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 medicinal chemistry antacids and antihistamines therapeutic action of different classes of drugs ph regulation of the stomach

An antacid tablet weighing 1 g containing aluminium hydroxide as the only basic substance and the rest of its components being neutral, was dissolved in 200 mL of 0.1 M HCI. The excess HCI was back titrated and required  90 mL of 0.1 N base for exact neutralization. Mill equivalents of aluminum hydroxide in the sample of antacid tablet is 

  1. 9

  2. 11

  3. 12

  4. 20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total milliequivalents of HCl = 200 mL * 0.1 M = 20 meq. The excess HCl neutralized by 90 mL of 0.1 N base = 9 meq. Therefore, the HCl consumed by the antacid = 20 - 9 = 11 meq. Since aluminum hydroxide is the only base, it must have neutralized 11 meq of HCl.

Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

Aqueous solution of ortho-boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of :

  1. trans-glycerol.

  2. catechol.

  3. cis-glycerol.

  4. both (b) and (c).
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Aqueous solution of ortho-boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of catechol and cis-glycerol.

Ortho-boric acid forms a stable cyclic complex with polyhydroxy compounds like catechol and cis-glycerol. This helps in the release of $H^+$ as $H _3O^+$ and therefore boric acid acts as a strong acid and hence can be titrated against NaOH with phenolphthalein.

Hence,option D is correct.

Multiple choice chemistry introduction to analytical chemistry different types of solutions various mixtures introduction to solutions

What volume would you dilute 0.2 L of a 15 M solution to obtain a 3 M solution?

  1. 1L

  2. 225L

  3. 10L

  4. 0.4L

  5. 0.1L
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that :

$\displaystyle M _1V _1 = M _2V _2 $ 

Given :

$\displaystyle  M _1 = 15 $  M
$\displaystyle V _1 = 0.2 $ L
$\displaystyle  M _2 =3$  M
$\displaystyle  V _2 =$  ????

$\displaystyle  15 \times 0.2 = 3 V _2$
Thus, $\displaystyle  V _2 = 1$  L

Hence, the correct option is A.
Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

What is the pH number of hydrochloric acid?

  1. Below 7

  2. Above 7

  3. 7

  4. Between 8 and 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pH of acids is given by the pH scale from 1-7. Strength of acids increases by moving down the scale from 7 to 1. High acidity ranks as 1 on the pH scale and 7 are neutral. HCl is highly acidic so it has pH less than 7 while on this scale basicity increases going from 7 to 14. So all the acids have pH under 7.


Option A is correct.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

Equations relating to acidic properties of an aq. solution of HCl gas is/are :

  1. $HCl\left( g \right) +{ H } _{ 2 }O\rightleftharpoons { H } _{ 3 }{ O }^{ + }+{ Cl }^{ - }$
  2. $Zn\left( OH \right) _{ 2 }+2{ H }Cl\rightarrow { ZnCl } _{ 2 }+2{ H } _{ 2 }O$
  3. $ { MnO } _{ 2 }+4HCl\rightarrow { MnCl } _{ 2 }+{ Cl } _{ 2 }+2{ H } _{ 2 }O$

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

1) $HCl\left( g \right) +{ H } _{ 2 }O\rightleftharpoons { H } _{ 3 }{ O }^{ + }+{ Cl }^{ - }$
2) ${ 2KMnO } _{ 4 }+16HCl\rightarrow { 2MnCl } _{ 2 }+{ 8H } _{ 2 }O+2KCl+{ 5Cl } _{ 2 }$
3) $ { MnO } _{ 2 }+4HCl\rightarrow { MnCl } _{ 2 }+{ Cl } _{ 2 }+2{ H } _{ 2 }O$
4) $Zn\left( OH \right) _{ 2 }+{ H }Cl\rightarrow { ZnCl } _{ 2 }+{ H } _{ 2 }O$

All of the above equation shows the acidic properties of hydrochloric acid.

Multiple choice chemistry substances in common use preparation, properties and uses of baking soda chemical from common salt compounds of carbon

'Sodium bicarbonate solution has $pH$ less than 7.'
State True or False:

  1. True

  2. False

  3. Cannot be determined

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sodium bicarbonate is amphoteric, reacting with acids and bases. It reacts violently with acids, releasing CO2 gas as a reaction product. Because of $NaHCO _3$ is a weak base the pH value should be greater than 7.

Reaction with sulfuric acid: 
$2 NaHCO _3 + H _2SO _4 \rightarrow Na _2SO _4 + 2 H _2O + 2 CO _2$ 

With sodium hydroxide: 
$NaHCO _3 + NaOH \rightarrow Na _2CO _3 + H _2O$

Hence, statement is false.

Multiple choice common laboratory equipments common laboratory apparatus and equipments laboratory equipments know about some common gases chemistry

A solution of glucose received from some research laboratory has been marked mole fraction x and molality (m) at $1{ 0 }^{ \circ  }C$. When you will calculate its molality and mole fraction in your laboratory at $24^{ \circ  }C$ you will find:

  1. mole fraction (x) and molality (m)

  2. mole fraction (2x) and molality (2m)

  3. mole fraction (x/2) and molality (m/2)
  4. mole fraction (x) and (m$ _{ - }^{ + } $dm) molality
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molality and mole fractions both are independent of temperature . Therefore they will remain same after change of temperature.

Multiple choice chemistry endothermic and exothermic reactions endothermic reactions what are enthalpy changes energy change in chemical reactions

A solution of 500 ml of 0.2 M KOH and 500 ml of 0.2 M HCl is mixed and stirred; the rise in temperature is $T _1$. The experiment is repeated using 250 ml of each solution, the temperature raised is $T _2$. Which of the following is true :

  1. $T _1=T _2$
  2. $T _1=2T _2$
  3. $T _1=4T _2$
  4. $T _2=9 T _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The neutralization reaction between a strong acid (HCl) and a strong base (KOH) is exothermic. The temperature rise depends on the heat released per mole of water formed; since the concentration and molar ratios remain the same in both experiments, the temperature rise T1 and T2 should be identical.

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

The concentration of acetate ions in $1 M$ acetic acid $(K _{a} = 2 \times 10^{-5})$ solution containing $0.1 M - HCl$ is

  1. $2 \times 10^{-1}$
  2. $2 \times 10^{-3}$
  3. $2 \times 10^{-4}$
  4. $4.4 \times 10^{-3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the presence of a strong acid (HCl), the dissociation of acetic acid is suppressed by the common ion effect. The concentration of H+ is dominated by HCl (0.1 M). Ka = [H+][CH3COO-] / [CH3COOH]. 2e-5 = (0.1 * [CH3COO-]) / 1.0. Thus, [CH3COO-] = 2e-4 M.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The % of hydrogen in water and hydrogen peroxide is $11.2$ % and $5.94$ % respectively. This illustrates the law of:

  1. constant proportions

  2. conservation of mass

  3. multiple proportions

  4. law of gaseous volume.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The law of multiple proportions say that if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. 

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Hydrogen peroxide and water contain $5.93$% and $11.2$% of hydrogen respectively. The data illustrates the law of:

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For hydrogen peroxide, 100 g of sample will contain 5.93 g hydrogen and  $\displaystyle 100 - 5.93 = 94.07$ g oxygen respectively.

The ratio of the mass of oxygen to the mass of hydrogen in hydrogen peroxide is  $\displaystyle \dfrac {94.07}{5.93} = 15.86$

For water, 100 g of sample will contain 11.2 g hydrogen and  $\displaystyle 100 - 11.2 = 88.8$ g oxygen respectively.

The ratio of the mass of oxygen to the mass of hydrogen in hydrogen peroxide is  $\displaystyle \dfrac {88.8}{11.2} = 7.93$

The two ratios are in the proportion $\displaystyle \dfrac {15.86}{7.93} = 2:1$

Hence, this illustrates the Law of multiple proportions. According to this law, if two elements chemically combine with each other forming two or more compounds with different compositions by mass then the ratios of masses of two interacting elements in the two compounds are small whole numbers.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

$100 ml$ of $0.2\ M\ H _{2}SO _{4}$ is reacted with $100\ ml$ of $0.5\ M\ NaOH$ solution. what is the normality of the solution 

  1. 0.3N

  2. 0.8N

  3. 0.1N

  4. 1N
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$1M-H _2SO _4=2N-H _2SO _4$

100ml of 0.2M 0.2M $H _2SO _4\equiv 100 \times 0.2$ml of 1M $H _2SO _4$

$\equiv 20ml$ of 2N $H _2SO _4$

$\equiv 40ml$ of 2N $H _2SO _4$

$1M NaOH=1N NaOH$

100ml of 0.2M 0.2M $NaOH\equiv 100 \times 0.2$ml of 1M $NaOH$

$\equiv 20ml$ of 1N $NaOH$

neutralisation occurs when acid and base are mixed due to the formation of salt and water.

20ml of 1N $NaOH\equiv $ 20ml of 1N $H _2SO _4$

$20ml \times 1N=200ml \times $ final strength of acid

therefore the normality of solution is $0.1N$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The first and second dissociation constant of an acid ${ H } _{ 2 }A$ are $1.0\ \times \ { 10 }^{ -5 }$ and $5.0\ \times \ { 10 }^{ -10 }$ respectively. The over all dissociation constant of the acid will be:

  1. $5.0\ \times \ { 10 }^{ -5 }$
  2. $5.0\ \times \ { 10 }^{ 15 }$
  3. $5.0\ \times \ { 10 }^{ -15 }$
  4. $0.2\ \times \ { 10 }^{ 5 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$H _2A\overset {K _1}{\rightleftharpoons} HA^-+H^+$


$\Rightarrow K _1=\cfrac {[HA^-][H^+]}{[H _2A]}$  $\longrightarrow (1)$


$HA^-\overset {K _2}{\rightleftharpoons} H^++A^{2-}$

$\Rightarrow K _2=\cfrac {[H^+][A^{2-}]}{[HA^-]}$    $\longrightarrow (2)$

Overall dissociation constant $K$

$\Rightarrow K=\cfrac {[H^+]^2[A^{2-}]}{[H _2A]}=K _1\times K _2$

$=1\times 10^{-5}\times 5\times 10^{-10}$

$=5\times 10^{-15}$ .

Multiple choice separation of components of mixtures methods of separation elements, compounds and mixtures chemistry

The volume of water that would convert $10\ ml$ of decamolar $HCl$ solution to decamolar solutions is 

  1. $10^{3}\ ml$
  2. $10^{2}\ ml$
  3. $9.9\times 10^{3}\ ml$
  4. $1.2\ L$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Milli moles of conc.HCl=milli moles of dil HCl
10×10=V×11010×10=V×110⇒V=1000ml⇒V=1000ml
Thus 990ml of water should be added to 10ml on conc.HCl to get decinormal solution.
Hence (a) is the correct answer.