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

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

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

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

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i think so it’s will be b yes i believe its

it is the most commonly accepted one