Tag: chemical equilibrium and acids-bases

Questions Related to chemical equilibrium and acids-bases

${H _3}B{O _3}$ 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

Show answer
Correct Option: C

The order of basicity of halides 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. $\displaystyle Cl^{-}< Br^{-}< I^{-}< F^{-}$

  2. $\displaystyle F^{-}< I^{-}< Br^{-}< Cl^{-}$

  3. $\displaystyle I^{-}< Br^{-}< Cl^{-}< F^{-}$

  4. $\displaystyle Cl^{-}< F^{-}< I^{-}< Br^{-}$

Show answer
Correct Option: C
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 $F^->Cl^->Br^->I^-$.  

In a $0.1\ M$ solution of $H _{3}PO _{4}$ the ionic species present are (besides $H _{3}O^+$ and $OH^{-}$):

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

  1. ${H _{2}PO _{4}}^{-}, {HPO _{4}}^{2-}, {PO _{3}}^{3-}$

  2. ${H _{2}PO _{2}}^{3-}, {H _{2}PO _{4}}^{-}, {PO _{4}}^{3-}$

  3. ${H _{2}PO _{4}}^{-}, {HPO _{3}}^{2-}, {PO _{2}}^{3-}$

  4. ${PO _{3}}^{4-}, {PO _{4}}^{3-}, {HPO _{4}}^{2-}, {H _{2}PO _{4}}^{-}$

Show answer
Correct Option: A
Explanation:

$H _3PO _4$$=$$H^+$$+$$H _2PO _4^-$

$H _2PO _4^-$$=$$H^+$$+$$HPO _4^{-2}$
$HPO _4^{-2}$$=$$H^+$$+$$PO _4^{-3}$
Thus $H _3PO _4$ will dissociate into the ions $H _2PO _4^-$, $HPO _4^{-2}$, $PO _4^{-3}$.
Option A is the correct answer.

How both the acids $\displaystyle HCl$ and $\displaystyle { HC } _{ 2 }{ H } _{ 3 }{ O } _{ 2 }$ are different?

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 first has less hydrogen in solution

  2. The second is highly ionized

  3. The second has more ionized hydrogen

  4. The first is highly ionized

Show answer
Correct Option: D
Explanation:

HCl undergoes complete ionization as it is a strong acid.
On the other hand, acetic acid  $HC _2H _3O _2$ is a weak acid and is ionized to a small extent.

The degree of dissociation of $PCl _{5(g)}$ at 16.8 bar and $127^{0}C$ is 0.4. The value of $K _{P}$ for the reaction is:
$PCl _{5} \leftrightharpoons PCl _{3(g)} +Cl _{2(g)}$ 

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

  1. $3.2 bar$

  2. $3.2 bar^{-1}$

  3. $12.8\ bar$

  4. $ 0.4$ x $16.8\ bar$

Show answer
Correct Option: A
Explanation:

total pressure P total = $16.8 \, bar$

degree of dissociation $\alpha = 0.4$
$PCl _5 \rightleftharpoons p Cl _3 + Cl _2$
$P _0 (1 - \alpha) \,\,\, P _0 \alpha \,\,\, P _0 \alpha$
$P \, total = P _0(1 - \alpha) + P _0 \alpha + P _0 \alpha$
$= P _0 (1 + \alpha)$
$P _0 (1 + \alpha) = 16.8$
$P _0 \times 1.4 = 16.8 \Rightarrow P _0 = \dfrac{16.8}{1.4} = 12 $ bar
$Kp = \dfrac{[PCl _3][Cl _2]}{[PCl _5]} = \dfrac{P _0 \alpha \times P _0 \alpha}{P _0 (1 - \alpha)}$
$= \dfrac{P _0 \alpha^2}{1 - \alpha}$
$= 12 \times \dfrac{0.4 \times 0.4}{0.6}$
$= 3.2 \, bar$

The ionisation constant of an acid, $K _a$ is the measure of strength of an acid. The $K _a$ values of acetic acid, hypochlorous acid and formic acid are $1.74 \times 10^{-5}, 3.0 \times 10^{-8}$ and $1.8 \times 10^{-4}$ respectively. Which of the following orders of pH of $0.1$mol $dm^{-3}$ solutions of these acids is correct?

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

  1. Acetic acid > Hypochlorous acid > Formic acid

  2. Hypochlorous acid > Acetic acid > Formic acid

  3. Formic acid > Hypochlorous acid > Acetic acid

  4. Formic acid > Acetic acid > Hypochlorous acid

Show answer
Correct Option: B
Explanation:

$K _a$ is a measure of the strength of the acid i.e., larger the value of $K _a$, the stronger is the acid.
Thus, the correct order of acidic strength is
$HCOOH > CH _3COOH > HClO$
Stronger the acid, lesser will be the value of pH. Hence, the correct order of pH is $HClO > CH _3COOH > HCOOH$.

The $[H^+]$ of a resulting solution that is $0.01\space M$ acetic acid $(K _a = 1.8\times10^{-5})$ and $0.01\space M$ in benzoic acid $(K _a = 6.3\times10^{-5})$:

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

  1. $9\times10^{-4}$

  2. $81\times10^{-4}$

  3. $9\times10^{-5}$

  4. $2.8\times10^{-3}$

Show answer
Correct Option: A

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.

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


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

  1. A

  2. B

  3. C

  4. D

Show answer
Correct Option: B
Explanation:

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


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

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