Chemistry

Solutions and pH Chemistry

400 Questions

Solutions and pH chemistry focus on the acidity or alkalinity of various substances, from human urine to acidic soils. Questions involve titration calculations, molarity determination, and understanding mole fractions. This topic is a staple in the chemistry sections of state and national level competitive exams.

Calculating solution molarityMole fraction calculationpH value interpretationAcidic soils and pHTitration and neutralization

Solutions and pH Chemistry Questions

Multiple choice biology properties of natural resources properties of soil soil pollution and its effects soil and soil pollution

A fertile soil is likely to have a pH value of

  1. 3.5-4

  2. 8.5-9

  3. 5.5-7

  4. 10.5-11
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Fertile soil contains sufficient minerals and nutrients needed for plant growth. Most minerals and nutrients are more soluble in slightly acidic soil than in neutral or slightly alkaline soils. The pH range of 5.5-7 is highly suitable for plant growth and hence such a soil is considered fertile. Thus the correct answer is option C.

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

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. $(IV)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(II)$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

$0.1\ N$ 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 ${ NaHCO } _{ 3 }=\dfrac { 0.5\times { 10 }^{ -3 } }{ 10 } \times { 10 }^{ 3 }\times 84=4.2gm$
Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases
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)
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

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

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)
Reveal answer Fill a bubble to check yourself
B Correct answer
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.