pH

[OU11BB]

if given a Ka value of 5.3x10-5 how can you determine the pH by estimation? Thanks

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[Extirpator]

If all you are given is a random Ka value... as far as I know, you can't get pH... You need to, at the very least, know what the Ka value belongs to (what rxn/equalib).

for an acid:

HA ----> A- + H+

 

[nze82]

That information alone is not enough to estimate the pH. They need to give you the concentration of the weak acid. For the purpose of answering this question, let's say they have told you that the concentration of the acid is 0.5M and the Ka = 5.3x10-5. Now let's solve for the pH:

HA <--> H+ + A-

Ka = [H+][A-]/[HA]

Let's see how the concentrations will change upon dissociation of this acid:

[HA] [H+] [A-]
*********************
0.5M 0 0 -->Initial concentrations
0.5-x x x -->Equilibrium concentrations after dissociation

Now, plug in the equilibrium concentrations in the formula for Ka:

Ka = (x)(x)/(0.5-x)

Since x is very small we can say that (0.5-x)~0.5 and rewrite the formula as:
Ka = x^2/0.5 = 5.3x10-5

Now let's solve for x:
x^2 = (5.3x10-5)(0.5)
x = Sqrt(5.3x10-5)(0.5)
x = 5.2x10^-3 = [H+]

Now that we have [H+] let's use it to calculate the pH:

pH = -log[H+]

pH = -log(5.2x10-3) = 3 - log(5.2) = 3 - 0.7 = 2.3

How do we estimate this?

-log(mx10-n) = n - logm
**Remember that log(1) = 0 and log(10) = 1. So, log(5.2) is somewhere between 0 and 1. So:
3 - log(5.2) must be smaller than 3 but larger than 2, which is what we see above.

 

[OU11BB]

Hey thanks for the help.