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BITSAT Chemistry Ionic Equilibrium 2022 BITSAT 2022

BITSAT Chemistry Question (2022) — Solution

Question

What is [ NH _4^ + ]in a solution that is 0.02 M NH _3 0.01 M KOH ? [K_b ( NH _3 )=1.8 10^ -5 ]

Options

  1. A. 3.6 10^ -5 M
  2. B. 1.8 10^ -5 M
  3. C. 0.9 10^ -5 M
  4. D. 7.2 10^ -5 M

Answer

A. 3.6 10^ -5 M

Step-by-step solution

NH _4 OH NH _4^ + + OH ^ - K_b= [ NH _4^ + ] [ OH ^ - ] [ NH _4 OH ] 1.8 10^ -5 = [ NH _4^ + ][0.01] [0.02] [ NH _4^ + ]=3.6 10^ -5 M

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