The basic ionisation constant for hydrazine- N2H4 is 9-6 - 10-7- What would be the percentage hydrolysis of 0-1M N2H5Cl -Given - -10-4-3-2 and -1-04 - 1-013

N_2H_5^+ + H_2O ⇌ N_2H_4 + H_3O^+ K_h = [N_2H_4][H_3O^+][N_2H_5^+] = nbsp;k_wk_b = 10^ 149.6 × 10^ 7 = 1.04 × 10^ 8 ∴ h= √(K_hc) ⇒ h= √(1.04 × 10^ 80.1) ⇒ ...

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Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Weak Base

The basic ion...

Question

The basic ionisation constant for hydrazine, $N_{2} H_{4}$ is $9.6 \times 10^{- 7}$ . What would be the percentage hydrolysis of $0.1 M N_{2} H_{5} C l$ ?
$(G i v e n : \sqrt{10.4} \approx 3.2 a n d \sqrt{1.04} = 1.013)$

0.01 %

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3.2 %

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1.013 %

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0.032 %

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Solution

The correct option is D $0.032 %$
$N_{2} H_{5}^{+} + H_{2} O ⇌ N_{2} H_{4} + H_{3} O^{+} K_{h} = \frac{[N_{2} H_{4}] [H_{3} O^{+}]}{[N_{2} H_{5}^{+}]} = \frac{k_{w}}{k_{b}} = \frac{10^{- 14}}{9.6 \times 10^{- 7}} = 1.04 \times 10^{- 8} ∴ h = \sqrt{\frac{K_{h}}{c}} \Rightarrow h = \sqrt{\frac{1.04 \times 10^{- 8}}{0.1}} \Rightarrow h = 3.2 \times 10^{- 4} Percentage of hydrolysis = 3.2 \times 10^{- 4} \times 100 = 0.032 %$

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Similar questions

Q. The basic ionisation constant for hydrazine,

N_{2} H_{4}

9.6 \times 10^{- 7}

. What would be the percentage hydrolysis of

0.1 M N_{2} H_{5} C l

(G i v e n : \sqrt{10.4} \approx 3.2 a n d \sqrt{1.04} = 1.013)

Q. The basic ionisation constant for hydrazine,

N_{2} H_{4}

9.6 \times 10^{- 7}

. What would be the percentage hydrolysis of

0.1 M N_{2} H_{5} C l

(G i v e n : \sqrt{10.4} \approx 3.2 a n d \sqrt{1.04} = 1.013)

Q. The base dissociation constant for hydrazine

N_{2} H_{4}

9.6 \times 10^{- 7}

. What would be the percent hydrolysis of

0.1 M N_{2} H_{5} C l

, a salt containing the acid ion conjugate to hydrazine base at

25^{\circ} C

Q. The base dissociation constant for hydrazine

N_{2} H_{4}

9.6 \times 10^{- 7}

. What would be the percent hydrolysis of

0.1 M N_{2} H_{5} C l

, a salt containing the acid ion conjugate to hydrazine base at

25^{\circ} C

Q. Hydrazine

(N_{2} H_{4})

can be produced according to the following reaction:

C l N H_{2} + 2 N H_{3} \to N_{2} H_{4} + N H_{4} C l

. 1 kg of

C l N H_{2}

is reacted with excess

N H_{3}

, find the actual number of molecules of

N_{2} H_{4}

produced if the percentage yield of the given reaction is 76%.

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The basic ionisation constant for hydrazine, N2H4 is 9.6×10−7. What would be the percentage hydrolysis of 0.1M N2H5Cl ? (Given:√10.4≈3.2 and √1.04=1.013)

The correct option is D 0.032%N2H+5+H2O⇌N2H4+H3O+Kh=[N2H4][H3O+][N2H+5]=kwkb=10−149.6×10−7=1.04×10−8∴h=√Khc⇒h=√1.04×10−80.1⇒h=3.2×10−4Percentage of hydrolysis=3.2×10−4×100=0.032%

The basic ionisation constant for hydrazine, $N_{2} H_{4}$ is $9.6 \times 10^{- 7}$ . What would be the percentage hydrolysis of $0.1 M N_{2} H_{5} C l$ ?
$(G i v e n : \sqrt{10.4} \approx 3.2 a n d \sqrt{1.04} = 1.013)$