The pH of a 0.05 M solution of acetic acid is 3. What is the value of the equilibrium constant for the dissociation of acetic acid?

k = 2,04x10⁻⁵

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺.

And the equilibrium constant is defined as:

k = [CH₃COO⁻] [H⁺] / [CH₃COOH] (1)

The equiibrium concentration of each specie if the solution of acetic acid is 0,05M is:

[CH₃COOH] = 0,05M - x

[CH₃COO⁻] = x

[H⁺] = x

-Where x is the degree of reaction progress-

As the pH is 3, [H⁺] = 1x10⁻³M. That means x =  1x10⁻³M

Replacing in (1):

k = (1x10⁻³)² / 0,05 - 1x10⁻³

k = 1x10⁻⁶ / 0,049

k = 2,04x10⁻⁵

I hope it helps!

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml