Tag: further aspects of equilibria

Questions Related to further aspects of equilibria

Purification of $NaCl$ by the passage of $HCl$ through brine solution is based on which of the following?

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

  1. Distribution coefficient

  2. Le Chatelier’s principle

  3. Displacement Law

  4. Common ion effect

Show answer
Correct Option: D
Explanation:

Purification of $NaCl$ by the passage of $HCl$ through brine is based on the common ion effect.
$HCl$ is a strong electrolyte and provides an ion $(Cl^-)$ that is common to that provided by the weak electrolyte. Thus, the ionization of weak electrolytes is suppressed.

What is the $[OH^-]$ in the final solution prepared by mixing $20.0\ mL$ of $0.050\ M$ $HCl$ with $30.0\; mL$ of $0.10 \;M\; Ba(OH) _2$?

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

  1. $0.12\ M$

  2. $0.10\ M$

  3. $0.40\ M$

  4. $0.0050\ M$

Show answer
Correct Option: B
Explanation:
$Ba(OH) _2 + 2HCl \rightarrow BaCl _2 + 2H _2O$

2 m mol of HCl neutralize 1 m mole of $Ba(OH) _2$

1 m mol of HCl neutralize 0.5 m mol of $Ba(OH) _2$

$Ba(OH) _2$ left = 3 - 0.5 m mol = 2.5 m mol

         $[Ba(OH) _2] = \frac{2.5}{50}\;M = 0.05\; M$

or      $[OH]^- = 2 \times 0.05  = 0.1\; M$

The solubility product constant $Ksp$ of $Mg(OH) _{2}$ is $9.0\times 10^{-12}$. If a solution is $0.010\ M$ with respect to $Mg^{2+}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH) _{2}$?

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

  1. $1.5\times 10^{-7}M$

  2. $3.0\times 10^{-7}M$

  3. $1.5\times 10^{-5}M$

  4. $3.0\times 10^{-5}M$

Show answer
Correct Option: D
Explanation:

$Ksp(Mg(OH) _2)=9.0\times 10^{-12}$

$(Mg(OH) _2 \leftrightharpoons Mg^{2+}+2[OH]^-$
$Ksp=[Mg^{2+}][OH^-]^2$
$9\times 10^{-12}=(10^{-12})(OH^-)^2$
$[OH^-]^2=3^2\times (10^{-5})^2$
$[OH^-]=3.0\times 10^{-5}M$
Maximum Hydroxide-ion concentration.

 8 gram of NaOH is mixed with 9.8 gram of $H _{2}SO _{4}$, the pH of the solution is:

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

  1. more than 7

  2. 7

  3. less than 7

  4. cant be said

Show answer
Correct Option: B
Explanation:

8 gram (0.2 mole) of NaOH (molecular weight 40 g/mol) completely neutralizes 9.8 gram (0.1 mole) of $H _2SO _4$ (molecular weight 98 g/mol).
Since the molar concentration of both the compound are approximately same, the resulting solution will be neutral. Its pH will be 7.

The $K _{sp}$ for AgCl is $2.8\times 10^{-10}$ at a given temperature. The solubility of AgCl in 0.01 molar HCl solution at this temperature 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. $2.8\times 10^{-12}mol L^{-1}$

  2. $2.8\times 10^{-8}mol L^{-1}$

  3. $5.6\times 10^{-8}mol L^{-1}$

  4. $2.8\times 10^{-4}mol L^{-1}$

Show answer
Correct Option: B
Explanation:

The chloride ion concentration in 0.01M HCl will be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
$K _{sp}=[Ag^+][Cl^-]$
Substitute values in the above expression.
$2.8 \times 10^{-10}=[Ag^+] \times 0.01$
Hence, $[Ag^+]= \frac {2.8 \times 10^{-10}} {0.01}=2.8 \times 10^{-8}$mol/L.

Solubility of AgCl will be minimum in ___________.

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

  1. 0.01 M $Na _{2} SO _{4}$

  2. 0.01 M $Ca Cl _{2}$

  3. Pure water

  4. 0.001 M $Ag NO _{3}$

Show answer
Correct Option: B
Explanation:
The solubility of insoluble substances can be decreased by the presence of a common ion. 

Present in silver chloride are silver ions $(Ag^+)$ and chloride ions $(Cl^-)$. $AgCl$ is not soluble in water. 

Silver nitrate (which is soluble) has silver ion in common with silver chloride. But the concentration of the common ion is low (0.001M)

Calcium chloride (also soluble) has chloride ion in common with silver chloride. In $CaCl _2$ has two moles of common ion (0.02M) that decreases solubility very rapidly.  

In $Na _2SO _4$ there is no common ion effect apply therefore is soluble in it. 

Solid $Ba{({NO} _{3})} _{2}$ is gradually dissoved in a $1\times {10}^{-4}M$ ${Na} _{2}{CO} _{3}$ solution. At what minimum conc. of ${Ba}^{-2}$ will a precipitate of $Ba{CO} _{3}$ begin to form? (${K} _{sp}$ for $Ba{CO} _{3}=5.1\times {10}^{-9}$)

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

  1. $4.1\times {10}^{-5}M$

  2. $8.1\times {10}^{-7}M$

  3. $5.1\times {10}^{-5}M$

  4. $8.1\times {10}^{-8}M$

Show answer
Correct Option: A

If $S _0, S _1, S _2$ and $S _3$ are the solubilities in water of $AgCl$, $0.01 \,M \,CaCl _2, 0.01 \,M \,NaCl$ and $0.5 \,M \,AgNO _3$ solutions, respectively, then which of the following is true?

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

  1. $S _0 > S _2 > S _1 > S _3$

  2. $S _0 = S _2 = S _1 > S _3$

  3. $S _3 > S _1 > S _2 > S _0$

  4. none of these

Show answer
Correct Option: A
Explanation:

The solubility of $AgCl$ or its ion formation will depend inversely on the concentration

$S _0=H _2O$
$S _1=0.01\ M\ CaCl _2$
$S _2=0.01\ M\ NaCl$
$S _3=0.05\ M\ AgNO _3$
$H _2O$ is dilute solution and least concentrated hence have maximum solubility of $AgCl$ in it. Out of $NaCl$ and $CaCl _2$, the solubility of $NaCl$ is high due to loess number of ions produced from $NaCl$
as compare to $CaCl _2$. More the ion produced lesser is the solubility of coming salt.
So, the Correct order is $S _0 > S _2 > S _1 > S _3$
We know that concentration of common ion $\alpha \dfrac{1}{solubility}$. The order of solubility of $AgCl : S _1 > S _3 > S _2 > S _4$

$Ag _3 PO _4$ would be least soluble at 25$^o$C in

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

  1. 0.1 M $AgNO _3$

  2. 0.1 M $HNO _3$

  3. pure water

  4. 0.1 M $Na _3PO _4$

  5. solubility in (a), (b), (c) or (d) is not different

Show answer
Correct Option: A
Explanation:

$Ag _3 PO _4$ is a weak electrolyte and $AgNO _3$ is a strong electrolyte containing common ion $(NO _3^-ion)$. Thus common ion effect is observed and the solubility of $Ag _3 PO _4$ is suppressed.
Hence, $Ag _3 PO _4$ is least soluble in 0.1M $AgNO _3$.

The solubility of $AgI$ in $NaI$ solution is less than that in pure water because

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

  1. $AgI$ forms complex with $NaI$

  2. Of common ion effect

  3. Solubility product of $AgI$ is less than that of $NaI$

  4. The temperature of the solution decreases

Show answer
Correct Option: B
Explanation:

The common ion presence in the solution decrease the solubility of a given sparingly soluble compound. So, solubility of AgI in NaI solution is less.