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A solution contains $Na _{2}CO _{3}$ and $NaHCO _{3}, 10\ mL$ of this solution required $2.5\ mL$ of $0.1\ M\ H _{2}SO _{4}$ for neutralisation using phenolphthalein indicator. Methyl orange is added after first end point, further titration required $2.5\ mL$ of $0.2\ M\ H _{2}SO _{4}$. The amount of $Na _{2}CO _{3}$ and $NaHCO _{3}$ in $1$ litre of the solution is:

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

  1. $5.3\ g$ and $4.2\ g$

  2. $3.3\ g$ and $6.2\ g$

  3. $4.2\ g$ and $5.3\ g$

  4. $6.2\ g$ and $3.3\ g$

Show answer
Correct Option: A
Explanation:
a) ${ 2Na } _{ 2 }{ CO } _{ 3 }+{ H } _{ 2 }{ SO } _{ 4 }\rightleftharpoons 2{ NaHCO } _{ 3 }+{ Na } _{ 2 }{ SO } _{ 4 }$
b) $2{ NaHCO } _{ 3 }+{ H } _{ 2 }{ SO } _{ 4 }\rightleftharpoons { Na } _{ 2 }{ SO } _{ 4 }+2{ H } _{ 2 }{ CO } _{ 3 }$
$2.5$ ml of $0.1M$ ${ H } _{ 2 }{ SO } _{ 4 }=2.5\times 0.1\times 2\times { 10 }^{ -3 }$ moles of ${ H }^{ + }$.
                                          $=0.5\times { 10 }^{ -3 }$ moles of ${ H }^{ + }$
$\therefore$   $0.5\times { 10 }^{ -3 }$ moles of ${ Na } _{ 2 }{ CO } _{ 3 }$ is present in the solution.
$2.5$ ml of $0.2M$ ${ H } _{ 2 }{ SO } _{ 4 }\equiv 2.5\times 0.2\times 2\times { 10 }^{ -3 }$ moles of ${ H }^{ + }$
                                          $=1.0\times { 10 }^{ -3 }$ moles
So total amount of ${ NaHCO } _{ 3 }$ after first end $=1\times { 10 }^{ -3 }$ moles
$\therefore$   The mixture contains $=\left( 1\times { 10 }^{ -3 }-0.5\times { 10 }^{ -3 } \right) $ moles of ${ NaHCO } _{ 3 }$.
The amount of ${ Na } _{ 2 }{ CO } _{ 3 }$ in $1$ litre solution $=\dfrac { 0.5\times { 10 }^{ -3 } }{ 10 } \times { 10 }^{ 3 }\times 106=5.3gm$
The amount of ${ NaHCO } _{ 3 }=\dfrac { 0.5\times { 10 }^{ -3 } }{ 10 } \times { 10 }^{ 3 }\times 84=4.2gm$
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 $0.1$ M $HCO _3^- \ +$ $50$ ml of $0.8$ M $CO _3^{2-}$.

[For $H _2CO _3$ : $K _{a _1}=4\times 10^{-7}$ and $K _{a _2}=2\times 10^{-11}$]
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(4.6 to 6.4)

  3. Phenolphthalein (8.3 to 10.1)

  4. Malachite green (11.4 to 13)

Show answer
Correct Option: D
Explanation:
Given that
[Salt] = 0.8M   and  [Acid] = 0.1M

The solution can be regarded as acidic buffer solution containing weak acid sodium bicarbonate and its salt sodium carbonate with strong base.

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

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

$pK _a=-logK _a=-log [2 \times 10^{-11}]= 10.7$

Substitute values in the above solution.

$pH=10.7+log \frac {0.8} {0.1}=11.6$

Hence, the suitable indicator is Malachite green with pH range from 11.4 to 13.

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.

How many zones are present in candle flame? 

chemistry combustion structure of flame flame combustion, fuels and flame

  1. 2

  2. 3

  3. 4

  4. 5

Show answer
Correct Option: B
Explanation:

A flame consist of three zones. These are the Innermost zone, middle zone, outer zone. The three zones of a flame have different colours and different temperature.


Hence, option B is correct.

Why the innermost zone of candle flame is the least hot?

chemistry combustion structure of flame flame combustion, fuels and flame

  1. Unburnt wax vapour

  2. No supply of oxygen

  3. Presence of wick

  4. All the above

Show answer
Correct Option: A
Explanation:

Dark zone is directly above the wick and wax gets vaporized at this region but does not undergo burning.


Hence the option A is correct.