What is the pH of a 0.0054 molal solution of C2H2O4. First ionization constant K1= 6.5 x 10-2, second K2= 6.1 x 10-5.
4 answers
That must be oxalic acid. I would write the ionization equation and use k1. I would ignore k2 since only 1 in 1000 of the bioxalate ions (hydrogen oxalate) ionize. Set up an ICE chart as you would do for a monoprotic acid, substitute into the k1 expression and solve. Post your work if you get stuck.
I'm not real sure about the K1 expressions. Is this correct?
C2H2O4 --> H+ + C2O4H-
K1 = (x + 0.0054)(x + 0.0054)/ (0.0054-x)
Solve for x then repeat for K2
C2H2O4 --> H+ + C2O4H-
K1 = (x + 0.0054)(x + 0.0054)/ (0.0054-x)
Solve for x then repeat for K2
Looks right to me.
Solve for x.
You can ignore the second ionization completely because K2 is significantly smaller than K1. It will minimally impact your answer only beyond the significant digits.
Solve for x.
You can ignore the second ionization completely because K2 is significantly smaller than K1. It will minimally impact your answer only beyond the significant digits.
Not quite right but close.
k1 = (x)(x)/(0.0054-x) and solve for x.
By the way, I have assumed all along that you made a typo when you wrote 0.0054 molal (and not 0.0054 molAR) solution. If it is molal you should convert molal to molar BUT you don't have a density; however, at this low concn molal and molar are essentially the same; therefore, it really won't make much difference.
k1 = (x)(x)/(0.0054-x) and solve for x.
By the way, I have assumed all along that you made a typo when you wrote 0.0054 molal (and not 0.0054 molAR) solution. If it is molal you should convert molal to molar BUT you don't have a density; however, at this low concn molal and molar are essentially the same; therefore, it really won't make much difference.
1 answer