A solution is 0.10 M in NH3 (K_b = 1.8x10^-5) and 0.10 M in CH3NH2 (K_b = 4.5x10^-4). Using [OH-] = sqrtK_1C_1+K_2C_2, the concentration of NH4+ is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia217

A solution is $0.10 M$ in $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and $0.10 M$ in $CH X_{3} NH X_{2}$ ( $K_{b} = 4.5 \times 1 0^{- 4}$ ). Using $[OH X^{-}] = K_{1} C_{1} + K_{2} C_{2}$ , the concentration of $NH X_{4} X^{+}$ is

[OH X^{-}] = 1.8 \times 1 0^{- 6} + 4.5 \times 1 0^{- 5} = 6.84 \times 1 0^{- 3} M

. Then

[NH X_{4} X^{+}] = K_{b} [NH X_{3}] / [OH X^{-}] = 1.8 \times 1 0^{- 5} \times 0.10/6.84 \times 1 0^{- 3} = 2.63 \times 1 0^{- 4} M

. The value

1.34 \times 1 0^{- 3}

is what ammonia alone would give.