NH4OH⇌NH4++OH−Keq=NH4+OH−NH4OH≃OH−2NH4OH0 ( since NH4+=OH−)or OH−=KeqNH4OH0OH−=1.8×10−5×0.1M2OH−=1.34×10−3MpOH=−log⁡OH−/M=2.87; pH=14−pOH=11.13

Download Pdf

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The pH of $0.1 M {NH}_{4} OH (K_{b} = 1.8 \times 10^{- 5} M)$ at 25 °C is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

1.8

11.13

12.2

2.87

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} {NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-} \\ K_{eq} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]} ≃ \frac{{[{OH}^{-}]}^{2}}{{[{NH}_{4} OH]}_{0}} \end{array}$ ( since $[{NH}_{4}^{+}] = [{OH}^{-}]$ )

or $\begin{matrix} [{OH}^{-}] = \sqrt{K_{eq} [{NH}_{4} {OH}_{0}} \\ [{OH}^{-}] = \sqrt{1 .8 \times 10^{- 5} \times 0 .1 M^{2}} \\ [{OH}^{-}] = 1 .34 \times 10^{- 3} M \\ pOH = - \log \{[{OH}^{-}] / M\} = 2 .87; \\ pH = 14 - pOH = 11.13 \end{matrix}$