MCQ
A certain weak base has a dissociation constant $2 \times 10^{-5}$. The equilibrium constant for its neutralisation reaction with strong acid is
- A$2 \times 10^{-5}$
- B$5 \times 10^{10}$
- C$2 \times 10^{-9}$
- ✓$2 \times 10^{9}$
$\therefore \mathrm{K}_{\mathrm{N}}=\frac{1}{\mathrm{K}_{\mathrm{h}}}=\frac{1}{\left(1 / 2 \times 10^{-9}\right)}=2 \times 10^{9}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text { A } \quad\quad\quad\quad\quad \text { B } \quad\quad\quad\text { C } \quad\quad\quad\quad\text { D }$
$1 \times 10^{-4} \quad 2 \times 10^{-4} \quad 0.1 \times 10^{-4} \quad 0.2 \times 10^{-4}$
(Where $E$ is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?
$(I)\, Co(III)$ is stabilised in presence of weak field ligands, while $Co(II)$ is stabilised in presence of strong field ligand.
$(II)$ Four coordinated complexes of $Pd(II)$ and $Pt(II)$ are diamagnetic and square planar.
$(III)\,[Ni (CN)_4]^{4-}$ ion and $[Ni (CO)_4]$ are diamagnetic tetrahedral and square planar respectively.
$(IV)\,Ni^{2+}$ ion does not form inner orbital octahedral complexes.
$2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})$
${\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}}$
(Given: $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )