you first have to know the number of moles of hydrogen present in oxalic acid from there progress it is not the same with my country ones so you can just try my formula if it works ph=-(log{h}^+)
make sure log is in base ten.
What is the pH of a 0.0054 molar solution of Oxalic Acid. K1= 6.5E-2, K2= 6.1E-5?
Here's what I have so far for K1:
6.5E-2 = (x)(x)/ (x-0.0054)
x= 0.0059
pH = -log(0.0059) = 2.23
K2 is pretty much the same, but I have to work it out.
6.1E-5 = (x)(x + 0.0059)/ (x-0.0059)
x = 0.0058 and pH = 2.24
How does this look? Am I close?
4 answers
Let's call oxalic acid H2Ox, then
H2Ox ==> H^+ + HOx^-
K1 = (H^+)(HOx^-)/(H2Ox) = 6.5 x 10^-2
then (H^+) = y which you have (I don't want to use x because we have x in the oxalate).
and (HOx^-) = y which you have.
and (H2Ox) = 0.0652-y
You can solve this quadratic for y. Then convert to pH by
pH = -log(H^+).
You can essentially ignore k2 since you get 1 H from HOx^- for every 1000 you get from H2Ox.
H2Ox ==> H^+ + HOx^-
K1 = (H^+)(HOx^-)/(H2Ox) = 6.5 x 10^-2
then (H^+) = y which you have (I don't want to use x because we have x in the oxalate).
and (HOx^-) = y which you have.
and (H2Ox) = 0.0652-y
You can solve this quadratic for y. Then convert to pH by
pH = -log(H^+).
You can essentially ignore k2 since you get 1 H from HOx^- for every 1000 you get from H2Ox.
When solving using the quadratic equation, I get two possible values. Which one do I use? Both are positive.
One of them won't make sense. If you have two positive values, one should be for y = a number larger than 0.0054 and that can't happen if the concn of the acid is 0.0054. That is, H^+ can't be larger than concn oxalic acid. Having said all of that, I solved the quadratic and obtained 0.005 and -0.07. Of course I took 0.005 as the value. The equation I have is
y^2 = 0.065(0.0054-y)
y2+0.065y - 0.00035 = 0
Check my arithmetic.
y^2 = 0.065(0.0054-y)
y2+0.065y - 0.00035 = 0
Check my arithmetic.
1 answer