Your solution assumes that 2-nitrophenol is a strong acid and dissociates completely (100%). But I found a pKa on the web listed as 7.2 which means Ka = 6.3E-8. If we call this compound HP, then
...........HP ==> H^+ + P^-
I.......0.0353....0......0
C.........-x......x.....x
E......0.0353-x...x......x
Then Ka = (H^+)(P^-)/(HP)
6.3E-8 = (x)(x)/(0.0353-x)
and solve for x = (H^+), then convert that to pH.
What is the pH of a solution prepared by dissolving 1.23g of 2-nitrophenol (FM 139.11) in 0.250 L?
Well so, i calculated the no.moles of 2-nitrophenol, and divided it by 0.250, to get the molarity which was 3.53E-2.
Since the molar ratio of dissociation is 1:1, i assumed [H+] to be 3.53E-2.
and then using pH=-log[H+], i got pH to be 1.45
Can you please confirm if my answer or method is good? I'll be very grateful
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