Wow, now this one is a hard one, took me quite a while to figure it out but here it is:

The abundances of the two isotopes is:
Antimony-121 = 57.2%
Antimony-123 = 42.8%
Answer:
2.3 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of calcium oxalate
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹
Answer : The hydroxide ion concentration of a solution is, 
Explanation :
As we know that
dissociates in water to give hydrogen ion
and carbonate ion
.
As, 1 mole of
dissociates to give 1 mole of hydrogen ion 
Or, 1 M of
dissociates to give 1 M of hydrogen ion 
So, 0.200 M of
dissociates to give 0.200 M of hydrogen ion 
Now we have to calculate the hydroxide ion concentration.
As we know that:
![[H^+][OH^-]=1\times 10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D1%5Ctimes%2010%5E%7B-14%7D)
![0.200\times [OH^-]=1\times 10^{-14}](https://tex.z-dn.net/?f=0.200%5Ctimes%20%5BOH%5E-%5D%3D1%5Ctimes%2010%5E%7B-14%7D)
![[OH^-]=5\times 10^{-14}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D5%5Ctimes%2010%5E%7B-14%7D)
Therefore, the hydroxide ion concentration of a solution is, 