JEE MainChemistryModeratePhysical Chemistry

pH of a solution obtained by mixing equal volumes of 0.1 M
Triethyl amine (K_b = 6.4 × 10^-5) & $\frac{4}{45}$ M NH_4OH (K_b =
1.8 × 10^-5) will be

Show answer and solution

\left\lbrack {OH}^{-} \right\rbrack = \sqrt{\left( 0.1 \times 6.4 + \frac{4 \times 1.8}{45} \right) \times 10^{- 5}}

= 0.002M

\therefore

pH = 11.3

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.