=== Model: grok-4.6-direct | Task: bench:gpqa_diamond | Rep: 1 ===
=== Score: 5/5 | Wall: 12.87s | Cost: $0.007272 ===
=== Pass: True | Finish: stop | In-reasoning-only: False ===

--- JUDGE ---
gpqa_diamond

--- RESPONSE ---
The Ca-EDTA complex is formed in a 1:1 stoichiometry according to the equilibrium
\[
\mathrm{Ca}^{2+}+\mathrm{EDTA}^{4-}\rightleftharpoons\mathrm{Ca\text{-}EDTA}^{2-},\qquad K=5\times10^{10}.
\]
The corresponding dissociation constant is therefore
\[
K_\mathrm{d}=1/K=2\times10^{-11}.
\]
In a 0.02 M solution of the stoichiometric complex the only source of free \(\mathrm{Ca}^{2+}\) and free EDTA is the slight dissociation of the complex. Letting \(x=[\mathrm{Ca}^{2+}]=[\mathrm{EDTA}]\) and noting that \(x\ll0.02\) gives
\[
K_\mathrm{d}=\frac{x^{2}}{0.02}=2\times10^{-11}\implies x^{2}=4\times10^{-13}\implies x=6.3\times10^{-7}.
\]
(The assumption of an “ideal” pH means that protonation of EDTA can be ignored, so the given formation constant may be used directly.)

The answer is (A)