MCQ
What is the $pH$ of $0.1\,M\,N{H_3}$
- A$11.27$
- B$11.13$
- C$12$
- D$9.13$
$Kb =\frac{\left[ NH _4^{+}\right]\left[ OH ^{-}\right]}{\left[ NH _3\right]}$
$1.8 \times 10^{-5}=\frac{ x ^2}{0.1}$
$\Rightarrow x =\sqrt{1.8 \times 10^{-6}}$
$=1.34 \times 10^{-3}$
$\therefore\left[ OH ^{-}\right]\left[ H ^{+}\right]=1.34 \times 10^{-3}\left[ H ^{+}\right]=10^{-14}$
$\Rightarrow pH =11.13$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$2 SO _2( g )+ O _2( g ) \rightleftharpoons 2 SO _3( g ), \Delta H =-190\,kJ$
The number of factors which will increase the yield of $SO _3$ at equilibrium from the following is $.............$.
$A.$ Increasing temperature
$B.$ Increasing pressure
$C.$ Adding more $SO _2$
$D.$ Adding more $O _2$
$E.$ Addition of catalyst