Download App

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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6.2. Equilibrium - II (icon Equilibrium)4 Marks
MCQ
Hydrogen ion concentration of an aqueous solution is $1 \times {10^{ - 4}}\,M$. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of $mol$ $d{m^{ - 3}}$ is
  • A
    $1 \times {10^{ - 8}}$
  • B
    $1 \times {10^{ - 6}}$
  • ✓
    $2 \times {10^{ - 10}}$
  • D
    $0.5 \times {10^{ - 10}}$
✓

Answer

Correct option: C.
$2 \times {10^{ - 10}}$
(c) $[{H^ + }] = 1 \times {10^{ - 4}}M \Rightarrow [{H^ + }]\,\,[O{H^ - }] = 1 \times {10^{ - 14}}$

$[O{H^ - }] = \frac{{1 \times {{10}^{ - 14}}}}{{0.5 \times {{10}^{ - 4}}}} = 2 \times {10^{ - 10}}$ $M$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The number of orbitals in $2p$ sub-shell is
→
In which of the following equilibrium $K_c$ and $K_p$ are not equal?
→
Which does not react with chlorine in dark
→
Solid $PC{l_5}$ exits as
→
Which one of the following compounds undergoes bromination of its aromatic ring (electrophilic aromatic substitution) at the fastest rate ?
→
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason $R$

Assertion $A$ : Zero orbital overlap is an out of phase overlap.

Reason $R$ : It results due to different orientation/ direction of approach of orbitals.

In the light of the above statements. Choose the correct answer from the options given below.

→
In a complexometric tritration of metal ion with ligand $M$ (Metalion) $+L$ (Ligand) $\to $ $C$ (Complex) end point is estimated spectrophotometrically (through light absorption) If $'M'$  and $'C'$  do not absorb light and only $'L'$  absorbs, then the titration plot between absorbed light $(A)$  versus volume of ligand $'L'$  $(V)$  would look like
→
Which is optically active
→
A set of solutions is prepared using $180$ $g$ of water as a solvent and $10$ $g$ of different non-volatile solutes $A, B$ and $C$. The relative lowering of vapour pressure in the presence of these solutes are in the order

[Given, molar mass of $A =100 \,g\, mol ^{-1} ; B =200 \,g\, mol ^{-1}$$\left. C =10,000 \,g\, mol ^{-1}\right]$

→
If equilibrium constant of

$C{H_3}COOH + {H_2}O \rightleftharpoons C{H_3}CO{O^ - } + {H_3}{O^ + }$ is $1.8 \times 10^{-5}$, equilibrium constant for

$C{H_3}COOH + O{H^ - } \rightleftharpoons C{H_3}CO{O^ - } + {H_2}O$

→