Tag: ionization constants of weak acids and weak bases

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Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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. $5 . 4\times { 10 }^{ -7 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\alpha=1.8\times10^{-7}$%

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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. 

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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. $99$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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=$1$%
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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 } } $
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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= $\cfrac {233}{98}\times 4.9=11.65g$ $BaSO _4$

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice chemistry chemical equilibrium and acids-bases dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

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 } } } } $
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
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 $\cfrac{[OH^-] _1}{[OH^-] _2}=\cfrac{C\alpha _1}{C\alpha _2}=\cfrac{\sqrt K _{b1}}{\sqrt K _{b2}}$

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

Which of the following solutions display common ion effect?

  1. $HCl+NaCl$
  2. ${ H } _{ 2 }S + KH$
  3. $NaHS+HCl$
  4. ${ H } _{ 2 }{ SO } _{ 4 }+{ K } _{ 2 }S$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.