Tag: ionisation of weak acids and weak bases

Questions Related to ionisation of weak acids and weak bases

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

  1. $0.12\ M$

  2. $0.10\ M$

  3. $0.40\ M$

  4. $0.0050\ M$


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 $Mg(OH) _{2}$?

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

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

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

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


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:

  1. more than 7

  2. 7

  3. less than 7

  4. cant be said


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 :

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


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:
$PCl _{5} \leftrightharpoons PCl _{3(g)} +Cl _{2(g)}$ 

  1. $3.2 bar$

  2. $3.2 bar^{-1}$

  3. $12.8\ bar$

  4. $ 0.4$ x $16.8\ bar$


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 ___________.

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

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

  3. Pure water

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


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 $Ba{CO} _{3}=5.1\times {10}^{-9}$)

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

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

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

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


Correct Option: A

The ionisation constant of an acid, $K _a$ is the measure of strength of an acid. The $K _a$ values of acetic acid, hypochlorous acid and formic acid are $1.74 \times 10^{-5}, 3.0 \times 10^{-8}$ and $1.8 \times 10^{-4}$ respectively. Which of the following orders of pH of $0.1$mol $dm^{-3}$ solutions of these acids is correct?

  1. Acetic acid > Hypochlorous acid > Formic acid

  2. Hypochlorous acid > Acetic acid > Formic acid

  3. Formic acid > Hypochlorous acid > Acetic acid

  4. Formic acid > Acetic acid > Hypochlorous acid


Correct Option: B
Explanation:

$K _a$ is a measure of the strength of the acid i.e., larger the value of $K _a$, the stronger is the acid.
Thus, the correct order of acidic strength is
$HCOOH > CH _3COOH > HClO$
Stronger the acid, lesser will be the value of pH. Hence, the correct order of pH is $HClO > CH _3COOH > HCOOH$.

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?

  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


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 $AgCl : S _1 > S _3 > S _2 > S _4$

The $[H^+]$ of a resulting solution that is $0.01\space M$ acetic acid $(K _a = 1.8\times10^{-5})$ and $0.01\space M$ in benzoic acid $(K _a = 6.3\times10^{-5})$:

  1. $9\times10^{-4}$

  2. $81\times10^{-4}$

  3. $9\times10^{-5}$

  4. $2.8\times10^{-3}$


Correct Option: A