My calculator is down but I think if pH = 4.62 then (H^+) = 2.39E-5. You should check that out from 4.62 = -log (H^+).
.................HA ==> H^+ + A^-
I............... X............0.........0
C .............-y.............y..........y
E.............X-y ...........y..........y
Ka = 0.0433 = (y)(y)/(X-y)
Substitute 2.39E-5 (or the value for H^+ of pH 4.62), for y, solve for X which the molarity of the HA solution. I suspect you will need to solve a quadratic equation.
Post your work if you get stuck.
An aqueous solution of a monoprotic acid HA (Ka = 4.33×10-2) has a pH of 4.62.
What is the molarity of the solution? (i.e. [HA]+[A-].)
1 answer