Tag: ionic equilibrium

Quizzes (3) Questions (39)

Questions Related to ionic equilibrium

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The dissociation constants of monobasic acids A,B,C and D are $6 \times 10^{4}, 5 \times 10^{5}, 3.6 \times 10^{6}\ and\ 7 \times 10^{10}$ respectively. The pH values of their 0.1 molar aqueous solutions are in the order:

  1. A < B < C < D

  2. A > B > C > D

  3. A = B = C = D

  4. A > B < C > D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the dissociation constant increases (A<B<C<D), dissociation of acid increases (A<B<C<D) and hence $[H^+]$ increases  (A<B<C<D) and hence pH value  decreases  (A>B>C>D). 

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The $K _a$ value for the acid $HA$ is $1.0 \times 10^{-6}$. What is the value of K for the  following reaction?
$A^+ + H _3O^+\rightleftharpoons HA + H _2O$

  1. $1.0 \times 10^{-8}$
  2. $1.0 \times 10^8$
  3. $1.0 \times 10^{-3}$
  4. $1.0 \times 10^6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$HA + H _2O\rightleftharpoons A^+ + H _3O^+$
$K _a=\dfrac{[A^-].[H _3O^+]}{[HA]}=10^{-6}$
As the given reaction is the reverse of above reaction $K$ will the reciprocal of the above reaction.
Therefore, $K=10^6$

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

$pK _a$ values of four acids are given below at $25^oC$. Indicate the strongest acid.

  1. 2.0

  2. 3.0

  3. 2.5

  4. 4.0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Ka$ represents the ionisation constant of an acid. It is a measure of the strength of the acid, provided the temperature remains constant.

    $pKa=-log(Ka)$

When $pKa=2$
           $2=-log(Ka)$
           $Ka=0.01$

When $pKa3$
           $3=-log(Ka)$
           $Ka=0.001$

Similarly, as $pKa $ values increase, $Ka$ value decrease and the acid becomes less strong. Hence smaller the $pKa$ value larger the $Ka$ value and hence stronger the acid.

So, the correct option is $A$

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Which of the following is/are ionization and not dissociation?

  1. $HCl \rightarrow H^+ + Cl^-$
  2. $Mg^+ \rightarrow Mg^{2+} + e^- $
  3. $Br _2 \rightarrow 2Br^.$
  4. $HBr \rightarrow H^+ + Br^-$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

In dissociation separation of ions take place which are already present as ions in the solid state. For example in the crystal structure of $NaCl$, $NaCl$ is present as $Na$ and $Cl$.

 In ionization, splitting of a neutral molecule into ions which are charged happen.
$HCl$ and $HBr$ are neutral $\mu $, there is no charge separation. But in solution they get ionised.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The process in which metal surface is made inactive is called:

  1. passivation

  2. galvanizing

  3. corrosion

  4. pickling
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Passivation involves creation of an outer layer of shield material that is applied as a microcoating, created by chemical reaction with the base material, or allowed to build from spontaneous oxidation in the air.

Galvanization is the process of applying a protective zinc coating to steel or iron, to prevent rusting

Corrosion is a natural process, which converts a refined metal to a more chemically-stable form, such as its oxide. 

Pickling is the process of preserving or expanding the lifespan of food by either anaerobic fermentation in brine or immersion in vinegar. 
Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The number of ions given by one molecule of $K {4}Fe(CN) _{6}$ after complete dissociation is___________.

  1. $5$
  2. $11$
  3. $2$
  4. $10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$K _4Fe(CN) _6\longrightarrow 4K^++[Fe(CN) _6]^{-4}$

Total number of ions on complete dissociation= $5$
As ions in co-ordination sphere will not dissociate so $Fe^{2+}$ and $CN^-$ ions will not dissociate into their respective ions.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Ionisation constant of each HA (weak acid) and BOH (weak base) are $3.0 x 10^{-7}$ each at 298K. The percentage degree of hydrolysis of BA at the dilution of 10L is :

  1. 25

  2. 50

  3. 75

  4. 40

  5. Data is insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\frac{h}{1-h}=\sqrt{KH}=\sqrt{\frac{kw}{K _{a}k _{b}}}=\sqrt{\frac{10^{-24}}{(3 \times 10^{-7})^{2}}}=\frac{1}{3}\Rightarrow h=0.25$

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The degree of dissociation of weak electrolyte increases by :

  1. increasing the concentration of weak electrolyte solution

  2. adding a common ion

  3. dissolution in solvent of low dielectric constant

  4. adding more amount of solvent into the solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When more amount of solvent is added to the solution, it helps in dissociating the molecules into ions of a weak electrolyte. Thus, degree of dissociation of weak electrolyte increases upon dilution.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Which of the following statements is not correct?

  1. Copper liberates hydrogen from acids

  2. In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine
  3. $Mn^{3+}$ and $Co^{3+}$ are oxidising agents in aqueous solution
  4. $Ti^{2+}$ and $Cr^{2+}$ are reducing agents in aqueous solution
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Copper has a positive standard reduction potential, meaning it cannot displace hydrogen from acids. Therefore, it does not liberate hydrogen gas from acids.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases
When sulphuric acid dissolves in water, the following reactions take place:

$H _2SO _4\longrightarrow H^+ +HSO _4^-$  ($100\%$ ionisation)

$H _2SO _4^-\longrightarrow H^+ +SO _4^{2-}$     ($10\%$ ionisation)

If $0.2 M$ aqueous solution of $H _2SO _4$ was taken, the concentration of $[SO _4^{2-}]$ will be:
  1. $0.1$ M
  2. $0.01$ M
  3. $0.2$ M
  4. $0.02$ M
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$Step \: I: \: H _2SO _4\longrightarrow H^+ +HSO _4^-$;      $100$% ionisation

$Step \: II: \: HSO _4^-\longrightarrow H^+ +SO _4^{2-}$;     $10$% ionisation

     $H^++H _2O\longrightarrow H _3O^+$

$SO _4^{2-} \: is \: from \: step \: II$.

$Step \: I:  \: 100$% and hence, $[H^+]= 0.2 \: M$.

and  $[HSO _4^-]=0.2 \: M$.

$Step \: II :\:  \: 10$% and hence, $[H^+]=0.2 \times  0.1= 0.02 \:M$

and  $[SO _4^{2-}]= 0.02 \:M$.

Total $[H _3O^+]=0.2 + 0.02 =0.22 \:M$, 

$[SO _4^{2-}]= 0.02 \:M$ and $[HSO _4^-] =0.20-0.02 =0.18 \:M$.

Hence, the correct option is $(D)$