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

Assertion:
Solubility of $AgCl$ in water decreases if $NaCl$ is added to it.
Reason:
$NaCl$ is soluble freely in water but $AgCl$ is sparingly soluble.

  1. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion

  3. Assertion is correct but Reason is not correct

  4. Assertion is not correct but Reason is correct

  5. Both Assertion and Reason are not correct
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$NaCl$ is a strong electrolyte. It completely dissociates to form ions. Hence, it is highly soluble in water.
$AgCl$ dissociates to little extent. Hence, it is sparingly soluble in water.
When $NaCl$ is added to a solution of $AgCl$, due to common ion effect (chloride ion is the common ion), the dissociation of $AgCl$ is suppressed. Also as the concentration of chloride ion (from $NaCl$) increases, the ionic product of $AgCl$ exceeds its solubility product and precipitation occurs.

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

(A) pH of $10^{-7}$ M NaOH solution exists between 7 to 7.3 at $25^o C$.
(R) Due to common ion effect ionization of water is suppressed. 

  1. Both (R) and (A) are true and reason is the correct explanation of assertion

  2. Both (R) and (A) are true but reason is not correct explanation of assertion

  3. Assertion (A) is true but reason (R) is false

  4. Assertion (A) and reason (R) both are false

  5. Assertion (A) is false but reason (R) is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A) $pH$ of ${ 10 }^{ -7 }$ $M$ $NaOH$ will be slightly greater than $7$. It is between $7$ to $7.3$ at ${ 25 }^{ 0 }C$ due to presence of very small quantity of excess ${ OH }^{ - }$.
R) Due to common ion effect ionisation of water is suppressed.
     ${ H } _{ 2 }O\rightleftharpoons { H }^{ + }+{ OH }^{ - }$
So in presence of common ion ${ H }^{ + }$ or ${ OH }^{ - }$ the ionisation of ${ H } _{ 2 }O$ will be decreased.
So both $(R)$ and $(A)$ are true and reason is the correct explanation of assertion.
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

By adding which of the following in 1 L 0.1 M solution of HA, $(Ka=10^{-5})$, the degree of dissociation of HA decreases appreciably?

  1. ${10^{-3} M \: HCl, 1\; L}$
  2. ${0.5 M \: HX (Ka=2\times 10^{-6}), 1 \: L}$
  3. ${0.1 M \: HNO _{3}, 1\: L}$

  4. All of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$HA,  \alpha =\sqrt{\displaystyle\frac{10^{-5}}{0.1}}=0.01$ 
(A) On adding $HCl, \left [ HA \right ]=\frac{0.1}{2},\left [ HCl \right ]=\displaystyle\frac{10^{-3}}{2}$
$HCl \rightarrow H^{+}+Cl^{-}$
$HA \rightleftharpoons H^{+}  +A^{-}$
$\displaystyle\frac{0.1}{2}\left ( 1- \alpha  \right )\left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) \displaystyle\frac{0.1 \alpha }{2} $
$\Rightarrow 10^{-5}=\displaystyle\frac{\displaystyle\frac{0.1\alpha }{2} \times \left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) }{\displaystyle\frac{0.1}{2}\left ( 1-\alpha  \right )}$
Neglecting $\alpha $
$10^{-5}=\alpha \times \left ( \displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1\alpha  }{2} \right )$
$\Rightarrow 0.1 \alpha ^{2}+10^{-3} \alpha -2\times 10^{-5}=0$
$\Rightarrow  \alpha ^{2}+10^{-2}\alpha -2\times 10^{-4}=0$
$\Rightarrow \alpha =\displaystyle\frac{-10^{-2}+\sqrt{10^{-4}+8\times 10^{-4}}}{2}=\displaystyle\frac{2\times 10^{-2}}
{2}=2$
Which is similiar to initial.
(B) In case of two weak acids
$\left [\mathrm H^{+}  \right ]=\sqrt{10^{-5}\times \frac{0.1}{2}+2\times 10^{-6}\times \displaystyle\frac{0.5}{2}}$
$=\sqrt{\displaystyle\frac{10^{-6}}{2}+\displaystyle\frac{10^{-6}}{2}}=10^{-3}$
$\alpha =\displaystyle\frac{10^{-5}}{10^{-3}}=10^{-2}$
which is same as earlier
(C) $HNO _{3}\rightarrow H^{+}+NO _{3}^{-}$
$\displaystyle\frac{0.1}{2}       \displaystyle\frac{0.1M}{2}$
$HA \rightleftharpoons H^{+}   +  A^{-}$
$\displaystyle\frac{0.1M}{2}\left ( 1-\alpha  \right )  \displaystyle\frac{0.1}{2}+\displaystyle\frac{0.1}{2}\alpha                
\displaystyle\frac{0.1}{2}\alpha$
$\Rightarrow 10^{-5}=\displaystyle\frac{\frac{0.1\alpha }{2}\times \displaystyle\frac{0.1 }{2} \left (1+\alpha  \right )  }{\displaystyle\frac{0.1}{2}\left (1-\alpha  \right )}$
Neglecting $\alpha $ due to common ion effect.
$\Rightarrow 10^{-5}=\displaystyle\frac{0.1 \alpha }{2}$
$\Rightarrow \alpha =2\times 10^{-4}$
Hence, it decreases from $0.01  to  2\times 10^{-4}$.

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

$\displaystyle { BaCl } _{ 2 }$ dissociates in water to give one $\displaystyle { Ba }^{ 2+ }$ ion and two $\displaystyle { Cl }^{ - }$ ions. If concentrated $\displaystyle HCl$ is added to this solution :

  1. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ increases
  2. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ remains constant
  3. $\displaystyle \left[ { OH }^{ - } \right] $ increases
  4. The number of moles of undissociated $\displaystyle { BaCl } _{ 2 }$ increases
  5. $\displaystyle \left[ { H }^{ + } \right] $ decreases
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.
$BaCl _2\rightarrow Ba^{2+} + 2Cl^{-}$
$HCl \rightarrow H^{+} + Cl^{-}$
As $Cl^{-} $ is the common ion so, the $ Ba^{2+}$and $ 2Cl^{-}$ combines ( associate ) to give the undissociated $BaCl _2$. 
Hence , there is increase in concentration of undissociated $BaCl _2$.

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 following reaction occurs in a beaker: $\displaystyle { Ag }^{ + }\left( aq \right) +{ Cl }^{ - }\left( aq \right) \rightarrow AgCl\left( s \right) $. If a solution of sodium chloride were added to this beaker,

  1. The solubility of the sodium chloride would decrease

  2. The reaction would shift to the left

  3. The concentration of silver ions in solution would increase

  4. The solubility of the silver chloride would decrease

  5. The equilibrium would not shift at all
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Ag^+ + Cl^- \rightarrow AgCl$
$NaCl \rightarrow Na^+ + Cl^-$
As $Cl^- $ is the common ion in both the solution ( i.e. $NaCl $ and $AgCl$ ) , hence $AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

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 has maximum $pK _a$ :-

  1. $CH _2FCOOH$
  2. $CH _3ClCOOH$
  3. $CH _3COOH$
  4. $HCOOH$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The correct answer is $(A)$.

Because,
$C{H _2}FCOOH$ is the strongest acid in the following:
Hence,
Fluorine is an electron withdrawing group the acidic strength increases.

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

Basic strength of $NH _{4}OH$   in presence of $NH _{4}CI$  

  1. Increase

  2. Remains unchanged

  3. Decreases

  4. Some times increases or sometimes decrease
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

While 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. 

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

A solution contains both $Mg^{2+}$(aq) and $Sr^{2+}$(aq) at the same concentration.
The solution is divided into two equal portions. Aqueous sodium hydroxide is added dropwise to one portion. Dilute sulfuric acid is added dropwise to the other portion.
Which row is correct?

precipitate seen first when NaOH(aq) is added precipitate seen first when $H _2SO _4$(aq) is added
A magnesium hydroxide magnesium sulfate
B magnesium hydroxide strontium sulfate
C strontium hydroxide magnesium sulfate
D strontium hydroxide strontium sulfate


  1. A

  2. B

  3. C

  4. D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{On reacting with NaOH(aq) the Mg+2 gives precipitation reaction.}$


$\text{Sr2+ forms strontium sulfate by reaction with sulphuric acid.}$

$\text{So option B is correct.}$