Chemistry-Ionic Equilibrium

NEET Chemistry Ionic Equilibrium MCQ Question

Type: MCQ-numerical-Hard-Class 11

What is the ionization constant KbK_b of a 0.004 M solution of hydrazine if the pH is 9.7?

A

00×10500 \times 10^{-5}

B

67×101067 \times 10^{-10}

C

5.98×1055.98 \times 10^{-5}

D

58×10858 \times 10^{-8}

Correct Answer

Option D

Detailed Explanation

Given pH = 9.7, we find [H+]=1.67×1010[H^+] = 1.67 \times 10^{-10}. Since [OH]=5.98×105[OH^-] = 5.98 \times 10^{-5}, KbK_b is calculated using [OH]2/0.004[OH^-]^2 / 0.004.

Found an issue with this question?