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secondary 4 | Chemistry
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Why is the answer B?
Date Posted: 3 years ago
Views: 264
lead(II)chloride (PbCl2) is a sparingly soluble salt (at O level, it is taken to be insoluble)
Reaction of PbCl2 with sulfuric acid (H2SO4, which is a strong acid that almost completely dissociates in water to form H+ ions and SO4²- ions) gives the sparingly soluble lead(II)sulfate (PbSO4) and hydrochloric acid (HCl).
PbCl2 + H2SO4 → PbSO4 + 2HCl
Since HCl is also a strong acid, the dissociation is also almost complete and so the H+ concentration in the solution remains unchanged.
(The volume of solution is also unchanged)
pH = -log[H+] (more correctly, -log[H3O+] )
So pH is unchanged. There is no neutralisation in this reaction.
Reaction of PbCl2 with sulfuric acid (H2SO4, which is a strong acid that almost completely dissociates in water to form H+ ions and SO4²- ions) gives the sparingly soluble lead(II)sulfate (PbSO4) and hydrochloric acid (HCl).
PbCl2 + H2SO4 → PbSO4 + 2HCl
Since HCl is also a strong acid, the dissociation is also almost complete and so the H+ concentration in the solution remains unchanged.
(The volume of solution is also unchanged)
pH = -log[H+] (more correctly, -log[H3O+] )
So pH is unchanged. There is no neutralisation in this reaction.
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