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Answer the following · Q4

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.

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Step 1. Write HCN's own equilibrium. HCN is a weak acid and ionizes only partially: HCN(aq)⇌H+(aq)+CN−(aq)HCN(aq)\rightleftharpoons H^+(aq)+CN^-(aq).

Step 2. Bring in HCl. HCl is a strong acid (section 3.4), so it dissociates almost completely: HCl(aq)→H+(aq)+Cl−(aq)HCl(aq)\rightarrow H^+(aq)+Cl^-(aq), 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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