What is the approximate concentration of free Cd2+ ion at equilibrium when 1.18x102 mol cadmium(II) nitrate is added to 1.00 L of solution that is 1.010 M in CN. For [Cd(CN)4]², Kf= 6.0×1018. [Ca²*] = [ M
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![**Question:**
What is the approximate concentration of free \( \text{Cd}^{2+} \) ion at equilibrium when \( 1.18 \times 10^{-2} \) mol cadmium(II) nitrate is added to \( 1.00 \) L of solution that is \( 1.010 \) M in \( \text{CN}^- \)? For \( [\text{Cd}(\text{CN})_4]^{2-} \), \( K_f = 6.0 \times 10^{18} \).
\[ [\text{Cd}^{2+}] = \_\_\_\_\_\_\_ \text{M} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F445c8372-a397-41f1-88ec-d7a9c1d0e786%2F0a24c1f8-c137-4b16-a019-21c4ac0ac7bf%2F5hkcp4l_processed.jpeg&w=3840&q=75)
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