Question

Help with Homework? :
Oxalic acid, C2O4H2, is a polyprotic acid with Ka1 = 5.9 × 10-2 and Ka2 = 6.4 × 10-5.
a. Write the equilibrium equations and Ka expressions for the two successive ionizations of oxalic acid.
b. For the amphoteric substance C2O4H-, predict whether it is more likely to act as an acid or a base

Answer 1

For (a) would it be in the format :
\[K= [C]^{c}[D]^{d}\div [A]^{a}[B ]^{b}\]

Answer 2

@fifib there are two Ka values because there are two protons on that acid. it's polyprotic. one ka value for each one.

Answer 3

can you write out what the two expressions would be?

\[\frac{ [PRODUCTS] }{ [REACTANTS] } = Ka \]

Answer 4

If I'm given oxalic acid, do i have to break it up into reactants and products? @Photon336

Answer 5

yeah, you have to write the dissociation equation for oxalic acid in order to write the equilibrium expression.

Answer 6

C2H2O4 ↔ C2O4-+ 2 H+ @aaronq like that?

Answer 7

The first dissociation is like this:

so the equation is: \(\sf C_2O_4H_2\rightarrow HC_2O^{-}_4+H^+\)

the second hydrogen (more accurately, proton) comes off next, make sure the charges are correct. Note that because the molecule is symmetric, it doesn't matter which proton comes off first.

Also, the first Ka is associated with the first dissociation and the second Ka with the second proton.

Answer 8

\[{[HC2O4^{-}][H ^{+}]\div [C2O4H2]}\] ? @aaronq

Answer 9

that's almost it, remember it has to be an equation, so you need an equal sign:

\(\sf K_A=\dfrac {[HC_2O_4^-][H ^{+}]} {[C_2O_4H_2]}\)

Answer 10

@aaronq Oh okay . I think i already asked this once, but do i have to substitute their molar masses?

Answer 11

Not their molar masses, the square brackets mean concentration in molarity units (mol/L).

The question itself doesn't ask you to solve for anything though, so thats not necessary (for the question).