Q.iv. Dissociation of HCN is suppressed by the addition of HCl. Explain.
[!NOTE]
In the textbook the printed numbering of this exercise jumps from iv. directly to vi. -- no item v. is printed. The numbering here follows the book exactly.
In the textbook the printed numbering of this exercise jumps from iv. directly to vi. -- no item v. is printed. The numbering here follows the book exactly.
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Start your 14-day free trial to unlock the full solution →Step 1. Write HCN's own equilibrium. HCN is a weak acid and ionizes only partially: .
Step 2. Bring in HCl. HCl is a strong acid (section 3.4), so it dissociates almost completely: , adding a large amount of H+ from a source entirely separate from HCN's own ionization.
Step 3. Identify the common ion. H+ is a PRODUCT of HCN's own dissociation equilibrium AND is supplied in bulk by HCl -- it is the ion the two acids have 'in common'.
Step 4. Apply Le Chatelier's principle. Increasing the concentration of a product of an equilibrium (from any source) is a stress on that equilibrium; the system responds by shifting toward the reactant side to partly counteract the added product. …
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