Question

Calculate the pH of a solution that is prepared by dissolving 0.500 mol of hydrocyanic acid (HCN, KA = 6.17x10-19) and 0.187 mol of acetylsalicylic acid (HC,H,04, K4 = 3.40x104) in water and diluting to 3.40 L. Also, calculate the equilibrium concentrations of HCN, CN, HC,H,O4, and C,H,O4. Do not make an approximation unless the initial acid concentration is greater than 1000 × K. (Hint: The pH will be determined by the stronger acid of this pair.) pH = [HCN]= м [CN] = м [HC,H,O4]= [C,H,04]=

Answer 1

@: (ka) = 6.17% 10-10 HON acetyl saliglic acid = 3.40 010-4 . (kal Acetyl salicylic acid > (ka) Hon HCN Stronger acid - pH will be decided by acetyl salicylic acid [acetyl Salicylic acid] = Moles Volume o.187 3.40 mol L = 0.055 M : Heg H704 ? Cg Hq Q4 + HO 0.055 +y + + y -Y 0.055 - y Ka 3.40 X 10 = [ Calzon] H+] = 1 [H Cg H₂o4] 3.40X10-4 = CY) (Y) (0.055 -Y)

- ypp = (3.40x1024) (0.0 55 - ) yo = 0.187 X10 -4- 3•40 x10 mly 3.40 x10 -4 y - 1.81x10 -5 =0 yP+ folve the periodic earing online. Solverr: Y = 0.0041577, Fo.0045] Not possible. -Acetyl T Jalicylic acid. deq = 0.055-Y = 0.055 -0.0041577. : = [000 5084 M) THCg Hq o y Ae Salicylate eg - meq = Y = 0.0041571 = [ cg +704-3ey. [H+] = y = 0.0041577 M PH = – 4 to •e 4/5 ++) pH = 9.38 HCN . * H ÆÐ + CNO Ka= 6.17x10-10 0.0041577 6.500 mol 3.4L = 0.147 -Y ty + Y 0.147-Y DIDO 41577 + Y Y

Ka is very very small for HEN lue of x will be small and hall and Can be neglected can be neal with diespect to o.0041577 and o.lLt. aleated -) :) Hệ -y o | | 666 415++ ty = b. O 04/17-• : Ra = DCN-1 CH 10 CHENI =) 6.17x10 -1° = (Y) (0.0041577) 0.147 y = 8. 182x10-IM Cvery Small) [on-leq = 0.182x10 -8 M [HON] = 6:147 M - 2:182x10 - = 0.147 MI + M