Question

-/0.1 points 33 0/4 Submissions Used (a) Calculate the percent ionization of 0.00720 M carbonic acid (Ka = 4.3e-07). % ionization = % (b) Calculate the percent ionization of 0.00720 M carbonic acid in a solution containing 0.0470 M sodium hydrogen carbonate. % % ionization = Submit Answer + -/0.1 points 34. 0/4 Submissions Used A buffer solution contains 0.76 mol of propionic acid (HC3H502) and 0.30 mol of sodium propionate (NaC3H502) in 7.30 L The Ka of propionic acid (HC3H502) is Ka = 1.3e-05. (a) What is the pH of this buffer? pH= (b) What is the pH of the buffer after the addition of 0.12 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.05 mol of HI? (assume no volume change) pH =

Answer 1

33.@ Initial (M) H2CO3 Caa) HCaa CO ag) O change CM) Equilibrium (M) c(i-a) oc = Degree of ionisation of H2cos] Coc Co Cot Ca.Co Co 2 1-ae eci-2) LH2CD3] 0.0072 a2 Or 4.3x10 5 5.97 x 1D OY 4.3x10 O 0072 5 d5.97 x1d 5.97X1D = O 5 +5.97x24 X1R 5197x 597 XID 2.x A -3 - 7-697X1D Lonisaion ofCarbonic aeid NaHcos Ls a shong elechnlyte an umdergoes complet diso cation in aqueous son So, LHCO3 'Na tcOs (a)- Naca)+ Hco= for the dissociationf Natcos- 0.047 (M) H2COs Cas) Hcag) HCOš Cag) O.0072. M) 0.00722CM) Now Change Now total Hc©3J = +0.00 72(M 0.00722M O O47 :0072 O 00722 0.00 720 0072 1x) (o 0 47+0 0072x) Ка H2C

xC0.0470.00722)=4.310 2 17 OYP Or o0072u 0.0412-43x10= O 5 6.528 -5.972x ID = O x2 OY -6.528 6.528)+4x1x 5.972 x5 2 -6 = 9.15 x1D -6 100 = 9.15X10 ionisation 9.15XID O 0009!5/