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.
Please reply me as soon as possible sir
Thank you
George.
Answers
DrBob222
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.
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.