A solution contains 0.10 M NH3 (K_b = 1.8 x 10^-5) and 0.10 M CH3NH2 (K_b = 4.4 x 10^-4). Use [OH-] = sqrtK_b1c_1 + K_b2c_2 and K_w = 1.0 x 10^-14. Find the pH (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ia099

A solution contains $0.10 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and $0.10 M$ $CH X_{3} NH X_{2}$ ( $K_{b} = 4.4 \times 1 0^{- 4}$ ). Use $[OH X^{-}] = K_{b 1} c_{1} + K_{b 2} c_{2}$ and $K_{w} = 1.0 \times 1 0^{- 14}$ . Find the pH (two decimals).

Numerical answer type, tolerance 0.01.

[OH X^{-}]^{2} = 1.8 \times 1 0^{- 6} + 4.4 \times 1 0^{- 5} = 4.58 \times 1 0^{- 5}

, so

[OH X^{-}] = 6.77 \times 1 0^{- 3} M

. pOH = 2.170, pH = 11.83.

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