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?
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.
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.
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.
Your approach looks correct, but there seems to be a mistake in your calculations. Let's go through the calculations step by step to determine the correct pH for a 0.0054 molar solution of Oxalic Acid.
For K1:
1. Start with the reaction equation: H2C2O4 ⇌ H+ + HC2O4-
2. Write the equilibrium expression: K1 = [H+][HC2O4-] / [H2C2O4]
3. Let x be the concentration of H+ and (0.0054 - x) be the concentration of HC2O4-. The initial concentration of H2C2O4 is 0.0054 M.
4. Substitute the values into the equilibrium expression and solve for x:
6.5E-2 = (x)(0.0054 - x) / (0.0054)
5. Rearrange the equation and solve for x:
0.0654 = x(0.0054 - x)
x^2 - 0.0054x + 0.0654 = 0
6. Solve the quadratic equation using the quadratic formula:
x = (-(-0.0054) ± √((-0.0054)^2 - 4(1)(0.0654))) / (2(1))
x ≈ 0.0052 or x ≈ -0.0002 (ignore the negative value as it is not physically meaningful)
7. Calculate the pH using the concentration of H+:
pH = -log[H+]
pH = -log(0.0052)
pH ≈ 2.28
For K2, you need to follow the same steps:
1. Start with the reaction equation: HC2O4- ⇌ H+ + C2O4^2-
2. Write the equilibrium expression: K2 = [H+][C2O4^2-] / [HC2O4-]
3. Let x be the concentration of H+ and (0.0054 + x) be the concentration of C2O4^2-. The initial concentration of HC2O4- is 0.0054 M.
4. Substitute the values into the equilibrium expression and solve for x:
6.1E-5 = (x)(0.0054 + x) / (0.0054)
5. Rearrange the equation and solve for x:
0.0054x + x^2 - 0.0061 = 0
x^2 + 0.0054x - 0.0061 = 0
6. Solve the quadratic equation using the quadratic formula:
x = (-0.0054 ± √(0.0054^2 - 4(1)(-0.0061))) / (2(1))
x ≈ -0.0049 or x ≈ 0.0009 (ignore the negative value as it is not physically meaningful)
7. Calculate the pH using the concentration of H+:
pH = -log[H+]
pH = -log(0.0009)
pH ≈ 3.05
Therefore, the pH of a 0.0054 molar solution of Oxalic Acid is approximately 2.28 for K1 and 3.05 for K2.