Determine the ph of a 0.461 m c6h5co2h m solution if the ka of c6h5co2h is 6.5 x 10-5.

The pH of 0.461 M C6H5CO2H is 2.27

pH and KapH of a solution is the negative logarithm to base 10 of hydrogen ion concentration of a solution Ka or acid dissociation constant is a constant used to express the ratio of products to reactant of a given weak acid.pH from dissociation of C6H5CO2H

Equation of dissociation of C6H5CO2 is given below:

C6H5CO2 > C6H5CO2- + H+

Using an ICE table

C6H5CO2 > C6H5CO2- + H+

Initial concentration of C6H5CO2H = 0.461

Initial concentration of C6H5CO2- = 0

Initial concentration of H+ = 0

Change in concentration of C6H5CO2H = -X

Change in concentration of C6H5CO2- = +X

Change in concentration of H+ = +X

Equilibrium concentration of C6H5CO2H = 0.461 - X

Equilibrium concentration of C6H5CO2- = X

Equilibrium concentration of H+ = X

Ka of C6H5CO2H = 6.5 x 10-5

Ka = [H+][C6H5CO2-]/[C6H5CO2H]

6.5 x 10-5 = X × X / 0.461 - X

X^2 = 6.5 × 10-5 (0.461 - X)

Assuming x is very small, 0.461 - X = 0.461

Solving for X:

X = 0.0054

Therefore [H+] = 0.0054

pH = -log[H+]

pH = - log 0.0054

pH = 2.27

Therefore, the pH of 0.461 M C6H5CO2H is 2.27

Learn more about pH and Ka at: link

By using ICE table:
        
               C6H5CO2H ↔ C6H5CO2-  +  H + 

intial         0.461                       0                    0 

change    -X                             +X              +X

Equ        (0.461 - X)                   X                  X

when Ka  = [H+] [C6H5CO2-] / [C6H5CO2H]
 
 6.5 x 10^-5 = X * X / (0.461-X)   by solving for X

∴ X = 0.0054 

∴[H+] = 0.0054 

∴ PH = -㏒[H+]

         = -㏒0.0054

         = 2.27

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