Tag: further aspects of equilibria

Quizzes (4) Questions (91)

Questions Related to further aspects of equilibria

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

If concentration of two acids are some, their relative strengths can be compared by:

  1. ${ \alpha } _{ 1 }/{ \alpha } _{ 2 }$
  2. $K _{ 1 }/K _{ 2 }$
  3. ${ \left[ { H }^{ + } \right] } _{ 1 }/{ \left[ { H }^{ + } \right] } _{ 2 }$
  4. $\sqrt { K _{ 1 }/K _{ 2 } } $
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Relative strength of two acids can be compared by their degree of dissociation.

$HA\rightleftharpoons H^++A^-$
$C$
$C-C\alpha$   $C\alpha$    $C\alpha$
If concentration of two acids are same so their relative strength can be compared by their $[H^{+}]$ concentration.
$K _a=C\alpha^2$
$\alpha=(K _a/C)^{0.5}$
If concentration of two acids are same so their relative strength can be compared by square root of their dissociation constants.

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

The solubility of $AgCl$ in $NaCl$ solution is less than that in pure water, because of the  ________.

  1. solubility product of $AgCl$ is less than of $NaCl$

  2. common ion effect
  3. both $A$ and $B$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, $NaCl$ is soluble to a very significant extent, when $AgCl$ is added to $NaCl$ solution, the common ion $[Cl^-]$ increases in the solution. To have the solubility product or $K _{sp}$ of $AgCl$ constant, $[Ag^+]$ will decrease or $AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of $AgCl$ in $NaCl$ solution will be less than that in pure water.

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

100 mL of 20.8% $BaCl _2$ solution and 50 mL of 9.8% $H _2SO _4$ solution will form $BaSO _4$
$(Ba=137, Cl=35.5, S=32, H=1, O=16)$
$BaCl _2+H _2SO _4\rightarrow BaSO _4+2HCl$

  1. 23.3 g

  2. 11.65 g

  3. 30.6 g

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$100ml$ of $20.8$% $BaCl _2$ solution= $20.8g$  $BaCl _2$

$50ml$ of $9.8$% $H _2SO _4$ solution= $4.9g$  $H _2SO _4$
Reaction: $BaCl _2+H _2SO _4\longrightarrow BaSO _4\downarrow +2HCl$
           $208 g mol^{-1}$   $98 g mol^{-1}$      $233 g mol^{-1}$
$\therefore 98g$ $H _2SO _4$ reacts with $208g$ $BaCl _2$
$4.9g$ $H _2SO _4$ reacts with $\cfrac {208}{98}\times 4.9=10.4 g$ $BaCl _2$
$98g$ $H _2SO _4$ will produce $233g$ $BaSO _4$
$\therefore 4.9g$ $H _2SO _4$ will produce= $\cfrac {233}{98}\times 4.9=11.65g$ $BaSO _4$

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

The addition of NaCl to AgCl decreases the solubility of AgCl because ________.

  1. Solubility product decreases

  2. Solubility product remains constant.

  3. solution becomes unsaturated

  4. solution becomes super saturated.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

NaCl is highly soluble and when it is added to AgCl it decreases the solubility of AgCl because of common ion $Cl^-$ and solution become super saturated.

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

Which of the following solutions display common ion effect?

  1. $HCl+NaCl$
  2. ${ H } _{ 2 }S + KH$
  3. $NaHS+HCl$
  4. ${ H } _{ 2 }{ SO } _{ 4 }+{ K } _{ 2 }S$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In common ion effect there should be common ions in the two or more chemicals. Here, $HCl+NaCl$ shows common ion effect. Because $Cl^-$ is common in both.

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

What is $[{ NH } _{ 4 }^{ + }]$ in a solution containing 0.02M ${ NH } _{ 3 }$ (${ K } _{ b }={ 1.8\times 10 }^{ -5 }$) and 0.01M KOH?



  1. ${ 1.8\times 10 }^{ -5 }$
  2. ${ 9\times 10 }^{ -6 }$
  3. ${ 3.6\times 10 }^{ -5 }$

  4. NONE OF THE ABOVE
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a solution of NH3 (weak base) and KOH (strong base), the concentration of OH- is dominated by the strong base (0.01M). Using the Kb expression, Kb = [NH4+][OH-] / [NH3], we get 1.8e-5 = NH4+ / 0.02. Solving for [NH4+] gives 3.6e-5 M.

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

$As _2S _3$ solution has negative charge, capacity to precipitate is highest in:

  1. $AlCl _3$
  2. $Na _3PO _4$
  3. $CaCl _2$
  4. $K _2SO _4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:- (A) $Al{Cl} _{3}$

According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence ${As} _{2}{S} _{3}$ precipitate the most in $Al{Cl} _{3}$.

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

At $100^\circ C$, value of $K _{w}$ is 

  1. $1.0\times 10^{-14}\quad m^{2}$
  2. less than $1.0\times 10^{-14}\quad m^{2}$
  3. greater than $1.0\times 10^{-14}\quad m^{2}$

  4. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At higher temperature the value of $kw$ increases.This is in according with le-chatelier principle.

At $100^o kw=51.3\times 106{-14}$
C is the correct answer.

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

The solubility of CaF$ _2$ (K$ _{sp} = 5.3\times 10^{-9}$) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

  1. $5.3 \times 10^{-10}$ M
  2. $5.3 \times 10^{-8}$ M
  3. $5.3 \times 10^{-7}$ M
  4. $5.3 \times 10^{-11}$ M
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(C)\ 5.3\times 10^{-7}m$

$CaF _2\rightleftharpoons Ca^{2+}+2F^-$

$K _{sp}=[Ca^{}2+][F^-]^2=S(S+0.1)^2=S\times 0.1^2=5.3\times 10^{-9}$

Note: $S<<0.1$ so, $S+ 0.1 \approx 0.1$ 

$\Rightarrow S=5.3\times 10^{-7}\ M$ 

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

The percentage of pyridine $\left( {{C _5}{H _5}N} \right)$ that forms pyridinum ion $\left( {{C _5}{H _5}{N^ + }H} \right)$ in a $0.10M$ aqueous pyridine solution $\left( Given - {{K _b}, for \  {C _5}{H _5}N = 1.7 \times {{10}^{ - 9}}} \right)$ is    

  1. $0.0060\% $
  2. $0.013\% $
  3. $0.77\% $
  4. $1.6\% $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a weak base B, Kb = C * alpha^2 / (1 - alpha). Since alpha is very small, Kb = C * alpha^2. Here, 1.7e-9 = 0.1 * alpha^2, so alpha^2 = 1.7e-8, and alpha = 1.3e-4. The percentage is 1.3e-4 * 100 = 0.013%.