Download App

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

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1\,N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$ )
  • A
    $3\times 10^{-5}$
  • B
    $1.86\times 10^{-6}$
  • ✓
    $1.75\times 10^{-5}$
  • D
    $2.86\times 10^{-5}$
✓

Answer

Correct option: C.
$1.75\times 10^{-5}$
c
$\quad\left[\mathrm{H}^{+}\right]=\cdot 10^{-11.27}=10^{-12}, 10^{0.73}=5.37 \times 10^{-12}$

$\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{H}^{+}\right]}=\frac{7.1 \times 10^{-15}}{5.37 \times 10^{-12}}=\frac{7.1}{5.37} \times 10^{-3}\, \mathrm{M}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}$

$\left(\frac{7.1}{5.37}\right)^{2} \times 10^{-6}=\mathrm{K}_{\mathrm{b}} \times 0.1$

$\boxed{{K_b} = 1.747 \times {{10}^{ - 5}}}$

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

Similar questions

For an electron, with $n = 3$ has only one radial  node. The orbital angular momentum of the electron will be
→
If the solubility product of $PbS$ is $8 \times 10^{-28}$, then the solubility of $PbS$ in pure water at $298\; K$ is $x \times 10^{-16}\; mol\; L ^{-1}$. The value of $x$ is $\dots$.

[Given $\sqrt2 = 1.41$]

→
The rate of $o-$ nitration of the above compounds, $(I)$ toluene, $(II)\, 2-D-$ toluene and $(III) \,2, 6-D_2$ -toluene is in the following order
→
The correct match between item $I$ and item $II$ is
Item $I$ Item $II$
$(a)$ Benzaldehyde $(p)$ Mobile phase
$(b)$ Alumina $(q)$ Adsorbent
$(c)$ Acetonitrile $(r)$ Adsorbate
→
The two structures $I$ and $II$ represents
→
Borax is known as :
→
Highest reactivity in $IA$ is of 
→
The rate constant for forward and backward reactions of hydrolysis of ester are $1.1 \times {10^{ - 2}}$ and $1.5 \times {10^{ - 3}}$ per minute respectively. Equilibrium constant for the reaction is $C{H_3}COO{C_2}{H_5} + {H_2}O$ $\rightleftharpoons$ $C{H_3}COOH$ $ + {C_2}{H_5}OH$
→
The correct order of heats of combustion of above compounds is
→
Reactant $P$ gives products $Q$ or $R$.

The possible reagents are :
$(I)\, 2Na/liq.NH_3$   $(II)\, H_2 /Pd/CaCO_3$ (quinoline) $(III)\, 2H_2 / Pd /C$
The correct statement with respect to the above conversion is/are

→