Question

A)Find the percent ionization of a 0.110 M solution of a weak monoprotic acid having Ka= 1.8×10−3.

B) Find the pH of a 0.110 M solution of a weak monoprotic acid having Ka= 0.18.

C)Find the percent ionization of a 0.110 M solution of a weak monoprotic acid having Ka= 0.18.

Answer 1

A) HA- ( H+) +(A-)

0.110M -0-0

-x+x+x

0.110x-x-x

(A-) =(H3O+) / (HA)

x2/0.110-x =1.8x10^-3

x2=0.000198-0.0018x

-x2+0.0018x-0.000198=0

Use the quadratic formula to find x

-b+sqrt(b2-4ac)/2a

= -0.0018+sqrt(0.0018)2-4x1x(-0.000198)/2x1

= 0.0018+sqrt (0.00000324+0.000792)/2

-0.0018+0.0282/2

=0.0132=1.32x10^-2

pH=-log (0.0132)

= 1.879

percent ionisation = (x/initial concentration)*100%

=0.0132/0.110 x 100

= 12 %

B)

HA- ( H+) +(A-)

0.110M -0-0

-x+x+x

0.110x-x-x

(A-) =(H3O+) / (HA)

x2/0.110-x =0.18

x2=0.0198-0.18x

-x2+0.18x-0.0198=0

Use the quadratic formula to find x

-b+sqrt(b2-4ac)/2a

= -0.18+sqrt(0.18)2-4x1x(-0.0198)/2x1

= 0.18+sqrt (0.0324+0.0792)/2

-0.18+0.33406/2

=0.154/2

=0.0770

pH=-log (0.0770)

= -(-1.1133)

pH = 1.113

C) percent ionisation = (x/initial concentration)*100%

=0.0770/0.110 x100

=70%