Tag: dissociation constants

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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

The value of observed and calculated molecular weight of silver nitrate are $92.64$ and 170 respectively. The degree of dissociation of silver nitrate is :

  1. $60\%$
  2. $83.5\%$
  3. $46.7\%$
  4. $60.23\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The van't Hoff factor i = (Calculated MW) / (Observed MW) = 170 / 92.64 = 1.835. For AgNO3, i = 1 + alpha(n-1). Since n=2, i = 1 + alpha. Thus, alpha = 0.835 or 83.5%.

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

In the dissociation of $NH _4OH$, if excess if $NH _4Cl$ is added before adding $NH _4OH$, the concentration of:

  1. $NH _4^+$ ions increases and $OH^-$ ions decreases
  2. both $NH _4^+$ ions and $OH^-$ ions increases
  3. $NH _4^+$ ions decreases and $OH^-$ ions increases
  4. both $NH _4^+$ ions and $OH^-$ ions decreases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the law of Mass action, the dissociation of $NH _4OH$ takes place and we have,
$\dfrac{[NH _4^+][OH^-]}{[NH _4OH]} =K$
Ammonium chloride, a strong electrolyte, ionises almost completely as follows:
$NH _4Cl \leftrightarrow NH _4^+ + Cl^-$
So, if excess of $NH _4Cl$ is added before adding $NH _4OH$, the concentration of $NH _4^+$ ions is increased and consequently the concentration of $OH^-$ ions is decreased.

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 will supress the ionisation of acetic acid in aqueous solution?

  1. $NaCl$
  2. $HCI$
  3. $KCI$

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

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

So addition of $CH _3COONa$ to $CH _3COOH$ increases the $pH$ of solution but addition of $NH _4Cl$ to $NH _4OH$ decreases the $pH$ of solution.
Here, the addition of $HCl$ acid will suppress the ionisation of acetic acid.

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: $pH$ value of $HCN$ solution decreases when $NaCN$ is added to it
Reason: $NaCN$ provides a common ion $CN^{\circleddash}$ to $HCN$

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

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

  3. Assertion is correct but Reason is incorrect

  4. Assertion is incorrect but Reason is correct

  5. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(A)$ is wrong because the addition of $NaCN$ to $HCN$, due to common ion $(CN^{\circleddash})$, the degree of dissociation of $HCN$ is suppressed and hence less $[H^{\oplus}]$ and increase in $pH.$

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

In the third group of qualitative analysis, the precipatitating reagent is $NH _4Cl / NH _4OH$. The function of $NH _4Cl$ is to 

  1. increase the ionisation of $NH _4OH$
  2. supress the ionisation of $NH _4OH$

  3. convert the ions of group third into their respective chlorides
  4. stabilise the hydroxides of group $III$ cations
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

So, addition of $CH _3COONa$ to $CH _3COOH$ increases the pH of solution but addition of $NH _4Cl$ to $NH _4OH$ decreases the pH of solution.
So, the function of $NH _4Cl$ to suppress the ionisation of $NH _4OH$ and to reduce conentration of hydroxide ion in solution.

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: Due to common ion effect, the solubility of $HgI _2$ is expected to be less in an aqueous solution of KI than in water. But $HgI _2$ dissolves in an aqueous solution of KI to form a clear solution.
Reason: $I^{\circleddash}$ ion is highly polarisable.

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

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

  3. Assertion is correct but Reason is incorrect

  4. Assertion is incorrect but Reason is correct

  5. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Due to common ion effect, the solubility of $HgI _2$ is expected to be less in an aqueous solution of KI than in water as
$HgI _2 + KI \longrightarrow K _2[HgI _4]$.
since $I^{\circleddash}$ ion is large sized and therefore is highly polarisable.
But (R) is not the correct explanation of (A)

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

In the presence of a common ion (incapable of forming complex ion), the solubility of salt _______  in solution.

  1. increases

  2. decreases

  3. remains the same

  4. cannot predict
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$AB \rightarrow A^+ + B^-$
$BC \rightarrow B^+ + C^-$
Since $B^+$ is incapable of forming a complex salt it tends to decrease the solubility by Le-Chatelier's principle.

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

The order of basicity of halides is:

  1. $\displaystyle Cl^{-}< Br^{-}< I^{-}< F^{-}$
  2. $\displaystyle F^{-}< I^{-}< Br^{-}< Cl^{-}$
  3. $\displaystyle I^{-}< Br^{-}< Cl^{-}< F^{-}$
  4. $\displaystyle Cl^{-}< F^{-}< I^{-}< Br^{-}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Fuorine is the most electronegative halogen element, but is the least stable, and the most basic. It turns out that when moving vertically in the periodic table, the size of the atom trump its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly  larger volume therefore, effective nuclear charge decreases. 

Overall order of basicity is $F^->Cl^->Br^->I^-$.  

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

Determine $[{OH}^{-}]$ of a $0.050\ M$ solution of ammonia to which has been added sufficient ${NH} _{4}Cl$ to make the total $[{NH} _{4}^{+}]$ equal to $0.100 M$. $[{K} _{b({NH} _{3})}=1.8\times {10}^{-5}]$

  1. $[{OH}^{-}]=9.0\times {10}^{-6}$
  2. $[{OH}^{-}]=9.0\times {10}^{-8}$
  3. $[{OH}^{-}]=9.0\times {10}^{-2}$
  4. $[{OH}^{-}]=9.0\times {10}^{-9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${NH} _{4}Cl\longrightarrow {NH} _{4}^{+}+{Cl}^{-}$
${NH} _{4}OH\longrightarrow {NH} _{4}^{+}+{OH}^{-}$
${K} _{b}=\cfrac { \left[ { NH } _{ 4 }^{ + } \right] \left[ OH \right]  }{ \left[ { NH } _{ 4 }OH \right]  } $
$[{NH} _{4}^{+}]=$ is due to salt because ${NH} _{4}OH$ ionise less amount due to common ions effect
$1.8\times {10}^{-5}=\cfrac{0.1\times [{OH}^{-}]}{0.05}$ 
$9\times {10}^{-6}=[{OH}^{-}]$

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

Why only ${As}^{+3}$ gets precipitated as ${As} _{2}{S} _{3}$ and not ${Zn}^{+2}$ as $ZnS$ when ${H} _{2}S$ is passed through an acidic solution containing ${As}^{+3}$ and ${Zn}^{+2}$?

  1. Solubility product of ${As} _{2}{S} _{3}$ is less than that of $ZnS$
  2. Enough ${As}^{+3}$ are present in acidic medium

  3. Zinc salt does not ionise in acidic medium

  4. Solubility product changes in presence of an acid
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The species having minimum value of ${K} _{sp}$ will get precipitated first of all because ionic product will exceed the solubility product of such species.

${K} _{sp}$ of ${As} _{2}{S} _{3}$ is less than $ZnS$. In acid medium ionisation of ${H} _{2}S$ is suppressed (common ion effect) and ${K} _{sp}$ of $ZnS$ does not exceed.