JEE Main Chemistry - Ionic Equilibrium in Aqueous Solution
Exam Duration: 60 Mins Total Questions : 30
The conjucgate base of HBr is
- (a)
Br+
- (b)
H+
- (c)
Br2
- (d)
H2Br+
Conjugate acid-base pair has a diference of H+
Ostald dilution law is applicable to the dissociation of
- (a)
ammonium hydroxide
- (b)
causctic soda
- (c)
hydrochloric acid
- (d)
sodium chloride
This law is applicable to the dissociation of weak electrolyte.
The ionic product of water at 1000C is 55 times that at 250C. The pH value of water at 1000C
- (a)
7.00
- (b)
6.13
- (c)
17.13
- (d)
8.00
Given that Kw( 100° C ) = 55 X Kw ( 25° C )
= 55 X 1.0 X 10-14
pH ( 100° C ) = - \({ { 1 }\over{ 2 } }\) log Kw = - \({ { 1 }\over{ 2 } }\) log ( 55 X 10-14 )
= 6.13
The pH of 0.1 M HCN is 5.2 The ka value of the acid is
- (a)
3.98 * 10-8
- (b)
3.98 * 10-10
- (c)
3.98 * 10-12
- (d)
3.98 * 10-6
pH = - log H+
5.2 = - log H+ or H+ = - 5.2 = 6.8
or H+ = 6.310 X 10-6
Now HCN \(\leftrightharpoons\) H+ + CN-
or H+ = \(C\alpha\)
\(\therefore\) \(\alpha={ { {H}^{+} }\over{ C } }={ { 6.310\times{10}^{-6} }\over{ 0.1 } }=6.31\times{10}^{-5}\)
And, Ka = \({ { {C}^{2}{\alpha}^{2} }\over{ C(1-\alpha) } }=C{\alpha}^{2}=0.1\times(6.31\times{10}^{-5})^{2}\)
= 3.98 X 10-10
Given 0.5 M CH3COOH. The volume to which one litre of this solution be diluted in order to double the hydroxide concentration :
- (a)
1 litre
- (b)
2 litre
- (c)
3 litre
- (d)
4 litre
\(\underset { C\left( 1-\alpha \right) }{ { CH }_{ 3 }COOH } \rightleftharpoons \underset { C\alpha }{ { CH }_{ 3 }COO } +\underset { C\alpha }{ { H }^{ + } } \)
H+ = \(C\alpha\) and \(\alpha=\sqrt{{K}_{a}/C}\)
From given value \(\alpha=\sqrt{{ { 1.8 \times{10}^{-5} }\over{ 0.5 } }}=6\times{10}^{-3}\)
\(\therefore\) H+ = \(C\alpha\) = 0.5 X 6 X 10-3 = 3 X 10-1 mol l-1
pH = - log H+ = - log ( 3 X 10-3 ) = 2.52
pH = - log H+ = - log ( 3 X 10-3 ) = 2.52
To double the pH, thus pH = 5.04
Since pH = - log H+ or H+ = 10-5.4 = 0.912 X 10-5
= 9.12 X 10-6 mol l-1
Thus Ca = 9.12 X 10-6
In dilution, \(\alpha\) will increase, and its valuw will not be negligible in comparison to one. Thus we shall have to use the expression:
Ka= \({ { {C}^{2}{\alpha}^{2} }\over{C(1-\alpha) } }={ { {C \alpha }^{2}}\over{ 1-\alpha } }={ {(9.12\times{10}^{-6})\alpha }\over{1-\alpha } }\)
or \(1.8\times{10}^{-5} \quad (1-\alpha)=9.12\times{10}^{-6}\alpha\)
which gives (9.12X10-6+1.8X10-5)\(\alpha\)=1.8X10-5
\(\alpha={ { 1.8\times{10}^{-5} }\over{ 27.17\times{10}^{-6} } }=0.6637\)
Since \(C\alpha\) = 9.12 X 10-6 and \(\alpha\) = 0.6637
C= \({ { 9.12\times{10}^{-6} }\over{0.6637 } }=1.374\times{10}^{-5}\) mol l-1
The volume to which the solution should be diluted = \({ { 0.5 }\over{ 1.374 \times{10}^{-5} } }=3.369\times{10}^{4}\) l
Given 0.5 M CH3COOH. The volume to which one litre of this solution be diluted in order to double pH (ka = 1.8 * 10-5)
- (a)
3.369 * 101 litre
- (b)
3.369 * 102 litre
- (c)
3.369 * 103 litre
- (d)
3.369 * 104 litre
To double the hydroxyl ion concentration [H+] in 0.5M CH3COOH=3\(\times\)10-3l-1
[OH-]=\(\frac { { 10 }^{ -14 } }{ 3\times { 10 }^{ -3 } } \)
In the present case the concentration of hydroxide ions becomes
[OH-]= \(\frac { { 2\times 10 }^{ -14 } }{ 3\times { 10 }^{ -3 } } \)which gives
[H+]=\(\frac { 3\times 10^{ -3 } }{ 2 } =1.5\times { 10 }^{ -3 }{ mol\quad l }^{ -1 }\)
For this concentration, we can use Ka=C\(\alpha^2\)
\(\alpha =\frac { { K }_{ a } }{ C\alpha } =\frac { 1.8\times { 10 }^{ -5 } }{ 1.5\times { 10 }^{ -3 } } =1.2\times { 10 }^{ -2 }\)
Thus C=\(\frac { 1.5\times { 10 }^{ -3 } }{ 1.2\times { 10 }^{ -2 } } =1.25\times { 10 }^{ -1 }=0.125{ mol\quad l }^{ -1 }\)
Volume to which this solution can be diluted
=\(\frac { 0.5 }{ 0.125 } =4litres\)
When sodium hydride is dissolved in water ,which of the following ions are produced?
- (a)
Na+,H-
- (b)
Na-,H+
- (c)
OH-
- (d)
H3+O
H(g) + H2O (l) \(\rightarrow\) OH + H2
pH of a 10-3 M solution of hydrochloric acid will be
- (a)
1.3
- (b)
2.0
- (c)
3.0
- (d)
4.5
If ka1 and ka2 of sulphuric acid are 1*10-2 1*10-6 respectively then concentration of sulphate ions 0.01 molar sulphuric acid solution will be
- (a)
1*10-2
- (b)
0.01*10-8
- (c)
1*10-6
- (d)
0.01*10-10
H2SO4 \(\rightarrow\) H+ + HS\(\bar{{O}_{4}}\) \({K}_{{a}_{1}}\) = 1 X 10-2
HSO4 \(\rightarrow\) H+ + \({S}_{{4}^{2-}}\) \({K}_{{a}_{2}}\) = 1 X 10-6
\({K}_{{a}_{1}}\times{K}_{{a}_{1}}={ { {[{H}^{+}]}^{2}[{SO}_{{4}^{2-}}]}\over{[{H}_{2}{SO}_{4}] } }\)
1 X 10-8 = \({ { (.01)^{2}\times({ SO }_{ 4 }^{ 2- }) }\over{ 0.01 } }=[{ SO }_{ 4 }^{ 2- }]=1\times{10}^{-6}\)
A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of
- (a)
3
- (b)
4
- (c)
3000
- (d)
10000
H+ = 10-2, H+ = 10-6
\(\therefore\) 104 times.
If a neutral solution has pKa=13.36 at 500C, the ph of the solution is
- (a)
6.68
- (b)
7
- (c)
7.63
- (d)
none
pKw = pH + pOH
pH = pOH
\(\therefore\) \({ { 13.36 }\over{ 2} }=6.68\)
The solubility product of mercurous chloride is 1.0*10-18 mol3 l-3. The solubility of the compound in formula weight per litre is about
- (a)
10-18
- (b)
10-12
- (c)
10-6
- (d)
10-4.5
Mercurous chloride is Hg2Cl2 giving 3 particles Hg2Cl2 \(\rightarrow\) \({ Hg }_{ 2 }^{ 2+ }\) + 2Cl-
Ksp = S3 = 1.0 X 10-18 mol3 L-3
\(\therefore
\) S = 10-6 mol L-1
The soubility product of silver carbonate be Ksp; its solubility is
- (a)
\(\sqrt [ 3 ]{ \frac { K_{ sp } }{ 8 } } \)
- (b)
\(\sqrt [ 3 ]{ { K{ sp } }{ } } \)
- (c)
\(\sqrt [ 3 ]{ \frac { K_{ sp } }{ 4 } } \)
- (d)
\(\sqrt [ 3 ]{ \frac { K_{ sp } }{ 2 } } \)
Let the solubility of Ag2CrO4 be S mol l-1
Ksp = 22 X 11 X (S)2+1 = 4 S3
so S = 3 \(\sqrt{{K}_{sp}/4}\)
Acidity of BF3 can be explained on the basis of which of the following concepts?
- (a)
Arrhenius concept
- (b)
Bronsted - Lowry concept
- (c)
Lewis concept
- (d)
Bronsted - Lowry as well as Lewis concept
Acidity of BF3 can be explained on the basis of Lewis concept.
The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be
- (a)
10-9
- (b)
10-7
- (c)
10-5
- (d)
10-3
C = 0.1 M, \(\alpha\) = 0.01 % = \({ { 0.01 }\over{ 100 } }=1\times{10}^{-4}\)
\(\because\) K = \({C\alpha}^{1}\)
\(\Rightarrow\) K = 0.1 X ( 1 X 10-4)2 = 1 X 10-9
pKa of weak acid is defined as
- (a)
\({ log }_{ 10 }{ K }_{ a }\)
- (b)
\(\frac { 1 }{ { InK }_{ a } } \)
- (c)
\({ log }_{ 10 }\frac { 1 }{ { K }_{ a } } \)
- (d)
\({ log }_{ 10 }\frac { 1 }{ { K }_{ a }^{ 2 } } \)
pKa of a weak acid is defined as \({ log }_{ 10 }\frac { 1 }{ { K }_{ a } } \)
If the solubility product of BaSO4 is \(15\times { 10 }^{ 10 }\) in water. Its solubility, in moles per litre, is
- (a)
\(2.5\times { 10 }^{ -9 }\)
- (b)
\(3.9\times { 10 }^{ -5 }\)
- (c)
\(7.5\times { 10 }^{ -5 }\)
- (d)
\(2.5\times { 10 }^{ -5 }\)
BaSO4 \(\rightleftharpoons\) Ba2+ + \({ SO }_{ 4 }^{ 2- }\)
Solubility product = S X S
15 X 10-10 = S2
S = \(\sqrt{15\times{10}^{-10}}=3.87\times{10}^{-5}\)
Let the solubility of an aqueous solution of Mg(OH)2 be x, then its KSP is
- (a)
\({ 4x }^{ 3 }\)
- (b)
\({ 108x }^{ 5 }\)
- (c)
\({ 27x }^{ 4 }\)
- (d)
\({ 9x }^{ 3 }\)
\(Mg(OH)_{ 2 }\rightleftharpoons { \underset { (x) }{ Mg } }^{ 2+ }+{ \underset { (2x) }{ 2OH } }^{ - }\\ { K }_{ SP }=[{ Mg }^{ 2+ }][{ OH }^{ - }]^{ 2 }=(x)\times { (2x) }^{ 2 }={ 4x }^{ 3 }\)
Match the following and choose the correct option.
| Species | Conjugate acid | ||
| A. | NH3 | 1. | CO32- |
| B. | HCO3 - | 2. | NH4+ |
| C. | H2O | 3. | H3O+ |
| D. | HSO4 - | 4. | H2SO4 |
| 5. | H2CO3 | ||
- (a)
A B C D 3 2 1 4 - (b)
A B C D 2 5 3 4 - (c)
A B C D 5 3 2 1 - (d)
A B C D 5 2 1 3
\(A\rightarrow 2,B\rightarrow 5,C\rightarrow 3,D\rightarrow 4\)
As conjugate acid \(\rightarrow Base+{ H }^{ + }\)
How many litres of water must be added to 1 L of an aqueous solution of HCI with a pH of 1, to create an aqueous solution with pH of 2?
- (a)
0.1 L
- (b)
0.9 L
- (c)
2.0 L
- (d)
9.0 L
\(\because\) pH = 1 \(\Rightarrow\) [H+] = 10-pH = 10-1 = 0.1 M
pH = 2 \(\Rightarrow\) [H+] = 10-2 = 0.01 M
For dilutionf of HCl,
M1V1=M2V2
0.1 X 1 = 0.01 X V2
V2 = \({ { 0.1 \times1 }\over{ 0.01 } }=10\) L
\(\therefore\) Volume of water to be added = 10 -1 = 9 L.
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionisation constant, Ka of the acid is
- (a)
\(3\times { 10 }^{ -1 }\)
- (b)
\(1\times { 10 }^{ -3 }\)
- (c)
\(1\times { 10 }^{ -5 }\)
- (d)
\(1\times { 10 }^{ -7 }\)
HQ = H+ + Q-
[H+] = \(\sqrt{{K}_{a}C}\) [ by Ostwald's dilution law ]
[H+] = 10-pH = 10-3 M
C = 0.1 M
Thus, 10-3 = \(\sqrt{{K}_{a}\times0.1}\)
10-6 = Ka X 0.1
\(\therefore\) Ka = 1 X 10-5
The solubility product of a salt having general formula \(MX_{2}\) in water is \(4\times10^{-12}\) . The concentration of \(M^{2+}\) ions in the aqueous solution of the salt is
- (a)
\(4.0\times 10^{-10} \ M\)
- (b)
\(1.6 \times 10^{-4} \ M\)
- (c)
\(1.0 \times 10^{-4} \ M\)
- (d)
\(2.0 \times 10^{-6}\ M\)
\({ MX }_{ 2 }\rightleftharpoons \underset { S }{ M } ^{ 2+ }+\underset { 2S }{ 2X } ^{ - }\)
Ksp=[M2+][X-]2
If solubility be S, then
Ksp=(S)(2S)2=4S3
4S3=4\(\times\)10-12
∴ S=1\(\times\)10-4M
∴ [M2+]=S=1\(\times\)10-4M
Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be
- (a)
\(3.98 \times 10^{-6}\)
- (b)
\(3.68\times 10^{-6}\)
- (c)
\(3.88\times 10^{6}\)
- (d)
\(3.98\times10^{8}\)
pH = 5.4
\(\therefore \quad \quad [H^{+}] = 10^{-5.4} = 10^{-6}.10^{0.6}\)
Antilog of 0.6 is = 3.98
\(\therefore \quad \quad [H^{+}]= 3.98\times 10^{-6}\ M\)
The conjugate base of \(H_{2}{PO}^{-}_{4}\) is
- (a)
\(PO^{3 \ -}_{4}\)
- (b)
\(P_{2}O_{5}\)
- (c)
\(H_{3}PO_{4}\)
- (d)
\(HPO^{2 \ -}_{4}\)
\(H_{2}{PO}^{-}_{4} \rightleftharpoons HPO^{2-}_{4} + H^{+} \\ Acid \quad \quad conjugate\ base\)
The molar solubility \((in \ mol \ L^{-1})\) of a sparingly soluble salt \(MX_{4}\) is S. The corresponding solubility product is \(K_{sp}\) , S is given in terms of \(K_{sp}\) by the relation
- (a)
\(S = (K_{sp}/128)^{1/4}\)
- (b)
\(S = (128 \ K_{sp})^{1/4}\)
- (c)
\(S=(256 \ K_{sp})^{1/5}\)
- (d)
\(S=(K_{sp}/256)^{1/5}\)
\(MX_{4}\rightleftharpoons M^{4+} + 4X^{-}\)
\(\therefore \quad K_{sp} = (s)(4S)^{4} = 256\ S^{5} or \ s = ({k_{sp}\over256})^{1/5}\)
The solubility of a sparingly soluble salt \(AB_{2}\) in water is \(1\times10^{-5} \ mol\ L^{-1}\) . Its solubility product will be
- (a)
\(4\times 10^{-15}\)
- (b)
\(4\times 10^{-10}\)
- (c)
\(1\times 10^{-15}\)
- (d)
\(1\times10^{-10}\)
\(AB_{2}\rightleftharpoons A^{2+}_{5} +{2B}^{-}_{25}\)
\(K_{sp} = [A^{2+}][B^{-}]^{2}= (s) (25)^{2} = 4S^{3}\)
\(= 4(1\times10^{-5}) = 4\times10^{-15}\)
Which one of the following pairs of solution is not an acidic buffer?
- (a)
HClO4 and NaClO4
- (b)
CH3COOH and CH3COONa
- (c)
H2CO3 and Na2CO3
- (d)
H3PO4 and Na3 PO4
Strong acid with its salt cannot from buffer solution. Hence, HClO4 and NaClO4 is not an acidic buffer.
The solubility product of a sparingly soluble salt AX2 is 3.2 x 10-11 . Its solubility (in mol/L) is
- (a)
5.6 x 10-6
- (b)
3.1 x 10-4
- (c)
2 x 10-4
- (d)
4 x 10-4
AX2 is ionised as follows
\(\underset { S\quad mol\quad { L }^{ -1 } }{ { AX }_{ 2 } } \rightleftharpoons \underset { S }{ { A }^{ 2+ } } +\underset { 2S }{ 2{ X }^{ - } } \)
Solubility product of AX2,
Ksp = [A2+][X-]2 = S x (2S)2 = 4S2
Ksp of AX2 = 3.2 x 10-11
3.2 x 10-11 = 4S2
S3 = 0.8 x 10-11 = 8 x 10-12
Solubility = 2 x 10-4 mol/L
The solubility product of AgI at 25°C is 1.0 x 10-16 mol2 L-2 . The solubility of AgI in 10-4N. solution of KI at 25°C is approximately (in mol L-1)
- (a)
1.0 x 10-10
- (b)
1.0 x 10-8
- (c)
1.0 x 10-16
- (d)
1.0 x 10-12
\(AgI\longrightarrow { Ag }^{ + }+{ I }^{ - }\)
For binary electrolyte
Ksp = S2
where, S = solubility in mol/L
1.0 x 10-16 = S2
or S = 1 x 10-8 mol/L
Normality of KI solution = 10-4 N
Here change is one
M = 10-4 M [n=1]
or S for KI solution = 10-4 M
Solubility of AgI in KI solution
= 1 x 10-8 x 10-4
= 1 x 10-12 mol/L
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