Calculate the molar solubilty of CaC2O4 in a solution that has been buffered to a pH of 4.00 Remember to account for reactions involving H2C2O4 acting as an acid in water. To account for the buffer, simply assume that H+ = 1.00X 10^-4

Answer 1

CaC2O4 is more soluble in the acidic buffered solution that it would normally be because of the formation of the bioxalate (hydrogen oxalate) ion.
CaC2O4 ==> Ca^+2 + C2O4^=
And H^+ + C2O4^= ==> HC2O4^-
And H^+ + HC2O4^- ==> H2C2O4

Look up Ksp for CaC2O4.
(1) Ksp = (Ca^+2)(C2O4^=) = ??
(2) Solubility = (Ca^+2) = (C2O4^=) + (HC2O4^-) + (H2C2O4)

(3) k1 H2C2O4 = you can do this.
(4) k2 H2C2O4 = you can do this.

First, the (H2C2O4) is quite small because the hydrolysis of HC2O4^- to form H2C2O4 is of the order of 10^-12 so we can neglect that.
Use k2 (equation 4) and plug in 1 x 10^-4 for H^+ and solve for (C2O4^=)/(HC2O4^-) = xx
and rearrange that to solve for (HC2O4^-) in terms of C2O4^=,
Now use equation 2 to obtain (Ca^+2) in terms of oxalate and rearrange that to solve for oxalate in terms of Ca^+2.

Finally, use equation 1 (Ksp) to plug in for Ca^+2 and for C2O4^= (in terms of Ca^+2). Solve for the only unknown of Ca^+2.
I have an answer of 8.09 x 10^-5 M but check my thinking and check my arithmetic. Post your work if you get stuck.

Answer 2

I would round the 8.09 to 8.1 x 10^-5 since the Ksp I found was to only two significant figures.