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CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 13 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Equilibrium
Quiz 13
When $$0.1\ mol$$ arsenic acid, $$H_{2}AsO_{4}$$ is dissolved in $$1L$$ buffer solution of $$pH = 4$$, which of the following hold good? $$K_{1} = 2.5\times 10^{-4}, K_{2} = 5\times 10^{4}, K_{3} = 2\times 10^{-23}$$ for arsenic acid $$['< <'$$ sign denotes that the higher concentration is at least $$100$$ times more than the lower one].
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$$[H_{2}AsO_{4}] < < [H_{2}AsO_{4}^{-}]$$
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$$[H_{2}AsO_{4}] < < [HAsO_{4}^{2-}]$$
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$$[HAsO_{4}^{2-}] < < [HAsO_{4}^{-}]$$
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$$[AsO_{4}^{2-}] < < [HAsO_{4}^{2-}]$$
Explanation
$$ \begin{array}{l} p H=4 \\ -\log \left[H^{+}\right]=4 \\ \text { } \quad\left[H^+\right]=10^{-4} \end{array} $$
$$ \frac{\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right]}{\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right]}=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{K}_{1}}=\frac{10^{-4}}{2.5 \times 10^{-4}} $$
$$ \therefore \frac{1}{2.5} $$
$$ \Rightarrow \quad\left[H_{3} As O_{4}\right]<\left[H_{2} As O_{4}^{-}\right] $$
$$ \begin{aligned} \frac{\left[H_{2} A_{S} O_{4}^{-}\right]}{\left[H As O_{4}^{2-}\right]}=\frac{\left[H^{+}\right]}{K_{2}} &=\frac{10^{-4}}{5 \times 10^{4}} \\ &=\frac{10^{-4-4}}{5} \\ &=\frac{1}{5 \times 10^{8}} \end{aligned} $$
$$ \Rightarrow\left[\mathrm{H}_{2} \mathrm{~A}{\mathrm{s}} \mathrm{O}_{4}^{-}\right]<<\left[\mathrm{H} \mathrm{A}{\mathrm{s}} \mathrm{O}_{4}^{2-}\right] $$
$$ \begin{array}{c} {\left[\mathrm{HAsO}_{4}{ }^{2-}\right]<<\left[\mathrm{H}_{2} \mathrm{~A}{\mathrm{s}} \mathrm{O}_{4}^{-}\right]} \\ \text {not correct proof above. } \end{array} $$
$$ \frac{\left[As O_{4}^{2-}\right]}{\left[H As O_{4}^{2-}\right]}=\frac{K_{3}}{\left[H^{+}\right]}=\frac{2 \times 10^{-23}}{10^{-4}} $$
$$ =\frac{2}{10 \times 10^{18}} $$
$$=\frac{1}{5 \times 10^{18}}$$
$$ \begin{array}{l} \Rightarrow\left[\mathrm{A}s \mathrm{O}_{4}{ }^{2-}\right]<<\left[\mathrm{H} \mathrm{A}_{\mathrm{S}} \mathrm{O}_{4}{ }^{2-}\right] \\ \end{array} $$
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: $$[pK_{a}(NH_{4}^{+}) = 9.26]$$
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$$0.05\ mole$$
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$$0.025\ mole$$
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$$0.10\ mole$$
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$$0.005\ mole$$
Percentage ionisation of water at certain temperature is $$3.6\times { 10 }^{ -7 }$$%. Calculate $${K}_{w}$$ and $$pH$$ of water.
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$${ 10 }^{ -14 }, \ pH=6/7$$
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$$4\times { 10 }^{ -14 }, \ pH=6.7$$
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$$2\times { 10 }^{ -14 },\ pH=7$$
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$${ 10 }^{ -14 },\ pH=7$$
Explanation
Given that
$$\alpha=\cfrac {3.6\times 10^{-7}}{100}$$
No. of ionized species $$=c\times \alpha$$
Molarity of water, C $$=\dfrac{1000}{18}$$ $$M$$
$$C \times \alpha =\dfrac{1000}{18} \times \cfrac{3.6\times 10^{-7}}{100}=2\times 10^{-7}M$$
$$[H^+][OH^-]=2\times 10^{-7}\times 2\times 10^{-7}=4\times 10^{-14}$$
$$\therefore K_w=4\times 10^{-14}$$
$$pkW=14-\log 4$$
$$pH=\cfrac{pkW}{2}=7-\log 2=6.7$$
$$1$$ litre buffer solution $$\left( HA+NaA \right) pH=4$$ , is mixed with $$2$$ litre buffer solution $$\left( HA+NaA \right) pH=4.3$$. $$pH$$ of resulting solution if both solutions are initially $$0.1M$$ $$MA$$.
[Given $${ \left( { K }_{ a } \right) }_{ HA }={ 10 }^{ -4 };\log { 2 } =0.3;\log { 3 } =0.48;\log { 5 } =0.7\quad $$]
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$$4.2$$
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$$4.22$$
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$$4.15$$
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$$6.4$$
What is the concentration of unionized acetic acid in the solution?
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$$0$$
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$$4.9\times 10^{-10}\ M$$
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$$7\times 10^{-8}\ M$$
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$$2.45\times 10^{-10}\ M$$
The relationship between the molar solubility(z) of $$MgF_2$$ in 0.10 M $$Mg(NO_3)_2$$ and of $$K_{sp}$$ of $$MgF_2$$ is:
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$$(K_{sp}/0.10)^{1/2}$$
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$$(K_{sp}/0.40)^{1/2}$$
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$$(K_{sp}/4)^{1/2}$$
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$$(K_{sp})^{1/2}$$
Which of the following mixtures will be a buffer solution when dissolved in $$500.00\ ml$$ of water?
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$$0.200$$ mol of aniline and $$0.200$$ mol of $$HCl$$
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$$0.200$$ mol of aniline and $$0.400$$ mol of $$NaOH$$
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$$0.200$$ mol of $$NaCl$$ and $$0.100$$ mol of $$HCl$$
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$$0.200$$ mol of aniline and $$0.100$$ mol of $$HCl$$
An aqueous solution contains $$0.01 M \ RNH_2 (K_b=2×10^{−6})$$ and $$10^{−4} M\ NaOH.$$The concentration of $$OH^−$$ is nearly:
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$$2.414\times 10^ {-4}M$$
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$$10^ {-4}M$$
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$$1.41\times 10^ {-4}M$$
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$$2\times 10^ {-4}M$$
One litre of water contains $$1.0\times { 10 }^{ -7 }$$ moles of $${ H }^{ + }$$ ions. The degree of ionization of water is:
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$$1.8\times { 10 }^{ -9}%$$
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$$0.8\times { 10 }^{ -9 }%$$
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$$3.6\times { 10 }^{ -9 }%$$
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$$3.6\times { 10 }^{ -7 }%$$
Explanation
Volume=$$1L$$
Density of water=$$1g/ml$$
Mass of water=$$1000g$$
Moles of water =$$\dfrac{1000}{18}$$
$$H_2O\rightleftharpoons H^{+}+OH^-$$
$$t=O$$ $$C$$ $$O$$ $$O$$
$$t.teq$$ $$c(1-\alpha)$$ $$ e\alpha$$ $$ e\alpha$$
$$c\alpha=10^{-7}$$
$$\Rightarrow \dfrac{1000}{8}\times \alpha=10^{-2}$$
$$\Rightarrow \alpha=1.8\times 10^{-9}$$
Solid $$Ba{({NO}_{3})}_{2}$$ is gradually dissoved in a $$1\times {10}^{-4}M$$ $${Na}_{2}{CO}_{3}$$ solution. At what minimum conc. of $${Ba}^{-2}$$ will a precipitate of $$Ba{CO}_{3}$$ begin to form? ($${K}_{sp}$$ for $$Ba{CO}_{3}=5.1\times {10}^{-9}$$)
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$$4.1\times {10}^{-5}M$$
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$$8.1\times {10}^{-7}M$$
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$$5.1\times {10}^{-5}M$$
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$$8.1\times {10}^{-8}M$$
Explanation
Alc to the problem
$$\begin{aligned} \mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow & 2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \\ \left[\mathrm{cO}_{3}^{2-}\right] &=1 \times 10^{-4} \mathrm{M} \end{aligned} $$
For, $$\mathrm{BaCO}_{3} \rightleftharpoons\left[\mathrm{Ba}^{2+}\right]+\left[\mathrm{CO}_{3}^{2-}\right]$$
$$ \mathrm{KG}_{p}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{CO}_{3}^{2-}\right] $$
$$ \begin{aligned} \Rightarrow \quad 5.1 \times 10^{-9} &=\left[B a^{2+}\right] \cdot\left[1 \times 10^{-4}\right] \\ &\left[\begin{array}{ll} {\left[B a^{2+}\right]=5.1 \times 10^{-9+4}} \\ {\left[B a^{2 +}\right]} =5.1 \times 10^{-5} \mathrm{M} \end{array}\right. \end{aligned} $$
$$\text { (c) is correct option }$$
Which of the following is a true statement:
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The ionisation constant and ionic product of water are same.
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Water is a strong electrolyte.
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The value of ionic product of water is less than that of its ionisation constant
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At $$298\ K$$, the number of $${ H }^{ + }$$ ions in a litre of water is $$6.023\times { 10 }^{ 16 }$$.
Explanation
Let's check the options one by one:
(1) False. They are not the same.
Ionisation constant of water is
$${ K }_{ a\left( { H }_{ 2 }O \right) }=\dfrac { \left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ { H }_{ 2 }O \right] } $$
Ionic product of water is
$${ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right] \\ { K }_{ w }=\left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] $$
(2) False. Water is a weak electrolyte. In fact, pure water is a non-electrolyte.
(3) False. $${ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right] $$
Clearly, $${ K }_{ w }>{ K }_{ a\left( { H }_{ 2 }O \right) }$$
value of ionic product is more than value of ionisation constant.
(4) At 298K, pH of water is 7
$$\therefore conc\quad of\quad \left[ { H }^{ + } \right] ={ 10 }^{ -7 }$$moles per litre
1 mole contains $$6.023\times { 10 }^{ 23 }$$ ions
$$\therefore conc\left[ { H }^{ + } \right] ={ 10 }^{ -7 }\times 6.023\times { 10 }^{ 23 }=6.023\times { 10 }^{ 16 }$$
The ionization constant of benzoic acid is $$6.46 \times 10^{-5}$$ and $$K_{sp}$$ for silver benzoate is $$2.5\times 10^{-13}$$. How many times is silver benzoate more soluble in a buffer of $$pH = 3.19$$ compared to its solubility in pure water?
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$$4$$
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$$3.32$$
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$$3.01$$
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$$2.5$$
Explanation
Since $$pH=3.19$$
$$[H_3O^+]=6.46\times10^{-4}\ M$$
$$C_6H_5COOH+H_2O\leftrightharpoons C_6H_5COO^-+H_3O$$
$$K_a=\dfrac{[C_6H_5COO^-][H_3O^+]}{[C_6H_5COOH]}$$
$$\dfrac{[C_6H_5COOH]}{[C_6H_5COO^-]}=\dfrac{[H_3O^+]}{K_a}=\dfrac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$$
Let the solubility of $$C_6H_5COOAg$$ be $$x\ mol/L$$
Then,
$$[Ag^+]=x$$
$$[C_6H_5COOH]+[C_6H_5COO^-]=x$$
$$10[C_6H_5COO^-]+[C_6H_5COO^-]=x$$
$$[C_6H_5COO^-]=\dfrac{x}{11}$$
$$K_{sp}[AG^+][C_6H_5COO^-]$$
$$2.5\times10^{-13}=x\left(\dfrac{x}{11}\right)$$
$$x=1.66\times10^{-6}\ mol/L$$
Thus, the solubility of silver benzoate in a pH 3.19 solution is $$1.66\times10^{-6}\ mol/L$$
Now, let the solubility of $$C_6H_5COOAg$$ be $$x'\ M$$
Then, $$[Ag^+]=x' \ M$$ and $$[C_6H_5COO^-]=x'\ M$$
$$K_{sp}=[Ag^+][C_6H_5COO^-]$$
$$K_{sp}=(x')^2$$
$$x'=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\ mol/L$$
$$\therefore \dfrac{x}{x'}=\dfrac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$$
Hence, $$C_6H_5COOAg$$ is approximately $$3.32$$ times more soluble in low pH solution.
A sodium salt on treatment with $${MgCl}_{2}$$ gives white precipitate only on heating. The anion of the sodium salt is:
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$$ { HCO }_{ 3 }^{ - }$$
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$$ { CO }_{ 3 }^{ 2- }$$
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$$ { NO }_{ 3 }^{ - }$$
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$$ { SO}_{ 4 }^{ 2- }$$
Explanation
The salt will be sodium bicarbonate and the anion is $$HC{ O }_{ 3 }^{ \ominus }$$
When sodium bicarbonate react with $$Mg{ Cl }_{ 2 }$$ it produce Magnesium bicarbonate which is also soluble in water
But, on heating $$Mg{ \left( HC{ O }_{ 3 } \right) }_{ 2 }$$ it produce $$Mg\left( C{ O }_{ 3 } \right) $$ which gives white precipitate
$$Mg{ \left( HC{ O }_{ 3 } \right) }_{ 2 }\rightarrow Mg\left( C{ O }_{ 3 } \right)+C{ O }_{ 2 }+{ H }_{ 2 }O$$
If the ionic product of water varies with temperature as follows and the density of water is nearly constant for this range of temperature.
The given equilibrium process is:
$$ { H }^{ + }+OH\rightleftharpoons { H }_{ 2 }O$$
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Exothermic
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Endothermic
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Cant say
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Ionization
The $$K_{sp}$$ of $$Ag_{2}CrO_{4}, AgCl, AgBr$$ and $$AgI$$ are respectively, $$1.1\times 10^{-12}$$, $$1.8\times 10^{-10}$$, $$5.0\times 10^{-13}$$ and $$8.3\times 10^{-17}$$. Which of the following salts will precipitate last if $$AgNO_{3}$$ solution is added to the solution containing equal moles of $$NaCl, NaBr, NaI$$ and $$Na_{2}CrO_{4}$$?
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$$Ag_{2}CrO_{4}$$
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$$AgI$$
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$$AgCl$$
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$$AgBr$$
Explanation
1. $$Ag_2CrO_4\rightleftharpoons 2Ag^++{CrO_4}^{2-}$$
$$Ksp=(2s)^2\times s=4s^2$$
$$Ksp=(1.1\times 10^{-12})$$
$$S=3\sqrt {\cfrac {Ksp}{4}}=6.5\times 10^{-5}$$M
2. $$AgCl\rightleftharpoons Ag^++Cl^-$$
$$Ksp=S\times S$$ ; $$Ksp=1.8\times 10^{-10}$$
$$S=\sqrt {Ksp}=1.34\times 10^{-5}$$M
3. $$AgBr\rightleftharpoons Ag^++Br^-$$
$$Ksp=S\times S$$ ; $$Ksp=5\times 10^{-13}$$
$$S=\sqrt {Ksp}=0.71 \times 10^{-6}$$M
4. $$AgI\rightleftharpoons Ag^++I^-$$
$$Ksp=S\times S$$ ; $$Ksp=8.3\times 10^{-17}$$
$$S=\sqrt{Ksp}=0.9\times 10^{-8}$$M
$$\therefore$$ Solubility of $$Ag_2CrO_4$$ is highest, so it will precipitate last.
Which of the following statements is true?
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$${H}_{3}{PO}_{3}$$ is stronger acid than $${H}_{2}{SO}_{3}$$
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In aqueous medium $$HF$$ is a stronger acid than $$HCl$$
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$$HCl{O}_{4}$$ is a weaker acid than $$HCl{O}_{3}$$
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$$H{NO}_{3}$$ is a stronger acid than $$H{NO}_{2}$$
If $$S_0, S_1, S_2$$ and $$S_3$$ are the solubilities in water
of $$AgCl$$
, $$0.01 \,M \,CaCl_2, 0.01 \,M \,NaCl$$ and $$0.5 \,M \,AgNO_3$$ solutions, respectively, then which of the following is true?
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$$S_0 > S_2 > S_1 > S_3$$
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$$S_0 = S_2 = S_1 > S_3$$
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$$S_3 > S_1 > S_2 > S_0$$
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none of these
Explanation
The solubility of $$AgCl$$ or its ion formation will depend inversely on the concentration
$$S_0=H_2O$$
$$S_1=0.01\ M\ CaCl_2$$
$$S_2=0.01\ M\ NaCl$$
$$S_3=0.05\ M\ AgNO_3$$
$$H_2O$$ is dilute solution and least concentrated hence have maximum solubility of $$AgCl$$ in it. Out of $$NaCl$$ and $$CaCl_2$$, the solubility of $$NaCl$$ is high due to loess number of ions produced from $$NaCl$$
as compare to $$CaCl_2$$. More the ion produced lesser is the solubility of coming salt.
So, the Correct order is $$S_0 > S_2 > S_1 > S_3$$
We know that concentration of common ion $$\alpha \dfrac{1}{solubility}$$. The order of solubility of $$AgCl : S_1 > S_3 > S_2 > S_4$$
$$As_2S_3$$ solution has negative charge, capacity to precipitate is highest in:
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$$AlCl_3$$
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$$Na_3PO_4$$
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$$CaCl_2$$
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$$K_2SO_4$$
Explanation
Solution:- (A) $$Al{Cl}_{3}$$
According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence $${As}_{2}{S}_{3}$$ precipitate the most in $$Al{Cl}_{3}$$.
The number of $${ H }_{ 3 }{ O }^{ + }$$ ions present in 10 ml of water at $${ 25 }^{ 0 }C$$ is
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$$6.023\times { 10 }^{ -14 }$$
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$$6.023\times { 10 }^{ 14 }$$
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$$6.023\times { 10 }^{ -19 }$$
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$$6.023\times { 10 }^{ 19 }$$
The amount $${ CH }_{ 3 }{ NH }_{ 2 }$$ dissolved in $$2\ L$$ of water that it produces concentration $${ OH }^{ - }$$ equal to $$5\times { 10 }^{ -4 }\ M$$, is
[Given $${ K }_{ b }$$ of $${ CH }_{ 3 }{ NH }_{ 2 }=2\times { 10 }^{ -6 }$$]
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$$5.6\ gm$$
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$$3.88\ gm$$
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$$7.75\ gm$$
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$$8.3\ gm$$
On adding ammonia to water,
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ionic product will increase
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ionic product will decrease
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$$[H_{3}O^{+}]$$ will increase
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$$[H_{3}O^{+}]$$ will decrease
With increase in temperature, ionic production of water
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decreases
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increases
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remains same
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may increases or decreases
Which buffer solution has maximum $$pH?$$
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mixture which is $$0.1$$ M in $$ \mathrm { CH } _ { 3 } \mathrm { COOH } $$ and $$0.1$$ M in $$ \mathrm { CH } _ { 3 } \mathrm { COONa } \left[ \mathrm { pK } _ { \mathrm { a } } \left( \mathrm { CH } _ { 3 } \mathrm { COOH } \right) = 4.74 \right] $$
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mixture which is $$0.2$$ M $$ \mathrm { CH } _ { 3 } \mathrm { COOH } $$ and $$0.2$$ M in $$ \mathrm { CH } _ { 3 } \mathrm { COONa } $$
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mixture which is $$0.1$$ M in $$ \mathrm { NH } _ { 4 } \mathrm { Cl } $$ and $$0.1$$ M in $$ \mathrm { NH } _ { 4 } \mathrm { OH } \left[ \mathrm { pK } _ { \mathrm { a } } \left( \mathrm { NH } _ { 4 } ^ { + } \right) = 9.26 \right] $$
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all the solution have equal $$pH$$ which is $$4.74$$
At certain temperature $${ K }_{ w }$$ for water $$4\times { 10 }^{ -14 }$$. Which of the following is incorrect for impure water at given temperature?
[Given: $$log\ 2 = 0.3$$]
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pH=6.7 and water is acidic.
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pH=6.7 and water is neutral.
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pOH=6.7 and water is neutral.
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pH + pOH = 13.4
How many grams of $$NaOH$$ is to be added to one litre of $$1 M H_2CO_3$$ to get a $$HCO^{-}_3/ CO^{-2}_3$$ buffer of maximum capacity ?
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90 gm
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60 gm
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20 gm
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50 gm
Solubility of silver cyanide is maximum in
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Acidic buffer solution
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Basic buffer solution
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In pure water
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Equal in all
A sample of water containing some dissolved table sugar and common salt is passed through organic ion exchange resins. The resulting water will be
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Sweet
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Salty
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Tasteless
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None of these
At $$100^\circ C$$, value of $$K_{w}$$ is
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$$1.0\times 10^{-14}\quad m^{2}$$
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less than $$1.0\times 10^{-14}\quad m^{2}$$
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greater than $$1.0\times 10^{-14}\quad m^{2}$$
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Zero
Explanation
At higher temperature the value of $$kw$$ increases.This is in according with le-chatelier principle.
At $$100^o kw=51.3\times 106{-14}$$
C is the correct answer.
Calculate the degree of ionization of 0.04M HOCl solution having ionization constant $$1.25\times { 10 }^{ -4 }$$?
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0.025
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0.25
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0.5
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0.055
When a solution of silver nitrate is added to pure carbon tetrachloride :
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A light blue precipitate soluble in ammonia is obtained
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A curdy precipitate insoluble in ammonia is obtained
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A Pale yellow precipitate in ammonia is formed
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No precipitate is formed
Which of the following compound has the maximum degree of ionization?
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$$1 M$$ $$NH_3$$
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$$0.001 M$$ $$NH_3$$
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$$0.1 M$$ $$NH_3$$
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$$0.0001 M$$ $$NH_3$$
In which of the following , the solubility of $$AgCl$$ will be minimum ?
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$$0.01 \ M \ Na_2{ SO_4}$$
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Pure water
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$$0.01 \ M \ CaCl_2$$
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$$0.01 \ M\ NaCl$$
Which shows weak ionisation in water?
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$$H_2SO_4$$
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$$NaCl$$
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$$HNO_3$$
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$$NH_3$$
Explanation
Because it is a weak electrolyte. Since the weak electrolytes have fewer ions in the solution.
Option D is correct.
The percentage of pyridine $$\left( {{C_5}{H_5}N} \right)$$ that forms pyridinum ion $$\left( {{C_5}{H_5}{N^ + }H} \right)$$ in a $$0.10M$$ aqueous pyridine solution $$\left( Given - {{K_b}, for \ {C_5}{H_5}N = 1.7 \times {{10}^{ - 9}}} \right)$$ is?
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$$0.0060\% $$
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$$0.013\% $$
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$$0.77\% $$
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$$1.6\% $$
At any temperature, the proton concentration of water is
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$${10}^{-7}M$$
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< $${10}^{-7}M$$
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> $${10}^{-7}M$$
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$$\sqrt { { K }_{ w } } $$
Amongst the following hydroxides,the one which has the lowest value of$${ K }_{ SP }$$ is?
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$${ Mg(OH) }_{ 2 }\quad \quad \quad$$
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$$Ca(OH)_{ 2 }$$
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$$Ba(OH)_{ 2 }$$
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$$Be(OH)_{ 2 }$$
Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be
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$$10^{-3}$$
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$$10^{-5}$$
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$$10^{-7}$$
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$$10^{-9}$$
If the ionic product of water is $$1.96 \times 10 ^ { - 14 }$$
at $$35 ^ { \circ } \mathrm { C } .$$ What is its value at $$10 ^ { \circ } \mathrm { C }$$
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$$2.95 \times 10 ^ { - 14 }$$
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$$1.96 \times 10 ^ { - 7 }$$
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$$2.95 \times 10 ^ { - 15 }$$
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$$3.9 \times 10 ^ { - 12 }$$
Explanation
Given:
Ionic product of water = $$1.96 \times 10^{-14} $$
Initial temperature= $$35^{0}C$$
Solution:
Ionic product follows a direct relationship with the temperature, i.e., the ionic product is directly proportional to temperature. As the temperature decreases, the ionic product also decreases.
According to question, as the temperature is decreasing $$(10^{0}C)$$ so, ionic product of water also decreases.
From the given options it can be observed that $$2.95 \times 10^{-15}$$ is the only value less than the present value.
Hence, the correct option is (C).
Which of the following solution mixture will show resistance to the small addition of acids?
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$$500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.1N\ NaOH$$
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$$500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.1N\ HCl$$
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$$500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.2N\ NaOH$$
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$$500ml\ of\ 0.2N\ CH_{3}COOH+500ml\ of\ 0.1N\ NaOH$$
Which of the following salt solution will act as a buffer?
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$$C{H_3}COON{H_4}\left( {aq.} \right)$$
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$$N{H_4}Cl\left( {aq.} \right)$$
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$$C{H_3}COONa\left( {aq.} \right)$$
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$$NaCl\left( {aq.} \right)$$
$$\frac { N } { 10 }$$ acetic acid was titrated with $$\frac { N } { 10 }$$ NaOH.When $$25 \% , 50 \%$$ and $$75$$$$\%$$ of titration is over then the pH of the solution will be $$: \left[ \mathrm { K } _ { a } = 10 ^ { - 5 } \right]$$
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$$5 + \log 1 / 3,5,5 + \log 3$$
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$$5 + \log 3,4,5 + \log 1 / 3$$
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$$5 - \log 1 / 3,5,5 - \log 3$$
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$$5 - \log 1 / 3,4,5 + \log 1 / 3$$
Explanation
t CH_3COOH NaOH CH_3COO^-Na^+
0 0.1 0.1
25% 0.1-0.025 0.1-0.025 0.025
50% 0.1-0.050 0.1-0.050 0.050
75% 0.1-0.075 0.1-0.075 0.075
$$pH=- \log K_a+\log \dfrac{[salt]}{[acid]}$$
t= 25%
$$pH=5+\log \dfrac{0.025} {0.075}$$
$$pH=5+\log \dfrac{1} {3}$$
t= 50%
$$pH=5+\log \dfrac{0.050} {0.050}$$
$$pH=5+\log 1=5$$
t= 75%
$$pH=5+\log \dfrac{0.075} {0.025}$$
$$pH=5+\log 3$$
In decimolar solution, CH$$_{3}$$COOH is ionised to the extent of 1.3 %. If $$log 1.3 = 0.11$$, what is the pH value of the solution?
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0%
3.89
0%
4.89
0%
2.89
0%
Unpredictable
On addition of sodium acetate, the ionization of acetic acid:
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0%
cannot be predicted
0%
decreases
0%
increases
0%
remains unaffected
The concentration of hydronium $$(H_{3}O^+)$$ ion in water is
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0%
$$Zero$$
0%
$$1\times 10^{-14}gm\ ion/litre$$
0%
$$1\times 10^{7}gm\ ion/litre$$
0%
$$1\times 10^{-7}gm\ ion/litre$$
Explanation
as,
$$H_{3}O^{+}\rightleftharpoons OH^{-}+H_{2}$$
and, $$pOH+pH=14$$
$$\Rightarrow [H^{+}]+[OH^{-}]=10^{-14}$$
$$\Rightarrow 10^{-7}+10^{-7}=10^{-14}$$
$$\Rightarrow [H_3O^+]=[OH^{-}]=10^{-7}\ gm\ ion/litre$$.
Which may be added to one litre water to act as a buffer.
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0%
1 mole of $$HC_{2}H_{3}O_{2}$$ and 1 mole of Hcl
0%
1 mole of $$NH_{4}OH$$ and 0.5 mole of $$NaOH$$
0%
1 mole of $$NH_{4}Cl$$ and 1 mole of Hcl
0%
1 mole of $$HC_{2}H_{3}O_{2}$$ and 0.5 mole of $$NaOH$$
At certain temperature, the $${ H }^{ + }$$ ion concentration of water is $$4\times { 10 }^{ -7 }$$M then the value of $${ K }_{ w }$$ at the same temperature is
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0%
$${ 10 }^{ -14 }{ M }^{ 2 }$$
0%
$$4\times { 10 }^{ -14 }{ M }^{ 2 }$$
0%
$$1.6\times { 10 }^{ -13 }{ M }^{ 2 }$$
0%
$$4\times { 10 }^{ -7 }{ M }^{ 2 }$$
The mass in grams of potassium permanganate (MM = 158) crystals required to oxidize $$750\,\,c{m^3}$$ of 0.1 M Mohr's salt solution in acidic medium is about?
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0%
11.8
0%
5.8
0%
6.0
0%
2.4
Degree of hydrolysis for a salt of weak acid and strong base can be changed by
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0%
Increasing concentration of salt
0%
Increasing temperature
0%
Adding base
0%
All of these
Which of the following mixture can form buffer solution ?
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0%
$$C{H_3}COOH + C{H_3}COONa$$
0%
$$NaCl + NaOH$$
0%
$$HCl + N{H_4}Cl$$
0%
$$C{H_3}COOH + HCl$$
If pH of a saturated solution of $$Mg(OH)_2$$ is 12.
Find its $$K_{sp}$$
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0%
$$2\times 10^{-4}$$
0%
$$5\times 10^{-5}$$
0%
$$4\times 10^{-6}$$
0%
$$5\times 10^{-7}$$
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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