Tag: chemistry

Questions Related to chemistry

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 $[{ As } _{ 2 }S _{ 5 }]=S,[As^{ 5+ }]=2S, [S^{ 2- }]=5S\\ Ksp=(2S)^{ 2 }\times (5S)^{ 5 }=4S^{ 2 }\times 3125S^{ 5 }=12500S^{ 7 }$

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.

The degree of dissociation of $PCl _{5(g)}$ at 16.8 bar and $127^{0}C$ is 0.4. The value of $K _{P}$ for the reaction is:
$PCl _{5} \leftrightharpoons PCl _{3(g)} +Cl _{2(g)}$ 

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

  1. $3.2 bar$

  2. $3.2 bar^{-1}$

  3. $12.8\ bar$

  4. $ 0.4$ x $16.8\ bar$

Show answer
Correct Option: A
Explanation:

total pressure P total = $16.8 \, bar$

degree of dissociation $\alpha = 0.4$
$PCl _5 \rightleftharpoons p Cl _3 + Cl _2$
$P _0 (1 - \alpha) \,\,\, P _0 \alpha \,\,\, P _0 \alpha$
$P \, total = P _0(1 - \alpha) + P _0 \alpha + P _0 \alpha$
$= P _0 (1 + \alpha)$
$P _0 (1 + \alpha) = 16.8$
$P _0 \times 1.4 = 16.8 \Rightarrow P _0 = \dfrac{16.8}{1.4} = 12 $ bar
$Kp = \dfrac{[PCl _3][Cl _2]}{[PCl _5]} = \dfrac{P _0 \alpha \times P _0 \alpha}{P _0 (1 - \alpha)}$
$= \dfrac{P _0 \alpha^2}{1 - \alpha}$
$= 12 \times \dfrac{0.4 \times 0.4}{0.6}$
$= 3.2 \, bar$

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