2 Marks Questions

Question 12 Marks

The species: H₂O, $\text{HCO}_3^-,$ $\text{HSO}_4^-$ and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer

The table below lists the conjugate acids and conjugate bases for the given species.

$\text{Species Conjugate acid Conjugate base}$
$\text{H}_2\text{O}$	$\text{H}_3\text{O}^+$	$\text{OH}-$
$\text{HCO}_3^-$	$\text{H}_2\text{CO}_3$	$\text{CO}_3^{2-}$
$\text{HSO}^-_4$	$\text{H}_2\text{SO}_4$	$\text{SO}_4^{2-}$
$\text{NH}_3$	$\text{NH}_4^+$	$\text{NH}_2^-$

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Question 22 Marks

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer

For the concentration of pure solid or pure liquid,
$\text{Molar conc.}=\frac{\text{Molar of the substance}}{\text{Volume of the substance}}$
Since density of pure solid or liquid is constant at constant temperature and molar mass is also constant therefore, their molar concentrations are constant and are included in the equilibrium constant.

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Question 32 Marks

Calculate the pH of the resultant mixtures:
10mL of 0.1M H₂SO₄ + 10mL of 0.1M KOH

Answer

$\text{Moles of H}_3\text{O}^+=\frac{2\times10\times0.1}{1000}=.002\text{mol}$
$\text{Moles of OH}^-=\frac{10\times.01}{1000}=0.001\text{mol}$
Excess of $\text{H}_3\text{O}^+=.001\text{mol}$
Thus, $[\text{H}_3\text{O}^+]=\frac{.001}{20\times10^{-3}}=\frac{10^{-3}}{20\times10^{-3}}=.05$
$\therefore\ \text{pH}=-\log(0.05)$
$=1.30$

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Question 42 Marks

Assuming complete dissociation, calculate the pH of the following solutions:
0.005 M NaOH

Answer

0.005M NaOH
$\text{NaOH}_\text{(aq)}\leftrightarrow\text{Na}_\text{(aq)}++\text{HO}^-_\text{(aq)}$
$[\text{H}\text{O}^-]=[\text{NaOH]}$
$\Rightarrow[\text{H}\text{O}^-]=.005$
$\text{pOH}=-\log[\text{H}\text{O}^-]=-\log(.005)$
$\text{pOH}=2.30$
$\therefore\text{pH}=14-2.30$
$=11.70$
Hence, the pH of the solution is 11.70.

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Question 52 Marks

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
What happens when equilibrium is restored finally and what will be the final vapour pressure?

Answer

Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

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Question 62 Marks

Calculate the pH of the resultant mixtures:
10mL of 0.01M H₂SO₄ + 10mL of 0.01M Ca(OH)₂

Answer

$\text{Moles of H}_3\text{O}^+=\frac{2\times10\times0.1}{1000}=.0002\text{mol}$
$\text{Moles of OH}^-=\frac{2\times10\times.01}{1000}=.0002\text{mol}$
Since there is neither an excess of $\text{H}_3\text{O}^+$ or $\text{OH}^-,$the solution is neutral. Hence, pH = 7.

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Question 72 Marks

Assuming complete dissociation, calculate the pH of the following solutions:

0.002 M HBr

Answer

0.002 M HBr
$\text{HBr}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{Br}^-$
$[\text{H}_3\text{O}^+]=[\text{HBr]}$
$\Rightarrow[\text{H}_3\text{O}^+]=.002$
$\therefore\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log(0.002)$
$=2.69$
Hence, the pH of the solution is 2.69.

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Question 82 Marks

Assuming complete dissociation, calculate the pH of the following solutions:
0.003M HCl

Answer

0.003M HCl:
$\text{H}_2\text{O}+\text{HCl}\leftrightarrow\text{H}_3\text{O}^++\text{Cl}^-$
Since HCl is completely ionized,
$[\text{H}_3\text{O}^+]=[\text{HCl].}$
$\Rightarrow[\text{H}_3\text{O}^+]=0.003$
Now,
$\text{pH}=-\log[\text{H}_3\text{O}^+]=-\log(.003)$
$=2.52$
Hence, the pH of the solution is 2.52.

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Question 92 Marks

Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
$\text{PCl}_5\text{ (g)}\rightleftharpoons\text{PCl}_3\text{ (g) + }\text{Cl}_2\text{ (g)}$

Answer

The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

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Question 102 Marks

Assuming complete dissociation, calculate the pH of the following solutions:

0.002 M KOH

Answer

0.002 M KOH:
$\text{KOH}_\text{(aq)}\leftrightarrow\text{K}_\text{(aq)}+\text{OH}^-_\text{(aq)}$
$[\text{O}\text{H}^-]=[\text{KOH]}$
$\Rightarrow[\text{O}\text{H}^-]=.002$
Now, $\text{pOH}=-\log[\text{OH}^-]$
$=2.69$
$\therefore\text{pH}=14-2.69$
$=11.31$
Hence, the pH of the solution is 11.31.

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Question 112 Marks

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
How do rates of evaporation and condensation change initially?

Answer

On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

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Question 122 Marks

At what temperature the solid and liquid are in equilibrium under 1 atm pressure?

Answer

Melting point or freezing point is a temperature at which solid and liquid are in equilibrium.

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Question 132 Marks

Why do we sweat more on humid day?

Answer

It is because on humid day Water vapours are more in air, therefore sweat does not get evaporated easily and therefore we sweat more.

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Question 142 Marks

Glycine is an a-amino acid. It exists in the form of Zwitter ion as ⁺NH₃CH₂COO^-
Write the formula of its

Conjugate acid.
Conjugate base.

Answer

Conjugate acid ⁺NH₃CH₂COOH.
Conjugate base NH₂CH₂COO^-.

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Question 152 Marks

Why is ammonia termed as a base though it does not contain OH^- ions?

Answer

The basic nature of ammonia is due to its tendency to donate electron pair. Therefore it is a Lewis base.

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Question 162 Marks

The ionisation of hydrochloric in water is given below:

$\text{HCl}\text{ (aq)}+\text{H}_2\text{O}\text{ (l)}\rightleftharpoons\text{H}_3\text{O}^+\text{ (aq)}+\text{Cl}^-{\text{ (aq)}}$

Label two conjugate acid-base pairs in this ionisation.

Answer

HCl	(Acid)
Cl^–	(Conjugate base)
H₂O	(Base)
H₃O⁺	(Conjugate acid)

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Question 172 Marks

Classify the following as homogeneous or heterogeneous equilibria:

$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
$2\text{Mg(s)}+\text{O}_2(\text{g})\rightleftharpoons2\text{MgO(s)}$

Answer

Homogeneous equilibria because all reactant and products are gases.
Heterogeneous equilibria because Mg and MgC are solid whereas O₂ is gas.

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Question 182 Marks

Ionisation constant of a weak base MOH, is given by the expression.

$\text{K}_\text{b}=\frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}$

Values of ionisation constant of some weak bases at a particular temperature are given below:

Base	Dimethylamine	Urea	Pyridine	Ammonia
K_b	5.4 × 10^–4	1.3 × 10^–14	1.77 × 10^–9	1.77 × 10^–5

Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

Answer

Greater is the ionization constant (K_b) of a base, greater is the ionization of the base. Order of extent of ionization at equilibrium is dimethylamine > ammonia > pyridine > urea. Dimethylamine is the strongest base due to maximum value of K_b.

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Question 192 Marks

$\text{CaCl}_2(\text{s})+\text{aq}\rightleftharpoons\text{CaCl}_2(\text{aq})+\text{Heat}$
Discuss the solubility if temperature is increased.

Answer

The solubility will decrease with increase in temperature because dissolution of CaCl₂ is exothermic process.

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Question 202 Marks

Consider the following equilibrium:
$\text{CO}_2(\text{g})+\text{C (graphite)}\rightleftharpoons2\text{CO}(\text{g})$
Write the equilibrium expression for K_c and calculate its units.

Answer

$\text{K}_{\text{c}}=\frac{[\text{CO}]^2}{[\text{CO}_2]}=\frac{(\text{mol L}^{-1})^2}{\text{mol L}^{-1}}=\text{mol L}^{-1}$
$\therefore[\text{C (graphite)}]=1$
$\because$ Graphite is pure solid.

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Question 212 Marks

The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride?

Answer

Sugar being a non - electrolyte does not ionize in water, whereas NaCl ionizes completely in water and produces Na⁺ and Cl^– ions which help in the conduction of electricity.
When concentration of NaCl is increased, more Na⁺ and Cl ions will be produced. Hence, conductance increases.

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Question 222 Marks

BF₃ does not have proton but still acts as an acid and reacts with NH₃. Why is it so? What type of bond is formed between the two?

Answer

In BF₃, the octet of boron is incomplete, therefore in order to complete its octet, it accepts a lone pair of electron. Any species which is capable of accepting a lone pair of electron is acidic in nature and known as Lewis acid. Hence BF₃ is a lewis acid.

When NH₃, which is having a lone pair of electron act as lewis base, reacts with BF₃ by donating its lone pair of electron forms an adduct. The bond form between the two species is known as coordinate bond,

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Question 232 Marks

K_b for NH₄OH is 1.8 × 10^-5 and for CH₃NH₂ is 4.4 × 10^-4Which of them is strongest base and why?

Answer

CH₃NH₂ is strongest base because it has high value of base dissociation constant (K_b). Higher the value of K_b stronger will be the base.

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Question 242 Marks

$\text{N}_2+\text{3H}_2\rightleftharpoons2\text{NH}_3+\text{Heat}$ What is the effect of increasing temperature on value of K?

Answer

'K' will decrease with increase in temperature because reaction is exothermic.

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Question 252 Marks

Write the conjugate acid for $\text{NH}_2^-\text{ and }\text{NH}_3.$
What is the relationship between pK_a and pK_b values?

Answer

The conjugate acid of $\text{NH}_2^-\text{ is }\text{NH}_3.$ whereas conjugate acid of $\text{NH}_3\text{ and }\text{NH}_4^+.$

[Add H^- to get conjugate acid]

$\text{pK}_{\text{a}}+\text{pK}_{\text{b}}=14,\text{pK}_{\text{a}}=14-\text{pK}_{\text{b}}$

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Question 262 Marks

Write conjugate acid and conjugate base of H₂O.

Answer

Conjugate acid is H₃O⁺ and conjugate base is OH^-. Add H⁺ to get conjugate acid and remove H⁺ to get conjugate base.

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Question 272 Marks

The K_spof Ag₂CrO₄ AgCl, AgBr and AgI are respectively, 1.1 × 10^-12, 1.8 × 10^-10, 5.0 × 10^-13, 8.3 × 10^-17, which one of the following salts will precipitate first AgNO₃ solution is adding to a solution containing equal mole of NaCl, NaBs, NaI and Na₂CrO₄.

Answer

AgI will precipitate first because K_sp of AgI is lowest, therefore, ionic product will exceed the solubility products easily.

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Question 282 Marks

Define Lewis acids and bases with example.

Answer

Lewis acids are those which can accept a pair of electrons or negatively charged ions, e.g. BCl₃. Lewis bases can donate a pair of electrons or negatively charged ions, e.g. NH₃.

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Question 292 Marks

Write an expression for K_a, for ionisation of HCN in aqueous solution. Give equation also.

Answer

$\text{HCN}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^{\oplus}+\text{CN}^-$

$\text{K}_{\text{a}}=\frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]}$

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Question 302 Marks

Which will have CO₂ to more extent, hot cold drink bottle or chilled cold drink bottle, why?

Answer

Chilled cold drink will dissolve more CO₂ because solubility of gases in liquid increases with decrease in temperature because force of attraction between gas and liquid increases.

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Question 312 Marks

$\text{A}+\text{3B}\rightleftharpoons\text{2X,}\text{K}=\text{x}$
What will be the equilibrium constant for the decomposition of 1mol of X?

Answer

$\text{K}=\frac{[\text{X}^2]}{[\text{A}][\text{B}]^3}=\text{x}$
$\text{X}\rightleftharpoons\frac{1}{2}\text{A}+\frac{3}{2}\text{B}$
$\text{K}'=\frac{[\text{A}^{\frac{1}{2}}][\text{B}]^{\frac{3}{2}}}{[\text{X}]}$
$=\frac{1}{\sqrt{\text{K}}}=\frac{1}{\sqrt{\text{x}}}$

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Question 322 Marks

The ionisation constant of formic acid is 1.8 × 10^-4. Calculate the ratio of sodium formate and formic acid in a buffer of pH 4.25.

Answer

$\text{pK}_{\text{a}}=-\log(1.8\times10^{-4})=3.74$
$\log\frac{\text{[Salt]}}{[\text{Acid}]}=\text{pH}-\text{pK}_{\text{a}}$
$=4.15-3.74=0.51$
$\text{or }\frac{\text{[Salt]}}{[\text{Acid}]}=\text{Antilog }0.51=3.24$

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Question 332 Marks

All Bronsted acids are not Lewis acids. Explain.

Answer

Bronsted acids can donate H⁺ easily but they may not be able to donate electrons e.g. HCl, H₂SO₄ HNO₃. Therefore, all Bronsted acids are not Lewis acids. They can not accept electrons.

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Question 342 Marks

The concentration of hydrogen ion in a sample of soft drink is:

$3.8\times10^{-3}\text{M}.$

What is its $\text{pH}(\log 3.8 = 0.58).$

Answer

$\text{pH}=-\log[\text{H}^+]$
$=-\log3.8\times10^{-3}$
$\text{pH}=-\log3.8-\log10^{-3}$
$=-0.579+3.0000=2.4202$

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Question 352 Marks

pK_a values of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0.
Which of them is strongest acid? Give reason.

Answer

Acid 'A' with pK_a = 1.5 is strongest acid, lower the value of pK_a stronger will be the acid. Higher the value of K_a, lower will be value of pK_a.

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Question 362 Marks

Which of the following is Lewis acid but not Bronsted acid?
$\text{HBrO}_3, \text{SbCl}_3, \text{HSO}_4^-, \text{AlF}_3$

Answer

SbCl₃ and AlF₃ are Lewis acids but not Bronsted acids because they connot donate H⁺(protons).

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Question 372 Marks

A sparingly soluble salt having general formula $\text{A}^{\text{p+}}_{\text{x}}\text{B}^{\text{q-}}_{\text{y}}$ and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

Answer

A sparingly soluble salt having general formula $\text{A}^{\text{p+}}_{\text{x}}\text{B}^{\text{q-}}_{\text{y}}.$ Its molar solubility is S mol L^-1.

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Question 382 Marks

Arrange the following in increasing order of pH.
$\text{KNO}_3(\text{aq}),\ \text{CH}_3\text{COONa(aq)},\ \text{NH}_4\text{Cl(aq)},\ \text{C}_6\text{H}_5\text{COONH}_4\text{(aq)}$

Answer

$\text{NH}_4\text{Cl}<\text{C}_6\text{H}_5\text{COONH}_4<\text{KNO}_3<\text{CROONa}$

Salts of strong acid and strong base do not hydrolyse and form neutral solution thus, pH will be nearly 7 of KNO₃. In sodium acetate, acetic acid remains unionised this results in increase in OH' concentration and pH will be more than 7. NH₄Cl formed from weak base, NH OH and strong acid, HCl, in water dissociates completely, aq. ammonium ions undergo hydrolysis with water to form NH₄OH and H⁺ions resulting in less pH value.

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Question 392 Marks

$\text{H}_2(\text{g})+\text{I}_2(\text{g})\rightleftharpoons2\text{HI}(\text{g}),\text{K}=49$
What is the value of K, for the reaction:
$\text{H}_2(\text{g})\rightleftharpoons\frac{1}{2}\text{H}_2(\text{g})+\frac{1}{2}\text{I}_2(\text{g})?$

Answer

$\text{K}=\frac{[\text{HI}^2]}{[\text{H}_2][\text{I}_2]}$
$\text{K}=49$
$\text{K}'=\frac{[\text{H}_2]^{\frac{1}{2}}[\text{I}_2]^{\frac{1}{2}}}{[\text{HI}]}=\frac{1}{\sqrt{\text{K}}}$
$\Rightarrow\text{K}'=\frac{1}{\sqrt{49}}=\frac{1}{7}$

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Question 402 Marks

$\text{A}+\text{B}\rightleftharpoons\text{AB};\text{K}=1\times10^2$
$\text{E}+\text{F}\rightleftharpoons\text{EF};\text{K}=1\times10^{-3}$
Out of AB and EF, which one is more stable AB or EF?

Answer

AB is more stable. Higher the value of K more will be the stability of product formed. More stable product is formed to more extent.

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Question 412 Marks

Predict the nature of solution when NH₄NO₃ undergo hydrolysis.

Answer

NH₄NO₃ on hydrolysis gives NH₄OH (weak base), $\text{H}^+\text{ and }\text{NO}^-_3$ (HNO₃ is strong acid) since H⁺ ions are more than OH^-.
$\therefore$ Solution is acidic.

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Question 422 Marks

'All Lewis bases are also Bronsted bases'. Is it true? If yes, Why?

Answer

Yes, it is true. It is because Lewis bases are- vely charged or electron rich. They are Bronsted bases also because they can accept H⁺ easily, e.g., NH₃ can donate electron (Lewis base) and accept H⁺ (Bronsted base).

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Question 432 Marks

$\text{Hb(s)}+\text{O}_2\text{(g)}\rightleftharpoons\text{HbO}_2\text{(s)}$
Predict the direction in which equilibrium gets shifted it partial pressure of O₂ is lowered.

Answer

The reaction will shift to backward direction, e.g., in tissues partial pressure of O₂ is less, oxy-haemoglobin releases oxygen.

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Question 442 Marks

pK_a value of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0. Which of them is strongest acid?

Answer

Acid A with pK_a = 1.5 is strongest acid, lower the value of pK_a stronger will be the acid.

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Question 452 Marks

Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$\text{OH}^–,\text{ RO}^–,\text{ CH}_3\text{COO}^–,\text{ Cl}^–$

Answer

Conjugate acids of given bases are:
$\text{H}_2\text{O},\text{ ROH},\text{ CH3COOH},\text{ HCl}.$
Their acidic strength is in the order.
$\text{HCl}>\text{CH}_3\text{COOH}>\text{H}_2\text{O}>\text{ROH}$ Hence, basic strength is in the order $\text{RO}^->\text{OH}^->\text{CH}_3\text{COO} \xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{Cl}^- > $

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Question 462 Marks

Find the conjugate base for the species H₂O and NH₄⁺.

Answer

Conjugate base of H₂O is OH^- and of $\text{NH}^+_4$ is NH₃[Remove H⁺ to get conjugate base]

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Question 472 Marks

Why pH of our blood remains almost constant at 7.4 though we quite often eat spicy food?

Answer

Blood is a buffer containing carbonic acid (H₂CO₃) and bicarbonate ions (HCO^-₃). Small amounts of the acid or base produced from the spicy food do not disturb its pH.

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Question 482 Marks

For an exothermic reaction, what happens to the equilibrium constant if temperature is increased?

Answer

$\text{K}=\frac{\text{k}_{\text{f}}}{\text{k}_{\text{b}}}$ In exothermic reaction, with increase of temperature, k_b increases much more than k_f. Hence, K decreases.

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Question 492 Marks

Ice melts slowly at higher altitudes. Explain why?

Answer

Ice(s) → Water
The melting of ice is favoured at high pressure because there is decrease in volume in the forward reaction. Since at high altitudes, atmospheric pressure is low and therefore, ice melts slowly.

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Question 502 Marks

Write an expression of K_c for the following reaction:
$\text{CaCO}_3(\text{s})\rightleftharpoons\text{CaO(s)}+\text{CO}_2(\text{g})$
What is the effect of increasing concentration of CO₂ on direction of reaction?

Answer

$\text{K}_{\text{c}}=[\text{CO}_2]$

$\because[\text{CaCO}_3(\text{s})]=1$

The rate of backward reaction will increase with the increase in concentration of CO₂ because if we increase concentration of products, rate of backward reaction will increase.

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