Tag: ionization constants of weak acids and weak bases

Questions Related to 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?



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

  1. ${ 1.8\times 10 }^{ -5 }$

  2. ${ 9\times 10 }^{ -6 }$

  3. ${ 3.6\times 10 }^{ -5 }$

  4. NONE OF THE ABOVE

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Correct Option: C

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

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

  1. $AlCl _3$

  2. $Na _3PO _4$

  3. $CaCl _2$

  4. $K _2SO _4$

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Correct Option: A
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 $Al{Cl} _{3}$.

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

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

  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

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Correct Option: C
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.

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) .

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

  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

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Correct Option: C
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, $S+ 0.1 \approx 0.1$ 

$\Rightarrow S=5.3\times 10^{-7}\ M$ 

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    

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

  1. $0.0060\% $

  2. $0.013\% $

  3. $0.77\% $

  4. $1.6\% $

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Correct Option: B

A $40.0 ml$ solution of weak base, $BOH$ is titrated with $0.1 N - HCl$ solution. The $pH$ of the solution is found to be $10.0$ and $9.0$ after adding $5.0 ml$ and $20.0 ml$ of the acid, respectively. The dissociation constant of the base is ($log 2 = 0.3$)

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

  1. $2 \times 10^{-5}$

  2. $1 \times 10^{-5}$

  3. $4 \times 10^{-5}$

  4. $5 \times 10^{-5}$

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Correct Option: A

If a salt of weak acid or base is added to a solution of its acid or base respectively, the:

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

  1. dissociation of acid or base is diminished

  2. the $pH$ of the solution in case of acid increases and in case of base decreases

  3. mixing of two leads for precipitation

  4. none of the above

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Correct Option: A,B
Explanation:

Due to common ion effect, if a salt of weak acid or base is added to a solution of its acid or base respectively, the dissociation of acid or base is diminished.


As a result, concentration of hydrogen ions or hydroxide ion will change and pH of solution increases in case of acid and decreases in case of base.

As $pH = -log [ H^+]$

What will be the effect of adding $100 ml$ of $0.001 M - HCl$ solution to $100 ml$ of a solution having $0.1 M - HA$? The acid dissociation constant of $HA$ is $10^{-5}$.

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

  1. The degree of dissociation of $HA$ will decrease but the $pH$ of solution remains unchanged.

  2. The degree of dissociation of $HA$ remains unchanged but the $pH$ of solution decreases.

  3. Neither degree of dissociation nor $pH$ of solution will change.

  4. The degree of dissociation as well as $pH$ of solution will decrease.

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Correct Option: C

${H _3}B{O _3}$ 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. monobasic aand weak lewis acid

  2. monobasic and weak bronsted acid

  3. monobasic and strong lewis acid

  4. tribasic and weak bronsted acid

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Correct Option: C

The common ion effects is shown by which of the following sets of solutions:-

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

  1. $BaCl _{2}+Ba(NO _{3}) _{2}$

  2. $NaCl+HCl$

  3. $NH _{4}OH+NH _{4}Cl$

  4. $None$

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Correct Option: C
Explanation:

Common ion effect is normally applied in selective precipitation of ions which involves a technique of separating ions in an aqueous solution by using a reagent that precipitates one or more of the ions. addition of common ions in a solution that is already having that given ion normally leads to a formation of a precipitate. The common ion normally decreases the solubility of a slightly insoluble salt.

Its not (1) because they are both solids.

The remaining are solute solvent pairs. So all the other pairs will show common ion effect if the solute is soluble to some extent and exists in an equilibrium with the solvent. The answer given is (3).