Chemistry

Solutions and pH Chemistry

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

Addition of ${NH} _{4}Cl$ does not effect the $pH$ of solution of ${NH} _{4}OH$. 

  1. True

  2. False

  3. Ambigous

  4. None

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.


So addition of ${NH} _{4}Cl$ reduces dissociation of ${NH} _{4}OH$ and because of that its pH will decreases.

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

(A) pH of $10^{-7}$ M NaOH solution exists between 7 to 7.3 at $25^o C$.
(R) Due to common ion effect ionization of water is suppressed. 

  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

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A) $pH$ of ${ 10 }^{ -7 }$ $M$ $NaOH$ will be slightly greater than $7$. It is between $7$ to $7.3$ at ${ 25 }^{ 0 }C$ due to presence of very small quantity of excess ${ OH }^{ - }$.
R) Due to common ion effect ionisation of water is suppressed.
     ${ H } _{ 2 }O\rightleftharpoons { H }^{ + }+{ OH }^{ - }$
So in presence of common ion ${ H }^{ + }$ or ${ OH }^{ - }$ the ionisation of ${ H } _{ 2 }O$ will be decreased.
So both $(R)$ and $(A)$ are true and reason is the correct explanation of assertion.
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

By adding which of the following in 1 L 0.1 M solution of HA, $(Ka=10^{-5})$, the degree of dissociation of HA decreases appreciably?

  1. ${10^{-3} M \: HCl, 1\; L}$
  2. ${0.5 M \: HX (Ka=2\times 10^{-6}), 1 \: L}$
  3. ${0.1 M \: HNO _{3}, 1\: L}$
  4. All of these

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

$HA,  \alpha =\sqrt{\displaystyle\frac{10^{-5}}{0.1}}=0.01$ 
(A) On adding $HCl, \left [ HA \right ]=\frac{0.1}{2},\left [ HCl \right ]=\displaystyle\frac{10^{-3}}{2}$
$HCl \rightarrow H^{+}+Cl^{-}$
$HA \rightleftharpoons H^{+}  +A^{-}$
$\displaystyle\frac{0.1}{2}\left ( 1- \alpha  \right )\left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) \displaystyle\frac{0.1 \alpha }{2} $
$\Rightarrow 10^{-5}=\displaystyle\frac{\displaystyle\frac{0.1\alpha }{2} \times \left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) }{\displaystyle\frac{0.1}{2}\left ( 1-\alpha  \right )}$
Neglecting $\alpha $
$10^{-5}=\alpha \times \left ( \displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1\alpha  }{2} \right )$
$\Rightarrow 0.1 \alpha ^{2}+10^{-3} \alpha -2\times 10^{-5}=0$
$\Rightarrow  \alpha ^{2}+10^{-2}\alpha -2\times 10^{-4}=0$
$\Rightarrow \alpha =\displaystyle\frac{-10^{-2}+\sqrt{10^{-4}+8\times 10^{-4}}}{2}=\displaystyle\frac{2\times 10^{-2}}
{2}=2$
Which is similiar to initial.
(B) In case of two weak acids
$\left [\mathrm H^{+}  \right ]=\sqrt{10^{-5}\times \frac{0.1}{2}+2\times 10^{-6}\times \displaystyle\frac{0.5}{2}}$
$=\sqrt{\displaystyle\frac{10^{-6}}{2}+\displaystyle\frac{10^{-6}}{2}}=10^{-3}$
$\alpha =\displaystyle\frac{10^{-5}}{10^{-3}}=10^{-2}$
which is same as earlier
(C) $HNO _{3}\rightarrow H^{+}+NO _{3}^{-}$
$\displaystyle\frac{0.1}{2}       \displaystyle\frac{0.1M}{2}$
$HA \rightleftharpoons H^{+}   +  A^{-}$
$\displaystyle\frac{0.1M}{2}\left ( 1-\alpha  \right )  \displaystyle\frac{0.1}{2}+\displaystyle\frac{0.1}{2}\alpha                
\displaystyle\frac{0.1}{2}\alpha$
$\Rightarrow 10^{-5}=\displaystyle\frac{\frac{0.1\alpha }{2}\times \displaystyle\frac{0.1 }{2} \left (1+\alpha  \right )  }{\displaystyle\frac{0.1}{2}\left (1-\alpha  \right )}$
Neglecting $\alpha $ due to common ion effect.
$\Rightarrow 10^{-5}=\displaystyle\frac{0.1 \alpha }{2}$
$\Rightarrow \alpha =2\times 10^{-4}$
Hence, it decreases from $0.01  to  2\times 10^{-4}$.

Multiple choice biology soil : our life soil properties soil - properties basics of soil pollution

The pH of a fertile soil is usually around

  1. 2-3

  2. 6-7

  3. 8-10

  4. 11-12

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Plants grow best in neutral or slightly acidic pH. Slight acidity favours tree growth and forms forests. Slight alkalinity is however, helpful in growth of grasslands and some crop plants like legumes. Hence, for most plants a pH value of 6 to 7 is favourable.

Multiple choice biology soil : our life soil properties soil - properties basics of soil pollution

pH of a normal fertile soil is

  1. 4 - 5

  2. 6 - 7

  3. 7.2 - 9.0

  4. None of the above

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Soil pH is in the range of 6 to 7 for most plants but some prefer acid or alkaline conditions. At this pH soil organic matter and structure is very good. A lot of micro-organisms, that support plant growth can live in the soil at this pH.

Multiple choice biology soil : our life soil properties soil - properties basics of soil pollution

A fertile soil is likely to have a pH value of

  1. 3-4

  2. 8-9

  3. 6-7

  4. 10-11

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

Soil pH is important because it influences several soil factors affecting plant growth, such as (1) soil bacteria, (2) nutrient leaching, (3) nutrient availability, (4) toxic elements and (5) soil structure. Bacterial activity that releases nitrogen from organic matter and certain fertilizers is particularly affected by soil pH, because bacteria operate best in the pH range of 5.5 to 7.0. Plant nutrients leach out of soils with a pH below 5.0 much more rapidly than from soils with values between 5.0 and 7.5. Plant nutrients are generally most available to plants in the pH range 5.5 to 6.5. For most plants, the optimum pH range is from 5.5 to 7.0.

Therefore, the correct answer is option D. 

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

A sample of municipal water contains one part of urea (molecular wt $=60$) per million parts of water by weight. The number of urea molecules in a drop of water of volume $0.05\ ml$ is 

  1. $2.5\times 10^{14}$
  2. $5\times 10^{14}$
  3. $5\times 10^{13}$
  4. $5\times 10^{15}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1 \ ppm = 1 \ mg/L = 10^{-3} g/L$

Water sample contains $1 \ ppm$ urea concentration.

$\therefore \ 10^{-3} \ g$ of urea in $ 1 \ L$

$x \ g$ urea in $0.05 \times 10^{-3} \ L$

$x= 0.05 \times 10^{-6} \ g$

$60 \ g$ urea $=6.023 \times 10^{23} \ molecules$

$0.05 \times 10^{-6} \ g = n \ molecules$

$n = \cfrac {6.023 \times 10^{23} \times 0.05 \times 10^{-6}}{60}$

$=0.005 \times 10^{17}$

$=5 \times 10^{14} \ molecules$

Multiple choice business organisation and correspondence environmental pollution and industries air pollution and its effects the environment and us environment and conservation

Lead concentration of blood is considered alarming at

  1. $4-6\mu m/100ml$
  2. $10\mu g/100ml$
  3. $20\mu g/100ml$
  4. $30\mu g/100ml$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Lead is the major heavy metal pollutant released by the vehicular exhausts. It persists in the air and precipitates with the rains. In reaches the human body when consumed through water or inhaled as suspended particle. At the concentration of 30μg/100ml in blood, it is considered poisoning as it can cause aplastic anaemia, tumour and cancer.

Hence, the correct answer is '30μg/100ml'.

Multiple choice business organisation and correspondence environmental pollution and industries air pollution and its effects the environment and us environment and conservation

Lead concentration in blood is considered alarming, if it is

  1. $4-6\mu g/100ml$
  2. $10\mu g/100ml$
  3. $20\mu g/100ml$
  4. $30\mu g/100ml$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Lead is the major heavy metal pollutant released by the vehicular exhausts. It persists in the air and precipitates with the rains. In reaches the human body when consumed through water or inhaled as suspended particle. At the concentration of 30μg/100ml in blood, it is considered poisoning as it can cause aplastic anaemia, tumour and cancer.

Hence, the correct answer is '30μg/100ml'.

Multiple choice chemistry chemistry in daily life antimicrobials and antifertility drugs therapeutic action of different classes of drugs drugs and their classification

The $pK _{a}$ of acetylsalicylic acid (aspirin) is $3.5$. The $pH$ of gastric juice in human stomach is about $2$ to $3$ and the $pH$ in the small intestine is about $8$. Aspirin will be

  1. unionized in the small intestine as well as in the stomach.

  2. completely ionized in the small intestine as well as in the stomach.

  3. ionized in the stomach and almost unionized in the small intestine.

  4. ionized in the small intestine and almost unionized in the stomach.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Aspirin is a weak acid with pKa = 3.5. In the stomach (pH 2-3), pH < pKa, so it remains mostly unionized (protonated). In the intestine (pH 8), pH > pKa, so it becomes ionized (deprotonated).

Multiple choice bio-chemistry unit of living - cell eukaryotic cell kinds of cell introduction to eukaryotic cell

In a $50 g$ living tissue, the amount of water would be

  1. $15 - 25 g$
  2. $25 - 30 g$
  3. $35 - 45 g$
  4. $70 - 90 g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Water present in the human body supports the metabolic reactions in the body. The enzymes which act as a catalyst in the biological reaction acts efficiently in presence of water. The salts like sodium and potassium which dissolves in the water act as a transporter. It helps in the transmission of the nerve impulse. The intercellular transports are also supported by water.
The total amount of water present in the human body is 80% out of which 65-70% is present in the living cells. 35-45 g counts to 65% - 70% (35/50*100).
So, the correct answer is option C.
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the resulting solution will be

  1. $2.11$ g/cc
  2. $1.11$ g/cc
  3. $12.2$ g/cc
  4. $20.4$ g/cc
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density of Brine$=\rho _{b}=1.2g/cc$

Volume of Brine$=v _{b}=40cc$
Mass of Brine$=m _{b}=\rho _{b}\times v _{b}=1.2\times 40=48g$
Density of Water$=\rho _{w}=1g/cc$
Volume of Water$=v _{w}=30cc$

Mass of Water$=m _{w}=\rho _{w}\times v _{w}=1\times 30=30g$
Density of mixture$=\dfrac{\text{Mass of mixture}}{\text{ Volume of mixture}}=\dfrac{m _{b}+m _{w}}{v _{b}+v _{w}}=\dfrac{48+30}{40+30}=\dfrac{78}{70}=1.11g/cc$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Brine has a density of $1.2  {g}/{cc}$. $40  cc$ of it are mixed with $30  cc$ of water. The density of solution is

  1. $2.11 {g}/{cc}$
  2. $1.11 {g}/{cc}$
  3. $12.2 {g}/{cc}$
  4. $20.4 {g}/{cc}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
given,  ${ \rho } _{ a }=1.2\quad 9/cc\quad \quad \quad { V } _{ a }=40cc$
                 ${ \rho } _{ b }=1g/cc\quad \quad \quad { V } _{ b }=30cc$

$ \rho _{mixture} = \dfrac { { \rho } _{ a }{ V } _{ a }+{ \rho } _{ b }{ V } _{ b } }{ { V } _{ a }+{ V } _{ b } } $

So  $ \rho _{mixture} = \dfrac { 1.2\times 40+1\times 30 }{ 40+30 } =\dfrac { 78 }{ 70 } $

        $\boxed { \rho _{mixture}=1.11\quad g/cc } $