Tag: dissociation constants

Questions Related to dissociation constants

The addition of $HCl$ will not suppress the ionisation of:

  1. Acetic acid

  2. Sulphuric acid

  3. ${H} _{2}S$

  4. Benzoic acid


Correct Option: B
Explanation:

The addition of $HCl$ will not suppress the ionisation of $H _2SO _4$ because $H _2SO _4$ is stronger acid than $HCl$.

One litre of water contains ${ 10 }^{ -7 }$ mole of ${H}^{+}$ ions. Degree of ionisation of water is:

  1. $1.8\times { 10 }^{ -7 }$

  2. $0.8\times { 10 }^{ -9 }$

  3. $5. 4\times { 10 }^{ -9 }$

  4. $5 . 4\times { 10 }^{ -7 }$


Correct Option: A
Explanation:

$H _2O⇌H^{ + }   +       OH^{ - }\quad $

  $C$        $0$                  $0$
$C(1-\alpha )$        $C\alpha $    $C\alpha $

$C\alpha=10^{-7}$

[H2O] =$\dfrac{ 1000}{18}$= 55.55 M 

C= 55.5M

So $\alpha=18\times10^{-10}$

In percentage $\alpha=1.8\times10^{-7}$%

When ${ NH } _{ 4 }Cl$ is added to ${ NH } _{ 4 }OH$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

  1. common ion effect

  2. hydrolysis

  3. oxidation

  4. reduction


Correct Option: A
Explanation:

When $NH _4Cl$ is added to $NH _4OH$ solution, concentration of $NH _4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced. 

The weak acid, $HA$ has a ${K} _{a}$ of $1.00\times { 10 }^{ -5 }$. If $0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

  1. $1$%

  2. $99.9$%

  3. $0.1$%

  4. $99$%


Correct Option: A
Explanation:
0.1 mole of acid is dissolved in 1 litre of water means $[HA]=0.1M$
Let '$\alpha$' be degree of dissocition
$HA\rightleftharpoons { H }^{ + }+{ A }^{ - }$
 $0.1$
$0.1(1-\alpha)$   $0.1\alpha$     $0.1\alpha$
${ K } _{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right]  }{ \left[ HA \right]  } =\cfrac { { 0.1 }^{ 2 }{ \alpha  }^{ 2 } }{ 0.1(1-\alpha)  } $
Let $\alpha<<1$ so $1-\alpha=1$
$K _a=0.1\alpha^2=10^{-5}$
$\alpha=10^{-2}$
% of acid dissociated=$1$%

If concentration of two acids are some, their relative strengths can be compared by:

  1. ${ \alpha } _{ 1 }/{ \alpha } _{ 2 }$

  2. $K _{ 1 }/K _{ 2 }$

  3. ${ \left[ { H }^{ + } \right] } _{ 1 }/{ \left[ { H }^{ + } \right] } _{ 2 }$

  4. $\sqrt { K _{ 1 }/K _{ 2 } } $


Correct Option: A,C,D
Explanation:

Relative strength of two acids can be compared by their degree of dissociation.

$HA\rightleftharpoons H^++A^-$
$C$
$C-C\alpha$   $C\alpha$    $C\alpha$
If concentration of two acids are same so their relative strength can be compared by their $[H^{+}]$ concentration.
$K _a=C\alpha^2$
$\alpha=(K _a/C)^{0.5}$
If concentration of two acids are same so their relative strength can be compared by square root of their dissociation constants.

The solubility of $AgCl$ in $NaCl$ solution is less than that in pure water, because of the  ________.

  1. solubility product of $AgCl$ is less than of $NaCl$

  2. common ion effect

  3. both $A$ and $B$

  4. none of these


Correct Option: B
Explanation:

Since, $NaCl$ is soluble to a very significant extent, when $AgCl$ is added to $NaCl$ solution, the common ion $[Cl^-]$ increases in the solution. To have the solubility product or $K _{sp}$ of $AgCl$ constant, $[Ag^+]$ will decrease or $AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of $AgCl$ in $NaCl$ solution will be less than that in pure water.

100 mL of 20.8% $BaCl _2$ solution and 50 mL of 9.8% $H _2SO _4$ solution will form $BaSO _4$
$(Ba=137, Cl=35.5, S=32, H=1, O=16)$
$BaCl _2+H _2SO _4\rightarrow BaSO _4+2HCl$

  1. 23.3 g

  2. 11.65 g

  3. 30.6 g

  4. None of these


Correct Option: B
Explanation:

$100ml$ of $20.8$% $BaCl _2$ solution= $20.8g$  $BaCl _2$

$50ml$ of $9.8$% $H _2SO _4$ solution= $4.9g$  $H _2SO _4$
Reaction: $BaCl _2+H _2SO _4\longrightarrow BaSO _4\downarrow +2HCl$
           $208 g mol^{-1}$   $98 g mol^{-1}$      $233 g mol^{-1}$
$\therefore 98g$ $H _2SO _4$ reacts with $208g$ $BaCl _2$
$4.9g$ $H _2SO _4$ reacts with $\cfrac {208}{98}\times 4.9=10.4 g$ $BaCl _2$
$98g$ $H _2SO _4$ will produce $233g$ $BaSO _4$
$\therefore 4.9g$ $H _2SO _4$ will produce= $\cfrac {233}{98}\times 4.9=11.65g$ $BaSO _4$

The addition of NaCl to AgCl decreases the solubility of AgCl because ________.

  1. Solubility product decreases

  2. Solubility product remains constant.

  3. solution becomes unsaturated

  4. solution becomes super saturated.


Correct Option: D
Explanation:

NaCl is highly soluble and when it is added to AgCl it decreases the solubility of AgCl because of common ion $Cl^-$ and solution become super saturated.

The relative strength of two weak bases at same concentration may be given as:

  1. $\cfrac { { K } _{ { b } _{ 1 } } }{ { K } _{ { b } _{ 2 } } } $

  2. $\cfrac { { \left[ { OH }^{ - } \right] } _{ 1 } }{ { \left[ { OH }^{ - } \right] } _{ 2 } } $

  3. $\cfrac { { \alpha } _{ 1 } }{ { \alpha } _{ 2 } } $

  4. $\cfrac { \sqrt { { K } _{ { b } _{ 1 } } } }{ \sqrt { { K } _{ { b } _{ 2 } } } } $


Correct Option: B,C,D
Explanation:

Relative strength of two bases are measured by their degree of dissociation.

For weak base $[OH^-]=C\alpha$ and $K _b=C\alpha^2$
So relative strength of two weak bases at same concentration may be given as $\cfrac{[OH^-] _1}{[OH^-] _2}=\cfrac{C\alpha _1}{C\alpha _2}=\cfrac{\sqrt K _{b1}}{\sqrt K _{b2}}$

Which of the following solutions display common ion effect?

  1. $HCl+NaCl$

  2. ${ H } _{ 2 }S + KH$

  3. $NaHS+HCl$

  4. ${ H } _{ 2 }{ SO } _{ 4 }+{ K } _{ 2 }S$


Correct Option: A
Explanation:

In common ion effect there should be common ions in the two or more chemicals. Here, $HCl+NaCl$ shows common ion effect. Because $Cl^-$ is common in both.