100 mL of 0.2 M ammonia (K_b=1.8x10^-5) is mixed with 100 mL of 0.2 M of another weak base (K_b=3.0x10^-6). Using [OH-]=sqrtK_1c_1+K_2c_2 with the concentrations after mixing, find the pH at 298 K to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3ie102

$100 mL$ of $0.2 M$ ammonia ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) is mixed with $100 mL$ of $0.2 M$ of another weak base ( $K_{b} = 3.0 \times 1 0^{- 6}$ ). Using $[OH X^{-}] = K_{1} c_{1} + K_{2} c_{2}$ with the concentrations after mixing, find the pH at 298 K to two decimal places.

Numerical answer type, tolerance 0.01.

After mixing each base is

0.1 M

[OH X^{-}]^{2} = 0.1 (1.8 \times 1 0^{- 5} + 3.0 \times 1 0^{- 6}) = 2.1 \times 1 0^{- 6}

, so

[OH X^{-}] = 1.449 e - 03 M

pOH = 2.839

and

pH = 11.16