Question

consider a 1.00 M solution of a weak acid, HA. The pH of the solution is found to be 3.85. A) calculate the [H3O+] in the solution. This would be the equilibrium concentration of H3O+ in the solution. B) write out an ICE table as before. Here, we don’t know the numerical value of Ka but we know the [H3O+] at equilibrium which you should see from your ICE table easily relates to the value of “x” in your table and knowing “x” from [H3O+], one can easily determine the other equilibrium concentrations of all species and solve for value of Ka.

C) [H3O+] = __________________

Answer 1

A).

Given pH = 3.85

............[ H⁺ ] = antilog of ( -3.85 )

again , H₃ O⁺ is nothing but solvated hydrogen ion

[H₃O⁺ ] = antilog of (- 3.85 )

................ = 1.4125 x 10^-4 M

B).

The weak acid HA ionizes to yield H₃ O⁺  & A^- .

The equilibrium reaction for the ionization of acid is written as,

....................HA + H₂ O <=========> H₃ O ⁺   + A^-

.I..................1.0......................................................0 ................ 0...........

C............. (1.0 - x ).................................................x...................x..........

E........[ ( 1.0 - x) / ( 1 + x ) ].........................x / (1.0 + x).........[ x /( 1.0 + x ) ]

&

K_a   = { ( x / 1.0 + x ) ( x / 1.0 + x ) } / { ( 1.0 - x ) / ( 1.0 + x ) }

.......= x² / (1.0 - x² )  

Considering the acid as a weak acid the value of x = <<<<< 1.0 , hence it

can be ignored from the denominator.

K_a   = x²

substtuting the value of x = [H₃ O⁺ ] as 1.4125 x 10^-4 we get,

K_a   = 1.9951 x 10^-8

C.)

[ H₃ O⁺ ] = 1.4125 x 10^-4

=========================================================