Equilibrium Test

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Equilibrium Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The concentration of $$[H^{+}]$$ and concentration of $$[OH^{-}]$$ of a $$0.1\ M$$ aqueous solution of $$2$$M ionised weak acid is: [ionic product of water $$=1\times 10^{-14}]$$.

A
$$0.02\times 10^{-3}M$$ and $$5\times 10^{-11}M$$

B
$$1\times 10^{-3}M$$ and $$3\times 10^{-11}M$$

C
$$2\times 10^{-3}M$$ and $$5\times 10^{-12}M$$

D
$$3\times 10^{-2}M$$ and $$4\times 10^{-13}M$$

Solution

Degree of ionization,
$$\alpha = \cfrac{{\left[ {{H^ + }} \right] \times 100}}{{\left[ {HA} \right]}}$$

$$\left[ {{H^ + }} \right] = \cfrac{{\alpha \times \left[ {HA} \right]}}{{100}}$$

$$ = \cfrac{{2 \times 0.1}}{{100}}$$

$$\left[ {{H^ + }} \right] = 2 \times {10^{ - 3}}M$$.

$${K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$$

$${10^{ - 14}} = 2 \times {10^{ - 3}} \times \left[ {O{H^ - }} \right]$$

$$\left[ {O{H^ - }} \right] = \frac{{{{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}}$$

$$\left[ {O{H^ - }} \right] = 5 \times {10^{ - 12}}M$$.
Question 2
1 / -0

Buffer solution $$A$$ of a weak monoprotic acid and its sodium salt in the concentration ratio $$x : y$$ has $$pH = (pH)$$,. Buffer solution $$B$$ of the same acid and its sodium salt in the concentration ratio $$y : x$$ has $$pH = (pH)_{2}$$. If $$(pH)_{2} - (pH)_{1} = 1$$ unit and $$(pH)_{1} + (pH)_{2} = 9.5\ units$$, then:

A
$$pK_{a} = 4.75$$

B
$$\dfrac {x}{y} = 2.36$$

C
$$\dfrac {x}{y} = 3.162$$

D
$$pK_{a} = 5.25$$

Solution

$$(pH)_1= pK_a+\log \cfrac {y}{x}$$
$$\Rightarrow (pH)_2=pK_a+\log \cfrac {x}{y}$$
$$\Rightarrow (pH)_2-(pH)_1=1=\log \cfrac {x}{y}-\log \cfrac {y}{x}= 2 \log \cfrac {x}{y}$$
$$\therefore \log \cfrac {x}{y}=\cfrac {1}{2} \Rightarrow \cfrac {x}{y}= 3.162$$
$$(pH)_1+(pH)_2=9.5=2pK_a+\log \cfrac {y}{x}\times \cfrac {x}{y}=2pK_a+0$$
$$\therefore pK_a=4.75$$
Question 3
1 / -0

In a vessel containing $$SO_{3}$$, $$SO_{2}$$ and $$O_{2}$$ at equilibrium, some helium gas is introduced so that total pressure increases while temperature and volume remain the same. According to Le Chatelier's principle the dissociation of $$SO_{3}$$ _________.

A
increases

B
decreases

C
remains unaltered

D
changes unpredictably

Solution

At constant volume, the partial pressures of the $$3$$ components can remain constant & equilibrium will be unchanged. Adding $$He$$ gas will increase the pressure of the system.
Question 4
1 / -0

At a temperature under high pressure $$K_w(H_2O) \, = \, 10^{10}$$, a solution of pH 5.4 is said to be:

A
Acidic

B
Basic

C
Neutral

D
Amphitoric
Question 5
1 / -0

To prepare a buffer of pH 8.26 amount of $$({ NH }_{ 4 }{ ) }_{ 2 }{ SO }_{ 4 }$$ to be added to 500 mL of 0.01 M $${ NH }_{ 4 }OH$$ solution is : $$[pK_{ a }({ NH }_{ 4 }^{ + })\ =9.26]$$

A
0.05 mole

B
0.025 mole

C
0.10 mole

D
0.005 mole

Solution

Let $$(NH_4)_2SO_4$$ to be added to 500 mL solution = x mol

$$[NH^{+}_4]=4xM$$

$$pH=8.26$$

$$pOH=14−8.26=5.74$$

$$pK_b=14−9.26=4.74$$

$$pOH=pK_b+log\dfrac{[NH^{+}_4]}{[NH_4OH]}$$

$$5.74=4.74+log\dfrac{4x}{0.01}$$

$$log4x\times 10^2 = 1$$

$$4x\times 10^2=1$$

$$x=\dfrac{1}{40}\ mol = 0.025\ mol$$
Question 6
1 / -0

Solid ammonium carbamate $$(NH_{2}COONH_{4})$$ dissociates as:

$$NH_{2}COONH_{4}(s)\rightarrow 2NH_3(g) + CO_2(g)$$

In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of $$NH_3$$ at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.

A
$$\dfrac{31}{27}$$

B
$$\dfrac{27}{31}$$

C
$$\dfrac{2}{27}$$

D
$$\dfrac{4}{27}$$
Question 7
1 / -0

The equilibrium constant $$K_{P}$$ for the reaction $$H_{2}(g) + CO_{2}(g) \rightleftharpoons H_{2}O(g) + CO(g)$$ is $$4.0$$ at $$1660^{\circ}C$$. Initially $$0.80\ mole\ H_{2}$$ and $$0.80\ mole\ CO_{2}$$ are injected into a $$5.0\ litre$$ flask. What is the equilibrium concentration of $$CO_{2}(g)$$?

A
$$0.533\ M$$

B
$$0.0534\ M$$

C
$$5.34\ M$$

D
$$None\ of\ these$$

Solution

$$\underset { 0.8-x }{ H_{2}(g) } \underset { 0.8-x }{ CO_{2}(g) } \rightleftharpoons \underset { x }{ H_{2}O(g) }+ \underset { x }{CO(g)} $$
moles of eqsn
conc. at eqm $$\dfrac{0.8-x}{5} \left(\dfrac{0.8-x}{5} \right) \left(\dfrac{x}{5}\right) \left(\dfrac{x}{5}\right)$$
$$\because \Delta n_g=0$$
$$\therefore K_c=\dfrac{\left(\dfrac{x}{5}\right)^2}{\left(\dfrac{0.8-x}{5}\right)^2}$$
$$\Rightarrow 2=\dfrac{x}{0.8-x}$$
$$\Rightarrow x=0.533$$
$$[CO_2(g)]=\dfrac{0.8-0.533}{5}$$
Question 8
1 / -0

A buffer solution made up of $$BOH$$ and $$BCl$$ of total molarity 0.29 M has $$pH = 9.6$$ and $${ K }_{ b }=1.8\ \times \ { 10 }^{ -5 }$$. Concentration of salt and base respectively is:

A
0.09 M and 0.2 M

B
0.2 M and 0.09 M

C
0.1 M and 0.19 M

D
0.19 M and 0.1 M

Solution

$$pH+pOH=14$$

$$\therefore\ pOH=14-9.6=4.4$$

$${for\ basic\ buffer}=p{OH}=p{K_b}+\log \dfrac {[salt]}{[Base]}$$

$$4.4=-\log (1.8\times 10^{-5})+\log \dfrac {[salt]}{[Base]}$$

$$4.4=4.74+\log \dfrac {[salt]}{[Base]}$$

$$-0.34=\log \dfrac {[salt]}{[Base]}$$

$$ {[Base] =2.18\ [salt]}$$

Now,
Total molarity $$= [salt] +[Base]= 0.29\ M$$

$$[salt]\ 3.18=0.29\ M$$

$$salt=0.09\ M$$

$$\therefore \ Base =0.29-0.09=0.2\ M$$
Question 9
1 / -0

Which one of the following mixture does not act as a buffer solution?

A
Boric acid and borax

B
Sodium phosphate & disodium hydrogen phosphate

C
Sodium propionate and propionic acid

D
Sodium acetate and sodium propionate

Solution

An acidic buffer contains equimolar quantities of weak acid and its salt with strong base. A basic buffer contains equinolar quantities of weak base and its salt with strong acid. Sodium acetate is a salt with strong base but sodium propionate is not weak acid, it is also a salt.
Question 10
1 / -0

For preparing a buffer solution of $$pH=7.0$$, which buffer system you will choose:

A
$$H_{3}PO_{4},H_{2}PO_4{^-}$$

B
$$H_{2}PO_4{^-},\ HPO^{2-}_{4}$$

C
$$HPO^{2-}_{4},PO^{3-}_{4}$$

D
$$H_{3}PO_{4},PO^{3-}_{4}$$

Solution

Solution:- (B) $${{H}_{2}P{O}_{4}}^{-}, {HP{O}_{4}}^{2-}$$
$${{H}_{2}P{O}_{4}}^{-}$$ is a weak acid, thus it's conjugate base will be a strong base and thus will form a buffer of $$pH$$ approx $$7$$.

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Equilibrium Test - 35

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Equilibrium Test - 35

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Question 1

$$0.02\times 10^{-3}M$$ and $$5\times 10^{-11}M$$

$$1\times 10^{-3}M$$ and $$3\times 10^{-11}M$$

$$2\times 10^{-3}M$$ and $$5\times 10^{-12}M$$

$$3\times 10^{-2}M$$ and $$4\times 10^{-13}M$$

Solution

Question 2

$$pK_{a} = 4.75$$

$$\dfrac {x}{y} = 2.36$$

$$\dfrac {x}{y} = 3.162$$

$$pK_{a} = 5.25$$

Solution

Question 3

increases

decreases

remains unaltered

changes unpredictably

Solution

Question 4

Acidic

Basic

Neutral

Amphitoric

Question 5

0.05 mole

0.025 mole

0.10 mole

0.005 mole

Solution

Question 6

$$\dfrac{31}{27}$$

$$\dfrac{27}{31}$$

$$\dfrac{2}{27}$$

$$\dfrac{4}{27}$$

Question 7

$$0.533\ M$$

$$0.0534\ M$$

$$5.34\ M$$

$$None\ of\ these$$

Solution

Question 8

0.09 M and 0.2 M

0.2 M and 0.09 M

0.1 M and 0.19 M

0.19 M and 0.1 M

Solution

Question 9

Boric acid and borax

Sodium phosphate & disodium hydrogen phosphate

Sodium propionate and propionic acid

Sodium acetate and sodium propionate

Solution

Question 10

$$H_{3}PO_{4},H_{2}PO_4{^-}$$

$$H_{2}PO_4{^-},\ HPO^{2-}_{4}$$

$$HPO^{2-}_{4},PO^{3-}_{4}$$

$$H_{3}PO_{4},PO^{3-}_{4}$$

Solution

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