 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/4/a/iii/solution

The equilibrium molecular-acid concentration is $[H_{2} N_{2} O_{2}] = 0.162 mol L^{- 1}$ . From the supplied pH convention,

[H^{+}] = c^{\circ} 1 0^{- 4} = 1 0^{- 4} M .

With only these monovalent ions appreciably present, electroneutrality requires

[H N_{2} O_{2}^{-}] = [H^{+}]

. Substitution in the acid dissociation constant gives

[H_{2} N_{2} O_{2}] = \frac{( 1 0 ^{- 4} M ) ^{2}}{( 6.1690 \times 1 0 ^{- 8} ) ( 1 M )} = 0.16210 M .

H^{+}

, negligible water autoionization, negligible second dissociation, and no additional electrolyte. The acid's analytical concentration would include its dissociated fraction and be approximately

0.16220 M

, rather than exactly the molecular-acid concentration requested.

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