The overall stability constant of the complex ion [Cu(NH\(_3\))\(_4\)]\(^{2+}\) is 2.1 \(\times\) 10\(^{13}\). The overall dissociation constant is y\(\times\)10\(^{-14}\). Then y is __________. (Nearest integer)
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Remember the inverse relationship between stability and dissociation constants. A large stability constant (\(\beta \gg 1\)) means the complex is very stable and does not dissociate easily, hence it will have a very small dissociation constant (\(K_d \ll 1\)). This simple concept is often tested.
Step 1: Understanding the Question
We are given the overall stability constant (\(\beta\)) for a complex ion and asked to find its overall dissociation constant (K\(_d\)). Step 2: Key Formula or Approach
The stability constant (or formation constant) and the dissociation constant (or instability constant) are reciprocals of each other.
The formation reaction is: Cu\(^{2+}\) + 4NH\(_3\) \(\rightleftharpoons\) [Cu(NH\(_3\))\(_4\)]\(^{2+}\)
The stability constant is \(\beta_4 = K_{stability} = \frac{[[\text{Cu(NH}_3)_4]^{2+}]}{[\text{Cu}^{2+}][\text{NH}_3]^4}\).
The dissociation reaction is the reverse: [Cu(NH\(_3\))\(_4\)]\(^{2+}\) \(\rightleftharpoons\) Cu\(^{2+}\) + 4NH\(_3\)
The dissociation constant is \(K_d = \frac{[\text{Cu}^{2+}][\text{NH}_3]^4}{[[\text{Cu(NH}_3)_4]^{2+}]}\).
Therefore, the relationship is:
\[ K_d = \frac{1}{\beta_4} \]
Step 3: Detailed Calculation Substitute the given value of the stability constant:
\(\beta_4 = 2.1 \times 10^{13}\)
\[ K_d = \frac{1}{2.1 \times 10^{13}} \]
Calculate the value of K\(_d\):
\[ K_d \approx 0.476 \times 10^{-13} \]
Express K\(_d\) in the required format (y\(\times\)10\(^{-14}\)):
\[ K_d = 0.476 \times 10^{-13} = 4.76 \times 10^{-14} \]
Identify the value of y:
Comparing with \(y \times 10^{-14}\), we get \(y = 4.76\).
Step 4: Final Answer
The question asks for the nearest integer value of y. The nearest integer to 4.76 is 5.
