What is the solubility of silver bromide in a 0.1 mol/L solution of potassium cyanide?
For #"AgBr", K_text(sp) = 7.7 × 10^"-13"# .
For #"Ag(CN)"_2^"-", K_text(f) = 5.6 ×
10^8# .
For
For
The solubility is 0.002 mol/L.
The two equilibria are here:
The result of adding the two equations is
Regarding this balance,
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The solubility of silver bromide in a 0.1 mol/L solution of potassium cyanide is increased significantly due to the formation of a soluble complex ion, [Ag(CN)2]-, through the reaction: AgBr(s) + 2KCN(aq) ⇌ KAg(CN)2 + KBr(aq).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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