Question

An acid solution is 0.120 M in HCl and 0.210 M in H2SO4. What volume of a 0.140 M KOH solution would have to be added to 500.0 mL of the acidic solution to neutralize completely all of the acid?

Answer 1

0.120 M Hel lass HA Cars & Colags | Ý, CHệ] = (15) = 0.120 M be, concentration of HQ Porom Hell = 0.120 M He 804 (ag) -248 fag) + S022@fags So, (høj = 2% [ Hq S0] =2 x 0.910M = 0.420 M to, concentration of H@ from H, 604 = 0.420 M . So, total concentration of Hos = 0.120m+0.42011 0.540M KOH Nolume of acid solution, Nut I 500.0 mL liven, concentration of KOH Solution, Sw = 0140 M Holags * KOH lay - Kelas) + H₂O las So, of kott solution would have to be added, Yeon = West Sit 500. OmLX 0.54 04 00140 x = 1928-571429 ML KOH = 1.928571429 L

Answer 2