A solution of HClO is mixed and found to
have a pH of 4.53. Find what the initial
concentration of HClO was for this solution.
Ka = 3.00 × 10−8
for HClO.
Answer in units of mol/L
please help this is due in 1 hour
2 answers
ka= 3.00 * 10 -8
pH = 4.53 = -log(H^+). Solve for (H^+) and that's approx 1E-5 but that's just close so you need to do it more accurately.
.........HClO ==> H^+ + ClO^-
I........x........0......0
C.......-y........y.......y
E........x-y......y........y
The problem tells you (actually the (H^+) from above tells you y = 10^-5
Ka = (H^+)(ClO^-)/(HClO)
(H^+) = 1E-5
(ClO^=) = 1E-5
(HClO) = x-1E-5
Solve for x.
.........HClO ==> H^+ + ClO^-
I........x........0......0
C.......-y........y.......y
E........x-y......y........y
The problem tells you (actually the (H^+) from above tells you y = 10^-5
Ka = (H^+)(ClO^-)/(HClO)
(H^+) = 1E-5
(ClO^=) = 1E-5
(HClO) = x-1E-5
Solve for x.