Question
A 0.100 mol quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA?
Answers
DrBob222
HA ==>H^+ + A^-
Ka = (H^+)(A^-)/((HA)
You find (H^+) from the pH. The (A^-) = (H^+) and (HA) = 0.1-(H^+)
Solve for Ka.
Ka = (H^+)(A^-)/((HA)
You find (H^+) from the pH. The (A^-) = (H^+) and (HA) = 0.1-(H^+)
Solve for Ka.