Tag: further aspects of equilibria

Questions Related to further aspects of equilibria

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

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

  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 } } $

Show answer
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  ________.

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

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

  2. common ion effect

  3. both $A$ and $B$

  4. none of these

Show answer
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 $BaSO _4$
$(Ba=137, Cl=35.5, S=32, H=1, O=16)$
$BaCl _2+H _2SO _4\rightarrow BaSO _4+2HCl$

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

  1. 23.3 g

  2. 11.65 g

  3. 30.6 g

  4. None of these

Show answer
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 ________.

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

  1. Solubility product decreases

  2. Solubility product remains constant.

  3. solution becomes unsaturated

  4. solution becomes super saturated.

Show answer
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.

Which of the following solutions display common ion effect?

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

  1. $HCl+NaCl$

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

  3. $NaHS+HCl$

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

Show answer
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.

What is $[{ NH } _{ 4 }^{ + }]$ in a solution containing 0.02M ${ NH } _{ 3 }$ (${ K } _{ b }={ 1.8\times 10 }^{ -5 }$) and 0.01M KOH?



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

  1. ${ 1.8\times 10 }^{ -5 }$

  2. ${ 9\times 10 }^{ -6 }$

  3. ${ 3.6\times 10 }^{ -5 }$

  4. NONE OF THE ABOVE

Show answer
Correct Option: C

$As _2S _3$ solution has negative charge, capacity to precipitate is highest in:

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

  1. $AlCl _3$

  2. $Na _3PO _4$

  3. $CaCl _2$

  4. $K _2SO _4$

Show answer
Correct Option: A
Explanation:

Solution:- (A) $Al{Cl} _{3}$

According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence ${As} _{2}{S} _{3}$ precipitate the most in $Al{Cl} _{3}$.

At $100^\circ C$, value of $K _{w}$ is 

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

  1. $1.0\times 10^{-14}\quad m^{2}$

  2. less than $1.0\times 10^{-14}\quad m^{2}$

  3. greater than $1.0\times 10^{-14}\quad m^{2}$

  4. Zero

Show answer
Correct Option: C
Explanation:

At higher temperature the value of $kw$ increases.This is in according with le-chatelier principle.

At $100^o kw=51.3\times 106{-14}$
C is the correct answer.

The solubility of CaF$ _2$ (K$ _{sp} = 5.3\times 10^{-9}$) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

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

  1. $5.3 \times 10^{-10}$ M

  2. $5.3 \times 10^{-8}$ M

  3. $5.3 \times 10^{-7}$ M

  4. $5.3 \times 10^{-11}$ M

Show answer
Correct Option: C
Explanation:
$(C)\ 5.3\times 10^{-7}m$

$CaF _2\rightleftharpoons Ca^{2+}+2F^-$

$K _{sp}=[Ca^{}2+][F^-]^2=S(S+0.1)^2=S\times 0.1^2=5.3\times 10^{-9}$

Note: $S<<0.1$ so, $S+ 0.1 \approx 0.1$ 

$\Rightarrow S=5.3\times 10^{-7}\ M$ 

The percentage of pyridine $\left( {{C _5}{H _5}N} \right)$ that forms pyridinum ion $\left( {{C _5}{H _5}{N^ + }H} \right)$ in a $0.10M$ aqueous pyridine solution $\left( Given - {{K _b}, for \  {C _5}{H _5}N = 1.7 \times {{10}^{ - 9}}} \right)$ is    

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

  1. $0.0060\% $

  2. $0.013\% $

  3. $0.77\% $

  4. $1.6\% $

Show answer
Correct Option: B