Tag: ionisation of weak acids and weak bases

Questions Related to ionisation of weak acids and weak bases

Addition of $HCl$ will not suppress the ionization of:

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

  1. acetic acid

  2. benzoic acid

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

  4. sulphuric acid

Show answer
Correct Option: D
Explanation:

Any acid weaker than $HCl$ will be suppressed by $HCl$. Hence, among the given options, only sulphuric acid is an acid with comparable acidity strength to $HCl$. The same can also be verified using $K _a$ values from the data.

State whether the given statement is true or false:

$AgCl$ is less soluble in aqueous sodium chloride solution than in pure water.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$NaCl \rightarrow Na^+ + Cl^-$

Sodium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of AgCl. Hence, the solubility of AgCl decreases.

The solubility of $Ca{ F } _{ 2 }  \left( { K } _{ sp } = 3.4 \times { 10 }^{ -11 } \right)$ in $0.1  M$ solution of $NaF$ would be

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

  1. $3.4\times { 10 }^{ -12 }M$

  2. $3.4\times { 10 }^{ -10 }M$

  3. $3.4\times { 10 }^{ -9 }M$

  4. $3.4\times { 10 }^{ -13 }M$

Show answer
Correct Option: C
Explanation:
$CaF _2 \rightarrow Ca^{+2} + 2F^- $
                    $S$        $0.1$
$CaF _2: { K } _{ sp } = 3.4 \times { 10 }^{ -11 }   = [Ca^{+2}][F^-]^2$
$S = \displaystyle \frac {3.4 \times { 10 }^{ -11 }}{(0.1)^2} = 3.4 \times { 10 }^{ -9 }$
Assertion: AgCl is less soluble in aqueous sodium chloride solution than in pure water.
Reason: AgCl dissociates completely and more rapidly than NaCl.
chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false

Show answer
Correct Option: C
Explanation:

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$NaCl \rightarrow Na^+ + Cl^-$
Sodium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of $AgCl$. Hence, the solubility of $AgCl$ decreases.

When solid $KCl$ is added to a saturated solution of $AgCI$ in $H _2O$

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

  1. Nothing happens

  2. Solubility of $AgCl$ decreases

  3. Solubility of $AgCl$ increases

  4. Solubility product of $AgCl$ increases

Show answer
Correct Option: B
Explanation:

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$KCl \rightarrow K^+ + Cl^-$

Potassium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of $AgCl$. Hence, the solubility of $AgCl$ decreases.

In a $0.1\ M$ solution of $H _{3}PO _{4}$ the ionic species present are (besides $H _{3}O^+$ and $OH^{-}$):

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

  1. ${H _{2}PO _{4}}^{-}, {HPO _{4}}^{2-}, {PO _{3}}^{3-}$

  2. ${H _{2}PO _{2}}^{3-}, {H _{2}PO _{4}}^{-}, {PO _{4}}^{3-}$

  3. ${H _{2}PO _{4}}^{-}, {HPO _{3}}^{2-}, {PO _{2}}^{3-}$

  4. ${PO _{3}}^{4-}, {PO _{4}}^{3-}, {HPO _{4}}^{2-}, {H _{2}PO _{4}}^{-}$

Show answer
Correct Option: A
Explanation:

$H _3PO _4$$=$$H^+$$+$$H _2PO _4^-$

$H _2PO _4^-$$=$$H^+$$+$$HPO _4^{-2}$
$HPO _4^{-2}$$=$$H^+$$+$$PO _4^{-3}$
Thus $H _3PO _4$ will dissociate into the ions $H _2PO _4^-$, $HPO _4^{-2}$, $PO _4^{-3}$.
Option A is the correct answer.

The solubility of silver chloride ___________ in the presence of sodium chloride because of __________.

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

  1. increases; common ion effect

  2. increases; aldol condensation

  3. decreases; common ion effect

  4. decreases; aldol condensation

Show answer
Correct Option: C
Explanation:

The solubility of silver chloride decreases in the presence of sodium chloride because of common ion effect.

$\displaystyle AgCl \rightleftharpoons  Ag^+ + Cl^-$

$\displaystyle NaCl \rightarrow Na^+ + Cl^-$

Sodium chloride is a strong electrolyte and completely dissociates to provide chloride ions that are common ions. The chloride ions shift the equilibrium for dissociation of $AgCl$ towards left.

How both the acids $\displaystyle HCl$ and $\displaystyle { HC } _{ 2 }{ H } _{ 3 }{ O } _{ 2 }$ are different?

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

  1. The first has less hydrogen in solution

  2. The second is highly ionized

  3. The second has more ionized hydrogen

  4. The first is highly ionized

Show answer
Correct Option: D
Explanation:

HCl undergoes complete ionization as it is a strong acid.
On the other hand, acetic acid  $HC _2H _3O _2$ is a weak acid and is ionized to a small extent.

What is the net effect of the common ion?

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

  1. It reduces the solubility of the solute in the solution.

  2. It reduces the increases of the solute in the solution.

  3. It causes more product to be made.

  4. It causes more ions to form.

  5. There is no net effect overall.

Show answer
Correct Option: A
Explanation:

The net effect of the common ion is that it reduces the solubility of the solute in the solution.
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 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.
For example, ammonium hydroxide is a weak electrolyte and ionizes to a small extent to give ammonium ions and hydroxide ions.
$\displaystyle  NH _4OH \rightleftharpoons NH _4^+  +  OH^-$
A strong electrolyte NaOH is added which furnishes hydroxide ions that are common ions. This suppresses the equilibrium for the dissociation of ammonium hydroxide.
$\displaystyle  NaOH \rightarrow Na^+  + OH^-$

$Ca{SO} _{4}$ 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.