Tag: properties of acids and bases

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

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, $p{ K } _{ In }=12.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

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. $4\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Phenolphthalein reacts with NaOH and half of Na2CO3. Methyl orange reacts with the remaining half of Na2CO3. Calculations based on the given volumes and normalities confirm 1g of NaOH.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. Methyl orange

  2. Self indicator

  3. Phenolphthalein

  4. Diphenylamine
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $D$.

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

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. $(y - x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

With phenolphthalein, Na2CO3 is converted to NaHCO3 (half neutralization). With methyl orange, the NaHCO3 is converted to NaCl. The volume for the second half is equal to the first, so the total volume for Na2CO3 is 2x.

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

$40\ mL$ 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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A$.
Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

$0.1\ N$ 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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$.
Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

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 $\frac {1}{2}[pK _{a _1}+pK _{a _2}]$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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.
$pH=7-\cfrac{1}{2}[pK _b+\log C]=\cfrac{1}{2}[pK _w-pK _b-\log C]$

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

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 $3.3\ g$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ NaHCO } _{ 3 }=\dfrac { 0.5\times { 10 }^{ -3 } }{ 10 } \times { 10 }^{ 3 }\times 84=4.2gm$