Tag: further aspects of equilibria

Questions Related to further aspects of equilibria

$Ca{SO} _{4}$ is somewhat soluble in water.
$I$. When ${H} _{2}{SO} _{4}$ is added to a solution of $Ca{SO} _{4}$, the solubility of the $Ca{SO} _{4}$ will be increased.
$II$. The addition of ${H} _{2}{SO} _{4}$ will lower the $pH$ of the solution.

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

  1. Statement $I$ is true, Statement $II$ is true

  2. Statement $I$ is true, Statement $II$ is false

  3. Statement $I$ is false, Statement $II$ is true

  4. Statement $I$ is false, Statement $II$ is false

Show answer
Correct Option: C
Explanation:

When $H _2SO _4$ is added to a $CaSO _4$ solution, due to common ion effect, the solubility of $CaSO _4$ decreases. Further, as the acid concentration increases, the $pH$ of the solution lowers due to addition of $H _2SO _4$.

Hence, statement $I$ is false, statement $II$ is true.

What is the 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. When the addition of an ion common to two solutes causes precipitation or reduces ionization.

  2. When a molecule is added to a solution and causes precipitation or reduces ionization

  3. When the subtraction of an ion common to two solutes causes precipitation or reduces ionization.

  4. The effect of adding ions to a solution.

  5. When any ion is added to a solution and causes precipitation or reduces ionization.

Show answer
Correct Option: A
Explanation:

The common ion effect is observed when the addition of an ion common to two solutes causes precipitation or reduces ionization.
Thus, if to a solution of a weak electrolyte, a solution of strong electrolyte is added which furnishes an ion common to that furnished by the weak electrolyte, the ionization of the weak electrolyte is suppressed.

A monoprotic acid in $1.00M$ solution is $0.001$% ionised. The dissociation constant of acid is:

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

  1. ${ \alpha }^{ 2 }C+\alpha K-K=0$

  2. ${ \alpha }^{ 2 }C-\alpha K-K=0$

  3. ${ \alpha }^{ 2 }C-\alpha K+K=0$

  4. ${ \alpha }^{ 2 }C+\alpha K+K=0$

Show answer
Correct Option: A
Explanation:

According to Ostwald's dilution law


$K=\cfrac { { \alpha  }^{ 2 }C }{ \left( 1-\alpha  \right)  } $

On solving, we get


$\alpha^2C+\alpha K-K=0$


Hence, the correct option is $\text{A}$

The ionisation constant of acetic acid is $1.8\times { 10 }^{ -5 }$.The concentration at which it will be dissociated to $2$% is:

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

  1. $1M$

  2. $0.045M$

  3. $0.018M$

  4. $0.45M$

Show answer
Correct Option: B
Explanation:

When acetic acid is dissolved in water, it partially dissociates (2%).

Ionisation constant $ K _a=1.87\times 10^{-5}$
$K _a=\cfrac{C\alpha \times C\alpha}{C(1-\alpha)}$

We assume $\alpha\sim  0\Rightarrow 1-\alpha\sim 1$
$\Rightarrow 1.87\times 10^{-5}=\cfrac{C\alpha^2}{1}={C\times 0.02\times 0.02}$
$\Rightarrow C = \cfrac{1.87\times 10^{-5}}{0.02^2}$
$=0.0467M$
Correct answer is option-B.

Which pair will show 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. $Ba{ Cl } _{ 2 }+Ba{ \left( { NO } _{ 3 } \right) } _{ 2 }$

  2. $NaCl+HCl$

  3. ${ NH } _{ 4 }OH+{ NH } _{ 4 }Cl$

  4. $AgCN+KCN$

Show answer
Correct Option: C
Explanation:

Those ions will show common ion effect in which one or both them does not dissociate completely.


$NaCl$ and $HCl$ dissociate completely as one of them is acid and other is salt of strong acid and strong base.


$AgCN$ and $KCN$ dissociate completely as one of them is acid and other is salt of strong acid and strong base.

NH4OH is weak base to which NH4Cl is added which provides common $NH _4^{+}$ion dissociation of NH4OH is suppressed.

The dissociation constants of two acids $ H{ A } _{ 1 }$ and $H{ A } _{ 2 }$ are $3.0\times { 10 }^{ -4 }$ and $1.8\times { 10 }^{ -5 }$ respectively. The relative strengths of the acids will be:

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

  1. $1:4$

  2. $4:1$

  3. $1:16$

  4. $16:1$

Show answer
Correct Option: B
Explanation:
The dissociation constants of $HA _1$ and $HA _2$ are $3\times10^{-4} $ and $1.8\times10^{-5}.$

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:
$\dfrac { { (Acidic\ Strength }) _{ HA _1 } }{ { (Acidic\ Strength }) _{ HA _2} } =\dfrac { \sqrt { { (Dissociation\ Constant }) _{ HA _1} }  }{ \sqrt { { (Dissociation\ Constant }) _{ HA _2 } }  } $

 Relative Acidic Strength= $\dfrac { { (Acidic Strength }) _{ HA _1 } }{ { (Acidic Strength }) _{ HA _2 } } =\dfrac { \sqrt { 3.0\times { 10 }^{ -4 } }  }{ \sqrt { 1.8\times { 10 }^{ -5 } }  } =4.08=4(approx)$

relative strengths of acids will be $4:1.$

Which of the following pairs will show 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. Barium chloride + barium sulphate

  2. Silver cyanide + potassium nitrite

  3. Ammonium hydroxide + ammonium chloride

  4. Sodium chloride + hydrogen chloride

Show answer
Correct Option: C

When strong base $(NaOH)$ is added to the weak acid (acetic acid, ${ CH } _{ 3 }COOH$), then dissociation of acetic acid increases; this effect is known as:

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

  1. Common ion effect

  2. Reverse ion effect

  3. Saltation effect

  4. Solubility effect

Show answer
Correct Option: B
Explanation:

$CH _3COOH\rightleftharpoons CH _3COO^{-}+H^+$
When $NaOH$ is added $H^+$ combines with $OH^{-}$ so concentration of $H^+$ decreases so equilibrium will shift towards right.
${ CH } _{ 3 }COOH+NaOH\longrightarrow { CH } _{ 3 }COONa+{ H } _{ 2 }O\quad $
Ionization of acetic acid will increase with the progress of its neutralization. This effect is called reverse ion effect.

In the third group of qualitative analysis, the precipitating reagent is $NH {4}Cl + NH _{4} OH$. The function of $NH _{4}Cl$ is to______

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

  1. increase the ionization of $NH _{4} OH$

  2. suppress the ionization of $NH _{4} OH$

  3. stabilise the hydroxides of group cations

  4. convert the ions of group third into their respective chlorides

Show answer
Correct Option: B
Explanation:

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion common to that provided by weak electrolyte.


The NH4OH is weak base it does not ionises completely. Thus due to presence of common ion NH4+ in NH4Cl, it supresses the ionisation of weak base NH4OH in order to decrease the OH- concentration so that higher group cations will not get precipitated.

Thus the pair $NH _{4} OH + NH _{4} Cl$ shows common ion effect. 

Ammonium chloride suppresses the ionization of ammonium hydroxide.

Option B is correct.

The solubility of ${ A } _{ 2 }{ X } _{ 5 }$ is $x\ mol\ { dm }^{ -3 }$. Its solubility product is:

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

  1. $36{ x }^{ 6 }$

  2. $64\times { 10 }^{ 4 }{ x }^{ 7 }$

  3. $126{ x }^{ 7 }$

  4. $1.25\times { 10 }^{ 4 }{ x }^{ 7 }$

Show answer
Correct Option: D
Explanation:
Let the solubility be $S.$
${ As } _{ 2 }X _{ 3 }\leftrightharpoons 2As^{ 5+ }+5X^{ 2- }\\ \quad \quad \quad   \quad 2S\quad \quad \quad 5S$
Solubility product is 
$K _{ sp }=[As^{ 5+ }]^{ 2 }\times [S^{ 2- }]^{ 5 }\\$ 
Let $[{ As } _{ 2 }S _{ 5 }]=S,[As^{ 5+ }]=2S, [S^{ 2- }]=5S\\ Ksp=(2S)^{ 2 }\times (5S)^{ 5 }=4S^{ 2 }\times 3125S^{ 5 }=12500S^{ 7 }$