Question

3. What is the ka for a solution of choloracetic acid (C2H5ClO2) which has a pH of 1.52 in a 0.112M solution?

Please reply me as soon as possible sir

Thank you
George.

Answer 1

For simplicity, let's call cloroacetic acid HA.

Then HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
Convert pH to (H^+) from pH = -log(H^+). (A^-) is the same. After equilibrium is attained, (HA) - 0.112-(H^+).
Solve for Ka.