Download App

6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
The acid that results when a base accepts a proton is called
  • A
    Conjugate base of the acid
  • B
    Conjugate protonated base
  • C
    Lewis base
  • ✓
    Conjugate acid of the base
✓

Answer

Correct option: D.
Conjugate acid of the base
d
The conjugate bases of weak acids are strong. The weaker an acid, the greater is the strength of its conjugate base. The acid that results when a base accepts a proton is called a conjugate acid of the base.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 6.2. Equilibrium - II (icon Equilibrium)→6.2. Equilibrium - II (icon Equilibrium) — all question types→Chemistry STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

In which of the following molecular shape $d_{z^2}$ orbital must not be involved in  bonding ?
→
The size of the iso-electronic species $Cl^-$ , $Ar$ and $Ca^{2+}$ is affected by
→
For the system $A_{(g)} + 2B_{(g)}$ $\rightleftharpoons$ $C_{(g)}$, the equilibrium concentrations are $(A) 0.06$ mole/litre $(B) 0.12$ mole/litre $(C) 0.216$ mole/litre. The ${K_{eq}}$ for the reaction is
→
If in Bohr's model, for unielectronic atom, time period of revolution is represented as $T_{n,Z}$ where $n$ represents shell no. and $Z$ represents atomic number then the value of $T_{1,2} : T_{2,1}$ will be
→
Select the odd one
→
Which of the following expression for an irreverssible process.
→
The equilibrium constants of the following are

$N_2 + 3H_2 \rightleftharpoons  2NH_3 \,;$        $K_1$

$N_2 + O_2 \rightleftharpoons  2NO\,;$               $K_2$

$H_2 + 2 O_2 \rightleftharpoons  H_2O\,;$          $K_3$

The equilibrium constant $(K)$ of the reaction :

$2NH_3 + \frac{5}{2} \overset K \leftrightarrows 2NO + 3H_2O$

→
The single covalent radius of $P$ is $0.11\, nm.$ The single covalent radius of $Cl$ will be
→
The electronic configuration of four elements are given in brackets$L\,\left( {1{s^2},\,\,2{s^2}2{p^1}} \right);\,\,M\,\left( {1{s^2},\,\,2{s^2}\,2{p^5}} \right)$

$Q\,\left( {1{s^2},\,\,2{s^2}\,2{p^6},\,\,3{s^1}} \right);\,\,R\,\left( {1{s^2},\,\,2{s^2}\,2{p^2}} \right)$The element that would most readily form a diatomic molecule is

→
For the reaction $C{H_3} - CH = C{H_2} + HOCl \to A$ the product $A$ is
→