Question

Part C

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.7×10⁻³.

Express your answer to two decimal places.

Part D

Find the percent dissociation of this solution.

Express your answer using two significant figures

Part E

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 0.15.

Express your answer to two decimal places.

Part F

Find the percent dissociation of this solution.

Express your answer using two significant figures.

Answer 1

Part C monoprotic acid is represented by HA. Its dissociation can be represented as

HA----> H+ + A-

Ka= [H+] [A-]/[HA]

1.7*10-3 = [H+] [A-]/[HA]

x= concentration of H+ at equilibrium

initial         HA                 H+             A-

equilibium 0.120-x          x               x

1.7*10-3 =x²/(0.12-x)

assuming x to be smaler than 0.12

x²= 0.12*1.7*10^-3, x=0.0143 which is less than 0.12 ( the assumption is justified)

H+= 0.0143, pH= -log(0.0143)= 1.85

at equilibrium [HA] = 0.12-0.0143=0.1057

part D :degree of dissocation = 100*(1-(0.1057/0.12)= 11.91%

part E : proceeding as mentioned above x²/(0.12-x)=0.15

using solver x is obtained   by assuming some value of x and matching LHS with 0.15

x= 0.07871

pH=-log (0.07871)=1.10397  

concentraions at equilibrium HA= 0.15-0.07871=0.07129   [H+] =0.07871 [A-] =0.07871

F: percent dissociation =100*{1-(0.07129/0.12)}=94.06%