Question

-/0.1 points 23. 0/4 Submissions Used Calculate the percent ionization of carbonic acid (H2CO3) in solutions of each of the following concentrations (Ka = 4.3e-07.) (a) 0.148 M % (b) 0.436 M % (c) 0.718 M %

Answer 1

h5୦୧୪- Given that, ka = 4:3x107 @ 0.148m Here, Ka=ca2 4.38107 = 0.148 d² x² = 403x107 0.148 x² = 2.90548106 x=82-9054X106 X = 170458103 2 %=17045 x163x600 a= 0.1704 %. 00.436M Ka=ca? 4.3x107 - 01436x² x2 403x10 7 01436 2 = J 908 623 X107 X= 9.90958104 2=9.93095X20-4x100 -0.09917./

0 0.718M Ka=d² 403x10 70. 1882 ? 4:3x167 0.718 d=f5.98888107 2 = 77387x15kx100 2 = 0.07734.7 co -