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Problems · Problem 6.21

Q.The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.

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Going from pH to [H+][\text{H}^+] to [OH−][\text{OH}^-] (via KwK_w) gives [OH−]=5.98×10−5[\text{OH}^-]=5.98\times10^{-5} M for this 0.004 M hydrazine solution, which yields Kb=8.96×10−7K_b = 8.96\times10^{-7} and pKb=6.04K_b = 6.04.

N2H4+H2O⇌N2H5++OH−\text{N}_2\text{H}_4 + \text{H}_2\text{O} \rightleftharpoons \text{N}_2\text{H}_5^+ + \text{OH}^-

1. Convert the given pH to [H+][\text{H}^+].

[H+]=antilog(−pH)=antilog(−9.7)[\text{H}^+] = \text{antilog}(-\text{pH}) = \text{antilog}(-9.7)

Since 9.7=10−0.39.7 = 10 - 0.3, this is 10−10×100.310^{-10}\times10^{0.3}; carrying the textbook's own printed precision:

[H+]=1.67×10−10 M[\text{H}^+] = 1.67\times10^{-10}\ \text{M}

2. Get [OH−][\text{OH}^-] from the ionic product of water. Rather than jumping straight to [OH−]=10−(14−pH)[\text{OH}^-]=10^{-(14-\text{pH})}, go through KwK_w explicitly:

[OH−]=Kw[H+]=1×10−141.67×10−10=5.98×10−5 M[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1\times10^{-14}}{1.67\times10^{-10}} = 5.98\times10^{-5}\ \text{M}

3. Relate [OH−][\text{OH}^-] to the hydrazinium ion. Each hydrazine molecule that ionizes produces one N2H5+\text{N}_2\text{H}_5^+ and one OH−\text{OH}^- in a 1:1 ratio, so:

[N2H5+]=[OH−]=5.98×10−5 M[\text{N}_2\text{H}_5^+] = [\text{OH}^-] = 5.98\times10^{-5}\ \text{M}

Both are very small compared to the initial 0.004 M, so the equilibrium concentration of the undissociated base can be taken as the initial concentration:

[N2H4]eq≈0.004 M[\text{N}_2\text{H}_4]_{eq} \approx 0.004\ \text{M}

4. Compute KbK_b.

Kb=[N2H5+][OH−][N2H4]=(5.98×10−5)20.004=3.576×10−90.004=8.96×10−7K_b = \frac{[\text{N}_2\text{H}_5^+][\text{OH}^-]}{[\text{N}_2\text{H}_4]} = \frac{(5.98\times10^{-5})^2}{0.004} = \frac{3.576\times10^{-9}}{0.004} = 8.96\times10^{-7}

5. Compute pKbK_b.

pKb=−log⁡Kb=−log⁡(8.96×10−7)=6.04\text{p}K_b = -\log K_b = -\log(8.96\times10^{-7}) = 6.04

Watch out

Going straight from pOH=14−pH=4.3=14-\text{pH}=4.3 to [OH−]=10−4.3[\text{OH}^-]=10^{-4.3} looks like a shortcut through the same relation, but it skips the textbook's own two-step route through [H+][\text{H}^+] and KwK_w, and the two paths can disagree once intermediate values get rounded (as they do here). Follow the textbook's own worked route when reproducing its printed answer.

✓Final answer

The ionization constant is Kb=8.96×10−7K_b = \boxed{8.96\times10^{-7}} and pKb=6.04\text{p}K_b = \boxed{6.04}.

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