2 Marks Questions

Question 512 Marks

Find the conjugate base for the species $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_4{ }^{+}$.

Answer

Conjugate base of $\mathrm{H}_2 \mathrm{O}$ is $\mathrm{OH}^{-}$and of $\mathrm{NH}_4^{+}$is $\mathrm{NH}_3$ [Remove $\mathrm{H}^{+}$to get conjugate base]

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Question 522 Marks

What is the effect of reducing volume on the following system?
$2\text{C}(\text{s})+\text{O}_2\rightleftharpoons2\text{CO}(\text{g})$

Answer

On reducing volume, pressure will increase. According to Le-Chatelier principle, equilibrium will shift to the side where number of moles of gas can be increased. In the above reaction, equilibrium will shift in forward direction.

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Question 532 Marks

Will AgCl be more soluble in aqueous solution or NaCl solution and why?

Answer

In NaCl solution, the $\mathrm{Cl}^{-}$ions will increase. Since solubility product, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$remains constant. $\left[\mathrm{Ag}^{+}\right]$will decrease. Therefore, solubility of AgCl will be less in NaCl solution than in water.

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Question 542 Marks

Consider the following equilibrium:
$\text{CO}_2(\text{g})+\text{C (graphite)}\rightleftharpoons2\text{CO}(\text{g})$
Write the equilibrium expression for $\text{K}_{\text{c}}$ 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 552 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 $\left(\mathrm{H}_2 \mathrm{CO}_3\right)$ and bicarbonate ions $\left(\mathrm{HCO}_3^{-}\right)$. Small amounts of the acid or base produced from the spicy food do not disturb its pH .

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Question 562 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, $\text{k}_{\text{b}}$ increases much more than $\text{k}_{\text{f}}$. Hence, $\mathrm K$ decreases.

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Question 572 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 582 Marks

Write an expression of $\text{K}_c$ 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 $\text{CO}_2$ 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 $\text{CO}_2$ because if we increase concentration of products, rate of backward reaction will increase.

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Question 592 Marks

One millilitre solution of 0.01M HCl is added to 1L of sodium chloride solution. What will be the pH of the resulting solutions?

Answer

NaCl is neutral, it simply dilutes the HCl solution from 1mL to 1000mL so that $[\text{H}^+]=\frac{0.01}{1000}=10^{-5}\text{M}$
$\text{pH}=-\log(10^{-5})=5$

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Question 602 Marks

Give the Henderson's Hasselbalch equation for acidic buffer solution.

Answer

$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$[\text{pK}_{\text{a}}=-\log\text{K}_{\text{a}}]$

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Question 612 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 $\text{CaCl}_2$ is exothermic process.

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Question 622 Marks

The value of Kc for the reaction $\text{2HI(g)}\rightleftharpoons\text{H}_2\text{(g)}+\text{I}_2\text{(g)}$ is $1 \times 10^{-4}$
At a given time, the composition of reaction mixture is At a given time, the composition of reaction mixture is $[\mathrm{HI}]=2 \times 10^{-5} \mathrm{~mol},\left[\mathrm{H}_2\right]=1 \times 10^{-5} \mathrm{~mol}$ and $\left[I_2\right]=1 \times 10^{-5} \mathrm{~mol}$ In which direction will the reaction proceed?

Answer

$\text{Q}=\frac{[\text{H}^2][\text{I}^2]}{[\text{HI}]^2}=\frac{1\times10^{-5}\times1\times10^{-5}}{(2\times10^{-5})^2}$
$=\frac{1}{4}=0.25=2.5\times10^{-1}$
Value of $\text{K}_\text{c}=1\times10^{-4}$
Since $\text{Q}>\text{K}_\text{c},$ the reaction will proceed in backward direction.

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Question 632 Marks

Mention the conditions of temperature and pressure when gas will dissolve in liquid to maximum extent with decrease in volume and absorption of heat.

Answer

Since decrease in volume takes place, high pressure is favourable for gas to dissolve. Since absorption of heat takes place, high temperature will be suitable to dissolve maximum amount of gas.

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Question 642 Marks

How does a catalyst affect the equilibrium constant? Explain.

Answer

The equilibrium constant is not affected by catalyst. Catalyst increases the rate of forward as well as backward reaction equally and equilibrium is attained faster.

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Question 652 Marks

Write an expression for $\mathrm{K}_{\mathrm{a}}$, 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 662 Marks

What will be value of $\left(\mathrm{pK}_{\mathrm{a}}+\mathrm{pK}_{\mathrm{b}}\right)$ at $25^{\circ} \mathrm{C}$ ?

Answer

$\text{pK}_{\text{a}} + \text{pK}_{\text{b}} = \text{pK}_{\text{w}} = 14$
$[\because\text{pK}_{\text{w}}=-\log\text{K}_{\text{w}}$
$=-\log10^{-14}$
$=14\log10=14\times1=14]$

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Question 672 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 682 Marks

What will be effect on boiling point of liquid if pressure is increased?

Answer

The boiling point of liquid will increase because the vapour pressure of liquid will become equal to external pressure at higher temperature.

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Question 692 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 702 Marks

The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4 \mathrm{~AgCl}, \mathrm{AgBr}$ and Agl are respectively, $1.1 \times 10^{-12}, 1.8 \times 10^{-10}, 5.0 \times 10^{-13}, 8.3 \times 10^{-17}$, which one of the following salts will precipitate first $\mathrm{AgNO}_3$ solution is adding to a solution containing equal mole of $\mathrm{NaCl}, \mathrm{NaBs}$, NaI and $\mathrm{Na}_2 \mathrm{CrO}_4$.

Answer

Agl will precipitate first because $K_{\text {sp }}$ of Agl is lowest, therefore, ionic product will exceed the solubility products easily.

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Question 712 Marks

$\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ and for $\mathrm{CH}_3 \mathrm{NH}_2$ is $4.4 \times 10^{-4}$. Which of them is strongest base and why?

Answer

$\mathrm{CH}_3 \mathrm{NH}_2$ is strongest base because it has high value of base dissociation constant.

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Question 722 Marks

Which of the following is strongest acid?
$\text{HCl},\text{HClO}_3,\text{HNO}_3,\text{H}_2\text{SO}_4,\text{HClO}_4$

Answer

$\mathrm{HClO}_4$ is strongest acid because ' Cl ' is in +7 oxidation state. Higher the oxidation state, stronger will be acid.

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Question 732 Marks

A tank is full of water. Water is coming in as well as going out at same rate. What will happen to level of water in a tank? What is name given to such a state?

Answer

It will remain the same because rate of inflow is equal to rate of outflow. The state is called state of 'equilibrium'.

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Question 742 Marks

On the basis of the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, the pH of $10^{-8} \mathrm{~mol} \mathrm{~dm}^{-3}$ solution of HCl should be 8 . However, it is observed to be less than 7.0. Explain the reason.

Answer

Concentration $10^{-8} \mathrm{~mol} \mathrm{~dm}^{-3}$ indicates that the solution is very dilute. So, we cannot neglect the contribution of $\mathrm{H}_3 \mathrm{O}^{+}$ ions produced from $\mathrm{H}_2 \mathrm{O}$ in the solution. Total $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-8}+10^{-7} \mathrm{M}$. From this we get the value of pH close to 7 but less than 7 because the solution is acidic.
From calculation, it is found that pH of $10^{-8} \mathrm{~mol} \mathrm{~dm}^{-3}$ solution of HCl is equal to 6.96 .

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Question 752 Marks

$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g});$
$\text{K}=0.50\text{ at }673\text{K.}$
Write the equilibrium expression and equilibrium constant for reverse reaction.

Answer

$2\text{NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
$\text{K}'_{\text{c}}=\frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$ for reverse reaction
$\text{K}_{\text{c}}=0.50$
$\Rightarrow\text{K}'_{\text{c}}=\frac{1}{\text{K}_{\text{c}}}=\frac{1}{0.50}=2$

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Chemistry 2 Marks Questions