Tag: chemistry

Questions Related to chemistry

Determine $[{OH}^{-}]$ of a $0.050\ M$ solution of ammonia to which has been added sufficient ${NH} _{4}Cl$ to make the total $[{NH} _{4}^{+}]$ equal to $0.100 M$. $[{K} _{b({NH} _{3})}=1.8\times {10}^{-5}]$

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

  1. $[{OH}^{-}]=9.0\times {10}^{-6}$

  2. $[{OH}^{-}]=9.0\times {10}^{-8}$

  3. $[{OH}^{-}]=9.0\times {10}^{-2}$

  4. $[{OH}^{-}]=9.0\times {10}^{-9}$

Show answer
Correct Option: A
Explanation:

${NH} _{4}Cl\longrightarrow {NH} _{4}^{+}+{Cl}^{-}$
${NH} _{4}OH\longrightarrow {NH} _{4}^{+}+{OH}^{-}$
${K} _{b}=\cfrac { \left[ { NH } _{ 4 }^{ + } \right] \left[ OH \right]  }{ \left[ { NH } _{ 4 }OH \right]  } $
$[{NH} _{4}^{+}]=$ is due to salt because ${NH} _{4}OH$ ionise less amount due to common ions effect
$1.8\times {10}^{-5}=\cfrac{0.1\times [{OH}^{-}]}{0.05}$ 
$9\times {10}^{-6}=[{OH}^{-}]$

Why only ${As}^{+3}$ gets precipitated as ${As} _{2}{S} _{3}$ and not ${Zn}^{+2}$ as $ZnS$ when ${H} _{2}S$ is passed through an acidic solution containing ${As}^{+3}$ and ${Zn}^{+2}$?

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 ${As} _{2}{S} _{3}$ is less than that of $ZnS$

  2. Enough ${As}^{+3}$ are present in acidic medium

  3. Zinc salt does not ionise in acidic medium

  4. Solubility product changes in presence of an acid

Show answer
Correct Option: A
Explanation:

The species having minimum value of ${K} _{sp}$ will get precipitated first of all because ionic product will exceed the solubility product of such species.

${K} _{sp}$ of ${As} _{2}{S} _{3}$ is less than $ZnS$. In acid medium ionisation of ${H} _{2}S$ is suppressed (common ion effect) and ${K} _{sp}$ of $ZnS$ does not exceed.

The $pH$ of a dilute solution of acetic acid was found to be $4.3$ The addition of a small crystal of sodium acetate will cause $pH$ to:

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

  1. become less than $4.3$

  2. become more than $4.3$

  3. remain equal to $4.3$

  4. unpredictable

Show answer
Correct Option: B
Explanation:

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Due to this common ion effect, when we add sodium acetate dissociation of acetic acid decreases and solution will have less number of hydrogen ion and so, pH increases. (as $pH = -log [H^+]$)

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.