Determine the pH of each of the following solutions:?
b. 0.0085 M phenol (Ka = 1.3 X 10-10)
c. 0.095 M hydroxylamine (Ka = 1.1 X 10-8)
Answer:
a) HClO <=> H+ + ClO-
Ka = [H+][ClO-]/[HClO]
[H+] = [ClO-] = x
[HClO] = .125 - x
3e-8 = x^2/(.125 - x)
x = 6.122e-5
pH = -log[H+] = -log 6.122e-5 = 4.21
b) C6H5OH <=> H+ + C6H5O-
Ka = [H+][C6H5O-]/[C6H5OH]
1.3e-10 = x^2/(.0085 - x)
x = 1.05e-6
pH = -log 1.05e-6 = 5.98
c) use the same method as above
pH = 4.49
Okay. As we only are given one Ka value for each, we can assume they're all monoprotic (donate one hydrogen each) acids (they are anyway, but that's an easy way we can figure it out).
Now for any monoprotic acid, we can represent its ionisation as follows:
HA <=> H+ + A-
So therefore, Ka = [H+][A-] / [HA]
Now let x be the amount of HA that is consumed by the time the system reaches equilibrium:
[HA] = [HA]0 - x
[H+] = x
[A-] = x
And so, Ka = x^2 / ([HA]0 - x)
However, all of the acids we are dealing with have very very low values of Ka, so we can simplify this expression by noting that ([HA]0 - x) will be so close to [HA]0 that it won't matter if it's actually there or not (only a tiny amount of the acid will actually ionise, insignificant compared to the original concentration)
Ka = x^2 / [HA]0
x^2 = Ka x [HA]0
x = sqrt( [HA]0 x Ka )
As [H+] = x in the solution, pH = -log x, or:
pH = -log( sqrt( [HA]0 x Ka ) )
We can now apply this formula to each of the questions above, and obtain the following answers:
a. pH = -log ( sqrt ( 0.125 x 3.0 x 10^-8 ) )
= 4.21
b. pH = 5.98
c. pH = 4.49
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