...........HA ==> H^+ + A^-
initial..0.05M......0....0
change....-x........x....x
equil....0.05-x.....x....x
Ka = (H^+)(A^-)/(HA)
Start with pH and convert that to H^+ (which is x in the above), subsubstitute into the Ka expression and solve for Ka.
.050 mole of weak acid (HA) is dissolved in enough water to make 1.0 L of solution. The pH of this solution is 3.50. What is the value of Ka for this weak acid?
1 answer
1 answer