Convert pKa to Ka. pKa - -log Ka
Let's call saccharin just HS.
M HS = grams/molar mass which I've estimated to be about 0.03 but you need a closer answer than that.
.....HS ==> H^+ + S^-
I..0.03.....0.....0
C....-x.....x.....x
E..0.03-x...x.....x
Substitute the E line into the Ka expression and solve for x = (H^+). Then pH = -log(H^+)
Note that I expect you will need to use the quadratic formula as I don't think 0.03-x will be close to 0.03.
Saccharin has a pKa of 2.32
It has a formula of HNC7H4SO3
If you dissolve 4.6 grams of saccharin in 478 ml of water, what is the resulting pH ?
1 answer
1 answer