Q. What concentration of H
2Se is required to produce a pH of 1.93 ?
A. 0.81 kmol/m^3
H
2Se
HSe
- + H
+10
-1.93 = H
+0.0117 M = H
+Since H
2Se is a weak acid, I need to make I.C.E table.
The equilibrium concentration of H
2Se is x-0.0117.
So the original concentration of H
2Se must be x.
I do not have the K
a provided for H
2Se but many internet sources say the pKa is 3.89 which translates to 1.29 x 10
-4.
Substituting into the eq expression, I get:
x = 1.07
which doesn't match the answer in the worksheet. What am I doing wrong ?
*btw how do you typset fractions and tables in this forum?