (a) Aspirin, a common pain reliever, is a monoprotic acid with the molecular formula HC9H7O4. Ka for this weak acid is 3.27 x 10-4. Calculate the pH of a solution made by dissolving 0.325 g of Aspirin in 200.0 mL of water. 1 0.325g 0.00181mol 180gmol 0.00181mol 0.0090M 0.200L − = = [HC9H7O4(aq) ] [H3O+ (aq) ] [C9H7O4¯(aq) ] Initial 0.0090 0 0 Change -x +x +x Equilibrium 0.0090 - x x x Thus at equilibrium, x(x) 4 3.27 10 0.0090 x − = × − Note that although the acid is weak, we can not assume that x<<0.0090, since there is little acid dissolved in this solution. Thus we must use the quadratic formula to solve this one. x 2 = 3.27 x 10-4(0.0090) – 3.27 x 10-4 x x 2 +3.27 x 10-4 x – 2.94 x 10-6 = 0 2 4 4 2 b b 4ac 3.27 10 (3.27 10 ) 4(1)( 2.94 10 ) x 2a 2(1) 0.000327 0.00344 2 0.00156or 0.00188 thus,x 0.00156 − − −± − −×± × −−× = = − ± = = − = −6 Thus, [H3O+ (aq)] = 0.00156 M pH = -log10[H3O+ (aq)] = -log10(0.00156) = 2.81 3