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Indicators and acid-base titration - class-XII

Description: indicators and acid-base titration
Number of Questions: 26
Created by:
Tags: further aspects of equilibria equilibrium acids and bases chemistry
Attempted 0/26 Correct 0 Score 0

1 kg of NaOH is added to 10 ml of 0.1N HCl, the resulting solution will:

  1. turn blue litmus red

  2. turn phenolphthalein solution pink

  3. turn methyl orange red

  4. will have no effect on red or blue litmus paper


Correct Option: B
Explanation:

The final solution is strongly basic. Phenolphthalein is an indicator used in acid-base reactions which turns pink in basic solution. Therefore, phenolphthalein turns pink. 

From the following in which titration methyl orange is a best indicator?

  1. $CH _{3}COOH+NaOH$

  2. $H _{2}C _{2}O _{4}+NaOH$

  3. $HCl+NaOH$

  4. $CH _{3}COOH+NH _{4}OH$


Correct Option: A
Explanation:

In the titration of acetic acid with a strong base such as sodium hydroxide the indicator used in the titration is methyl orange changed to yellow.

Dissociation of ${ CH } _{ 3 }COOH$ is suppressed by adding

  1. HCL

  2. ${ H } _{ 2 }{ SO } _{ 4 }$

  3. $CH _{ 3 }COONa\quad$

  4. Any of the above


Correct Option: C

Which one of the following indicatprs works in the pH range $8-9.8$?

  1. Litmus

  2. Phenolphthalein

  3. Methyl red

  4. Methyl orange


Correct Option: B
Explanation:

Phenolphthalein works in $pH$ range of $8.3-10.0$

Methyl Red works in $pH$ range of $4.4-6.5$
Methyl Orange works in $pH$ range of $3.2-4.5$

For the titration between oxalic acid and sodium hydroxide, the indicator used is

  1. Potassium permanganate

  2. Phenolphthalein

  3. Litmus

  4. Methyl orange


Correct Option: B
Explanation:

Oxalic acid is titrated against NaOH. Phenolphthalein is chosen because it changes color in a pH range between 8 – 10. It appears pink in basic solutions and colourless in acidic solutions.

In the case of a strong acid-strong base titration, this pH transition would take place within a fraction of a drop of actual neutralization, since the strength of the base is high.

Select the best indicator from the given table for titration of $20\space mL$ of $0.02\space M\space CH _3COOH$ with $0.02\space M\space NaOH$. Given $pK _a\ \text{of}\ CH _3COOH= 4.74)$

Indicator $pH$ range
(I) Bromothymol blue $6.0 - 7.6$
(II) Thymolphathalein $9.3 - 10.5$
(III) Malachite green $11.4 - 13$
(IV) M-Cresol purple $7.4 - 9.0$
  1. $(I)$

  2. $(II)$

  3. $(III)$

  4. $(IV)$


Correct Option: B
Explanation:

Since $pH=\left( SW+WA \right) =7+1/2\left( pKa+logC \right) $    {at equilance pointer}

             $C=0.01$  after mixing
  $\therefore \quad pH=7+1/2\left( 4.74+log0.01 \right) $
  $\therefore \quad pH=8.37$
So, it lies in range of $(IV)$ and somenehat of $(II)$

The fraction of total volume occupied by by the atoms present in a simple cube is

  1. $\frac { \Pi }{ \sqrt [ 3 ]{ 2 } }$

  2. $\frac { \Pi }{ \sqrt [ 4 ]{ 2 } }$

  3. $\frac { \Pi }{ 4 }$

  4. $\frac { \Pi }{ 6 }$


Correct Option: D

Assertion: In general, phenolphthalein is used as an indicator for the titration of weak acid (HA) against strong base (NaOH).
Reason: At equivalent point, solution is basic.

  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


Correct Option: B
Explanation:

The phenolphthalein  indicator changes color over pH range 8.0 to 9.8.
During titration of weak acid (HA) against strong base (NaOH), at equivalence point complete neutralization occurs and a salt NA of weak acid (HA) with strong base (NaOH) is obtained.
The hydrolysis of this salt gives alkaline solution in the pH range 8.0 to 9.8.
Hence, in general phenolphthalein is used as an indicator for the titration of weak acid (HA) against strong base (NaOH).

A certain indicator (an organic dye) has $p{K} _{a} = 5$. For which of the following titrations it may be suitable?

  1. Acetic acid against $NaOH$

  2. Aniline hydrochloride against $NaOH$

  3. Sodium carbonate against $HCl$

  4. Barium hydroxide against oxalic acid


Correct Option: C
Explanation:

Give $pK _a$ of indicator $= 5$


This $pK _a$ value corresponds to methyl red indicator. Methyl red is a suitable indicator for titrating carbonates of weak acids (ex. $H _2CO _3$ salt $Na _2CO _3$) with strong acid $HCl$.


 Above titration reaction complated in following steps.

a) $Na _2 CO _3 (aq) + HCl (aq) \rightarrow NaHCO _3 (aq) + NaCl (aq)$

b) $NaHCO _3 (aq) + HCl(aq) \rightarrow NaCl (aq) + CO _2 (g) + H _2O(aq)$

Correct option (C)

The amount of sodium hydrogen carbonate, $NaH{ CO } _{ 3 }$, in an antacid tablet is to be determined by dissolving the tablet in water and titrating the resulting solution with hydrochloric acid. Which indicator is the most appropriate for this titration?
Acid                  ${K} _{a}$
${ H } _{ 2 }{ CO } _{ 3 }$          $2.5\times { 10 }^{ -4 }$
${H{ CO } _{ 3} }^{ - }$           $2.4\times { 10 }^{ -8 }$

  1. Methyl orange, $p{ K } _{ In }=3.7$

  2. Bromothylmol blue, $p{ K } _{ In }==7.0$

  3. Phenolphtalein, $p{ K } _{ In }=9.3$

  4. Alizarin yellow, $p{ K } _{ In }=12.5$


Correct Option: A
Explanation:
The indicator should be used such a way that it shows change in colour in the same $pH$ range as required around the equivalence point. Now when solution of ${ NaHCO } _{ 3 }$ is titrating against $HCl$ solution just after equivalence point there will be presence of very low amount of $HCl$ and $pH$ will be around $\sim 3.6$.
$pH={ pK } _{ a }+log\dfrac { \left[ { HCO } _{ 3 }^{ - } \right]  }{ \left[ { H } _{ 2 }{ CO } _{ 3 } \right]  } =3.6+log\dfrac { \left[ { HCO } _{ 3 }^{ - } \right]  }{ \left[ { H } _{ 2 }{ CO } _{ 3 } \right]  } $
$\therefore$  So Methyl orange having $pK$ in if $3.7$ is the most appropriate for this titration.

A solution containing $Na _{2}CO _{3}$ and $NaOH$ requires $300\ mL$ of $0.1\ N\ HCl$ using phenolphthalein as an indicator. Methyl orange is then added to the above-titrated solution when a further $25\ mL$ of $0.2\ N\ HCl$ is required. The amount of $NaOH$ present in the original solution is:

  1. $0.5\ g$

  2. $1\ g$

  3. $2\ g$

  4. $4\ g$


Correct Option: B

In the titration of ${ NH } _{ 4 }OH$ with $HCl$, the indicator which cannot be used is:

  1. Phenolphthalein

  2. Methyl orange

  3. Methyl red

  4. Both orange and methyl red


Correct Option: A
Explanation:

In case of strong acid($HCl$) and weak base($NH _4OH$) phenolphthalein cannot be used as an indicator because it can detect $pH$ only in range of 8-10.

In the titration of nitric acid against potassium carbonate, the indicator used is:

  1. Methyl orange

  2. Self indicator

  3. Phenolphthalein

  4. Diphenylamine


Correct Option: A
Explanation:

In short methyl orange is used as the indicator in the titration of nitric acid against potassium carbonateIt is frequently used in titrations because of its clear and distinct colour change.

A solution containing $Fe^{2+}$ ions is titrated with $KMnO _{4}$ solution. Indicator used will be:

  1. phenolphthalein

  2. methyl orange

  3. litmus

  4. none of the above


Correct Option: D
Explanation:

Titration of ${ Fe }^{ 2+ }$ with ${ KMnO } _{ 4 }$ is an redox titration.

${ Fe }^{ 2+ }+\underbrace { 7{ MnO } _{ 4 }^{ - } } +14{ H }^{ + }={ Fe }^{ 3+ }+\underbrace { 7{ Mn }^{ 2+ } } +7{ H } _{ 2 }O$
                  violet                                     colourless
So, phenolpthalein, methyl orange and litmus are all acid base indicators. They can't be used in this redox titration. ${ KMnO } _{ 4 }$ is a self-indicator changing from violet to colourless.
$\therefore$   Answer will be $D$.

In the mixture of $NaHCO _{4}$ and $Na _{2}CO _{3}$, volume of a given $HCl$ required is $x\ mL$ with phenolphthalein indicator and $y\ mL$ with methyl orange indicator in same titration. Hence, volume of $HCl$ for complete reaction of $Na _{2}CO _{3}$ present in the original mixture is

  1. $2x$

  2. $y$

  3. $x/2$

  4. $(y - x)$


Correct Option: A

$40\ mL$ of $0.05\ M\ Na {2}CO _{3}\cdot NaHCO _{3} \cdot 2H _{2}O$ (sesquicarbonate) is titrated against $0.05\ M\ HCl.\ x\ mL$ of $HCl$ is used when phenolphthalein is the indicator and $y\ mL\ HCl$ is used when methyl orange is the indicator in two separate titrations, hence $(y - x)$ is_______.

  1. $80\ mL$

  2. $30\ mL$

  3. $120\ mL$

  4. none of the above


Correct Option: A
Explanation:
Titration of ${ Na } _{ 2 }{ CO } _{ 3 }.{ NaHCO } _{ 3 }.2{ H } _{ 2 }O$ with $HCl$ involves following reactions :
a) ${ Na } _{ 2 }{ CO } _{ 3 }+HCl\rightleftharpoons { NaHCO } _{ 3 }+NaCl$
b) ${ NaHCO } _{ 3 }+HCl\rightleftharpoons NaCl+{ H } _{ 2 }O+{ CO } _{ 2 }$
In step $a$, $40$ ml of $0.05M$ $HCl$ will react with $40$ ml of $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$ using phenolpthalein.
$\therefore$   $x=40$ ml
Now, in a separate titration $40$ ml of $0.05M$ $HCl$ will need to react with $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$. Now in the second step $b$ total $80$ ml of $0.05M$ $HCl$ will need to neutralise ${ NaHCO } _{ 3 }$ completely.
$\therefore$   $y=40+40\times 2=120$
$\therefore$   $y-x=120-40=80$ ml
Answer will be $A$.

$0.1\ N$ solution of $Na _{2}CO _{3}$ is being titrated with $0.1\ N\ HCl$, the best indicator to be used is:

  1. potassium ferricyanide

  2. phenolphthalein

  3. methyl orange

  4. litmus


Correct Option: C
Explanation:
Titration of ${ Na } _{ 2 }{ CO } _{ 3 }$ with $HCl$ involves following two reaction :
a) ${ Na } _{ 2 }{ CO } _{ 3 }+HCl\rightleftharpoons { NaHCO } _{ 3 }+NaCl$
b) ${ NaHCO } _{ 3 }+HCl\rightleftharpoons NaCl+{ H } _{ 2 }O+{ CO } _{ 2 }\quad \left\{ { H } _{ 2 }{ CO } _{ 3 }-carbonic\quad acid \right\} $
We know phenolphthalein is an indicator and it works in the basic medium that is why it causes only $50$% of neutalisation of ${ Na } _{ 2 }{ CO } _{ 3 }$ because in the step $b$ the medium turns acidic due to formation of ${ H } _{ 2 }{ CO } _{ 3 }$ and phenolphatein will not work. On the other hand methyl orange is a basic indicator and works in the acidic medium and causes $100$% neutralisation of ${ Na } _{ 2 }{ CO } _{ 3 }$.
Answer will be $C$.

Select incorrect statement(s) among the following.

  1. Phenolphthalein is suitable indicator for the titration of HCl (aq) with $NH _4OH(aq)$

  2. An acid-base indicator in a buffer solution of $pH=pK _{ln}+1$ is ionized to the extent of $\frac {1000}{11}$%

  3. In the titration of a monoacidic weak base with a strong acid, the pH at the equivalent point is always calculated by $pH=\frac {1}{2}[pK _w-pH _b-logC]$

  4. When $Na _3PO _4(aq)$ is titrated with HCl(aq), the pH of solution at second equivalent point is calculated by $\frac {1}{2}[pK _{a _1}+pK _{a _2}]$


Correct Option: A,C
Explanation:

$A.$ Phenolphthalein gives a colour change when the $pH$ range from $8.3$ to $10$ i.e., in slightly basic solution. Titration of weak base $NH _4OH$ with strong acid $HCl$ will finally make the solution acidic, and phenolpthalein will not give colour change or denote the end point correctly.

$C.$ At equivalent point, all the weak base reacts with strong acid and the salt of this base with the strong acid is formed.
For a salt of weak base and strong acid.
$pH=7-\cfrac{1}{2}[pK _b+\log C]=\cfrac{1}{2}[pK _w-pK _b-\log C]$

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:

  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 $3.3\ g$


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 $K _{a _2}=2\times 10^{-11}$]
  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)


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:

  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


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?

  1. < $6.568$

  2. $6.568$ to $8.346$

  3. > $8.346$

  4. None of these


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.

  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)


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

  1. Methyl orange

  2. Methyl red

  3. Phenolphthalein

  4. Bromothymol blue


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:

  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


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.

  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


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.

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