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Study 3.1.10 (HL)—Acid/base strength constants conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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Two hydrides of nitrogen are ammonia and hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$ . One derivative of ammonia is methanamine whose molecular structure is shown below.

Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second $\mathrm{H}^{+}$ is so small it can be neglected.)

\begin{gathered} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ \mathrm{pK}_{\mathrm{b}}(\text { hydrazine })=5.77 \end{gathered}

Calculate the pH of a $0.0100 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of hydrazine.

$K_b=10^{-5.77}=1.698\times10^{-6}$

or:
$K_b=\frac{[\mathrm{N}_{2}\mathrm{H}_{5}^{+}][\mathrm{OH}^{-}]}{[\mathrm{N}_{2}\mathrm{H}_{4}]}$ $[\mathrm{OH}^{-}]^2 =1.698\times10^{-6}\times0.0100 =1.698\times10^{-8}$ $[\mathrm{OH}^{-}] =\sqrt{1.698\times10^{-8}} =1.303\times10^{-4}\ \mathrm{mol\ dm^{-3}}$ $\mathrm{pOH}=3.89,\qquad \mathrm{pH}=10.11$

Accept answers in the range 10.1-10.2.

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IB Chemistry HL 3.1.10 Hl Acid Base Strength Constants Topic Practice

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