Question
Using the value of the formation constant for the complex ion $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{2+},$ calculate the dissociation constant.
Step 1
The given value of $K_f$ is $1.310 \times 10^{5}$. Show more…
Show all steps
Your feedback will help us improve your experience
Aadit Sharma and 52 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Refer
Sometimes equilibria for complex ions are described in terms of dissociation constants, $K_{\mathrm{d}}$. For the complex ion AlF $_{6}^{3-}$ the dissociation reaction is: $$\mathrm{AlF}_{6}^{3-} \rightleftharpoons \mathrm{Al}^{3+}+6 \mathrm{F}^{-} \text {and } K_{\mathrm{d}}=\frac{\left[\mathrm{A} 1^{3+}\right]\left[\mathrm{F}^{-}\right]^{6}}{\left[\mathrm{AlF}_{6}^{3-}\right]}=2 \times 10^{-24}$$ Calculate the value of the formation constant, $K_{6}$, for $\mathrm{AlF}_{6}^{3-}$.
The formation constant, Kf, for the complex ion, [Ni(NH3)6]2+ is 1.0x10^8. Using this value and Ksp for nickel(II) hydroxide, calculate the value of the equilibrium constant for: Ni(OH)2(s) + 6NH3(aq) = [Ni(NH3)6]2+(aq) + 2OH-(aq)
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD