Tag: further aspects of equilibria

Questions Related to further aspects of equilibria

When $20\ mL$ of $\dfrac {M}{10}NaOH$ are added to $10\ mL$ of $\dfrac {M}{10}HCl$, the resulting solution will:

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

  1. turn blue litmus red

  2. turn phenolphthalein solution pink

  3. turn methyl orange red

  4. have no effect on either red or blue litmus

Show answer
Correct Option: B
Explanation:
$20$ ml of $\dfrac { M }{ 10 } NaOH=2\times { 10 }^{ -3 }$ moles of $NaOH$
$10$ ml of $\dfrac { M }{ 10 } HCl=1\times { 10 }^{ -3 }$ moles of $HCl$.
Excess amount of $NaOH=\left( 2\times { 10 }^{ -3 }-1\times { 10 }^{ -3 } \right) $ moles
                                             $=1\times { 10 }^{ -3 }$ moles
$\therefore$   $\left[ { OH }^{ - } \right] =\dfrac { { 10 }^{ -3 }\times 1000 }{ 30 } =0.033$
$\therefore$   $pH=14-pOH=12.49$
In this $pH$, Phenolphthalein solution turns into pink.
Answer will be $B$.

An indicator $HIn$ has a standard ionization constant of $9.0\times {10}^{-9}$. The acid colour of the indicator is yellow and the alkaline colour is red. The yellow colour is visible when the ratio of yellow form to red form is $30$ to $1$ and the red colour is predominant when the ratio of red form to yellow form is $2$ to $1$. What is the $pH$ range of the indicator?

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

  1. < $6.568$

  2. $6.568$ to $8.346$

  3. > $8.346$

  4. None of these

Show answer
Correct Option: B
Explanation:
$ \underset {Yellow}{HIn} + H _2O \rightleftharpoons H _3O^+ + \underset {Red}{In^-}$
$K _{In}= (\dfrac {[H _3O^+][In^-]}{HIn})$
Yellow colour is visible when the ratio of acid form to base form is 3 to 1.

$K _{In}= (\dfrac {[H _3O^+][1]}{30})$
$9 \times 10^{-9}= (\frac {[H _3O^+][1]}{30})$
$[H _3O^+]= 270 \times 10^{-9}$
$-log [H _3O^+]= pH= 6.569$
Red colour is predominant when the ratio of base form to acid form is 2 to 1.

$K _{In}= (\dfrac {[H _3O^+][2]}{1})$
${[H _3O^+]}= 4.5 \times 10^{-9}$
$-log [H _3O^+]= pH= 8.523$
 The pH range of the indicator is 6.569 to 8.523.

Find the $pH$ of the resulting solution and then mark the option in which $pH$ exists between color transition range of an indicator.

50 ml of 0.2M HA solution $(K _a=10^{-5})$ + 50ml of 0.1M HCl solution + 100 ml of 0.13 M NaOH solution.

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

  1. Phenol red (6.8 to 8.4)

  2. Propyl red (4.6 to 6.4)

  3. Phenolphthalein (8.3 to 10.1)

  4. Malachite green (11.4 to 13)

Show answer
Correct Option: B
Explanation:

Milimoles of HCl = Volume (ml) $\times$ Molarity = $50\times 0.1=5$ milimoles.
Milimoles of NaOH $=100 \times 0.13=13$ milimoles.

Some of the NaOH will be neutralized by HCl.
Milimoles of NaOH remaining unreacted $13-5=8$ milimoles.

Milimoles of HA$=50 \times 0.2 =10$ milimoles.
Out of this 8 milimoles will be neutralized to form a salt.

Thus the solution now contains 2 milimoles of weak acid HA and 8 milimoles of its salt with strong base.

It is an acidic buffer solution.
The expression for the pH of the acidic buffer solution is as given below.

$pH=pK _a+log \frac {[salt]} {[acid]}$

$pK _a=-log10^{-5}=5$

$pH=5+log \frac {8} {2}=5.6$.

Thus the suitable indicator is propyl red with pH range 4.6 to 6.4.

In the titration of $CH _{3}COOH$ against $NaOH$, we cannot use the

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

  1. Methyl orange

  2. Methyl red

  3. Phenolphthalein

  4. Bromothymol blue

Show answer
Correct Option: A
Explanation:

For the titration of weak acid and strong bases like $CH _3COOH+NaOH$ methyl orange indicator is used to observe the end point.
Hence option A is correct.

During the titration of mixture of $NaOH,{ Na } _{ 2 }{ CO } _{ 3 }$ and an inert substance against hydrochloric acid:

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

  1. Phenolphthalein is used to detect the end point when $NaOH$ is completely neutralized and half of ${ Na } _{ 2 }{ CO } _{ 3 }$ is neutralized

  2. Methyl orange is used to detect the final end point

  3. Methyl orange is used to detect the first end point

  4. Phenolphthalein is used to detect the final end point

Show answer
Correct Option: A,B
Explanation:

During the titration of mixture of $NaOH,{ Na } _{ 2 }{ CO } _{ 3 }$ and an inert substance against hydrochloric acid Phenolphthalein is used to detect the end point when $NaOH$ is completely neutralized and half of ${ Na } _{ 2 }{ CO } _{ 3 }$ is neutralized,but Methyl orange is used to detect the final end point.

Hence option A,B are correct.

(A) In general phenolphthalein is used as an indicator for the titration of weak acid $(CH _3COOH)$ and strong base (NaOH).
(R) At equivalence point solution is basic.

chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and 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:

If the titration is carried out for strong base and weak acid.At the end point the solution would be basic in nature .in this case phenolphthalein indicator.
So Phenolphthalein is used as an indicator for the titration of weak acid ($CH _3COOH$) and strong base ($NaOH$).Here the end point is basic in nature.

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

Hence option A is correct.

The addition of $HCl$ will not suppress the ionisation of:

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

  1. Acetic acid

  2. Sulphuric acid

  3. ${H} _{2}S$

  4. Benzoic acid

Show answer
Correct Option: B
Explanation:

The addition of $HCl$ will not suppress the ionisation of $H _2SO _4$ because $H _2SO _4$ is stronger acid than $HCl$.

One litre of water contains ${ 10 }^{ -7 }$ mole of ${H}^{+}$ ions. Degree of ionisation of water is:

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

  1. $1.8\times { 10 }^{ -7 }$

  2. $0.8\times { 10 }^{ -9 }$

  3. $5. 4\times { 10 }^{ -9 }$

  4. $5 . 4\times { 10 }^{ -7 }$

Show answer
Correct Option: A
Explanation:

$H _2O⇌H^{ + }   +       OH^{ - }\quad $

  $C$        $0$                  $0$
$C(1-\alpha )$        $C\alpha $    $C\alpha $

$C\alpha=10^{-7}$

[H2O] =$\dfrac{ 1000}{18}$= 55.55 M 

C= 55.5M

So $\alpha=18\times10^{-10}$

In percentage $\alpha=1.8\times10^{-7}$%

When ${ NH } _{ 4 }Cl$ is added to ${ NH } _{ 4 }OH$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

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

  1. common ion effect

  2. hydrolysis

  3. oxidation

  4. reduction

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Correct Option: A
Explanation:

When $NH _4Cl$ is added to $NH _4OH$ solution, concentration of $NH _4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced. 

The weak acid, $HA$ has a ${K} _{a}$ of $1.00\times { 10 }^{ -5 }$. If $0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

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

  1. $1$%

  2. $99.9$%

  3. $0.1$%

  4. $99$%

Show answer
Correct Option: A
Explanation:
0.1 mole of acid is dissolved in 1 litre of water means $[HA]=0.1M$
Let '$\alpha$' be degree of dissocition
$HA\rightleftharpoons { H }^{ + }+{ A }^{ - }$
 $0.1$
$0.1(1-\alpha)$   $0.1\alpha$     $0.1\alpha$
${ K } _{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right]  }{ \left[ HA \right]  } =\cfrac { { 0.1 }^{ 2 }{ \alpha  }^{ 2 } }{ 0.1(1-\alpha)  } $
Let $\alpha<<1$ so $1-\alpha=1$
$K _a=0.1\alpha^2=10^{-5}$
$\alpha=10^{-2}$
% of acid dissociated=$1$%