Tag: ionisation of weak acids and weak bases

Questions Related to ionisation of weak acids and weak bases

$Ag _3 PO _4$ would be least soluble at 25$^o$C in

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

  1. 0.1 M $AgNO _3$

  2. 0.1 M $HNO _3$

  3. pure water

  4. 0.1 M $Na _3PO _4$

  5. solubility in (a), (b), (c) or (d) is not different

Show answer
Correct Option: A
Explanation:

$Ag _3 PO _4$ is a weak electrolyte and $AgNO _3$ is a strong electrolyte containing common ion $(NO _3^-ion)$. Thus common ion effect is observed and the solubility of $Ag _3 PO _4$ is suppressed.
Hence, $Ag _3 PO _4$ is least soluble in 0.1M $AgNO _3$.

The solubility of $AgI$ in $NaI$ solution is less than that in pure water because

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

  1. $AgI$ forms complex with $NaI$

  2. Of common ion effect

  3. Solubility product of $AgI$ is less than that of $NaI$

  4. The temperature of the solution decreases

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Correct Option: B
Explanation:

The common ion presence in the solution decrease the solubility of a given sparingly soluble compound. So, solubility of AgI in NaI solution is less.

The ionic strength of $C{H} _{3}CO{O}^{-}$ ion in $0.1\ M$ $C{H} _{3}COOH$ solution having ${K} _{a}= 1.8\times {10}^{-5}$ is

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

  1. $0.1$

  2. $0.05$

  3. $6.7\times {10}^{-4}$

  4. $1.34\times {10}^{-3}$

Show answer
Correct Option: C
Explanation:

$C{H} _{3}COOH\rightleftharpoons C{H} _{3}CO{O}^{-} + {H}^{+}$


$[C{H} _{3}CO{O}^{-}]= C\alpha =C\sqrt {\frac{K _a}{C}}=\sqrt {K _aC} =\sqrt {1.8\times {10}^{-5}\times 0.1} =1.34\times {10}^{-3}$

Ionic strength of $C{H} _{3}CO{O}^{-}=\dfrac{1}{2}n\ [CH _3COO^-]{Z}^{2}=\dfrac{1}{2}\times 1\times 1.34\times {10}^{-3}\times (-1)^2=6.7\times {10}^{-4}$

where, n = no. of ions and Z= charge on the ion.

Option C is correct.

Simultaneous solubility of $AgCNS\ (a)$ and $AgBr\ (b)$ in a solution of water will be

${K} _{{sp} _{(AgBr)}}=5\times {10}^{-13}$ and ${K} _{{sp} _{(AgCNS)}}={10}^{-12}$ 

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

  1. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$

  2. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

  3. $a=8.16\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Suppose solubility of $AgCNS$ and $AgBr$ in a solution are $a$ and $b$ respectively.

$AgCNS(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + }(aq) } +\underset { a }{ { CNS }^{ - }(aq) } $

$AgBr(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + } } +\underset { b }{ { Br }^{ - } } $

$\therefore$ $[{Ag}^{2+}]=(a+b); [CN{S}^{-}]=a$ and $[{Br}^{-}]=b$

For $AgCNS:{ K } _{ { sp } _{ AgCNS }\quad  }=\left[ { Ag }^{ + } \right] \left[ { CNS }^{ - } \right] $

or $1\times {10}^{-12}=(a+b)(a)......(i)$

For $AgBr: { K } _{ { sp } _{ AgBr }\quad  }=\left[ { Ag }^{ + } \right] \left[ { Br }^{ - } \right] $

or $5\times {10}^{-13}=(a+b)(b)........(ii)$

By Eqs $(i)$ and $(ii)$, we get

$\cfrac{a}{b}=\cfrac{{10}^{-12}}{5\times {10}^{-13}}=2$  ($a=2b$)

$\therefore $ By Eq $(i)$

$(2b+b)(2b)=1\times {10}^{-12}$

$\therefore$ $6{b}^{2}=1\times {10}^{-12}$

or $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

Similarly by Eq $(ii)$

$[a+(a/2)]a=1\times {10}^{-12}$

$a=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$

Addition of ${NH} _{4}Cl$ does not effect the $pH$ of solution of ${NH} _{4}OH$. 

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

  1. True

  2. False

  3. Ambigous

  4. None

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Correct Option: B
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 ${NH} _{4}Cl$ reduces dissociation of ${NH} _{4}OH$ and because of that its pH will decreases.

The solubility of $Hg{I} _{2}$ in water decreases in presence of $KI$.

State whether the given statement is true or false.


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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:
$Hg{I} _{2}$ forms soluble complex with $KI$
$2KI+Hg{I} _{2}\longrightarrow {K} _{2}Hg{I} _{4}$

So solubility of $Hg{I} _{2}$ in water increases in presence of $KI$.

Assertion: AgCl will not dissolve in a concentrated solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.

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

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

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

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false

Show answer
Correct Option: A
Explanation:

Assertion: AgCl will not dissolve in a concentrated NaCl solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.
This is an example of a common ion effect. The chloride ions are common ions.
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

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.

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

  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

Show answer
Correct Option: B
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.

(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. 

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

  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

Show answer
Correct Option: A
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.

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?

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

  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

Show answer
Correct Option: C
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 $0.01  to  2\times 10^{-4}$.