IISER Chemistry - Ionic Equilibrium in Aqueous Solution
Exam Duration: 45 Mins Total Questions : 30
According to Lowry-Bronsted concept the relative strength of the bases CH3COO-,OH- and CI- follows the order
- (a)
OH->CH3COO->CI-
- (b)
CH3COO->OH->CI-
- (c)
CH3COO->CI->OH-
- (d)
OH->CI->CH3COO-
According to Lowry Bronted a base is a substance molecule or an ion which can accept a proton . And the strength of the base is determined by firmness ,it can hold the proton given off by an acid .OH- holds H+ ion so firmly that only one molecule out of 550 million molecules is ionised .In CH3COOH there is a state of equilibrium between the ionised and non-inoised molecule .Whereas HCI is completely ionised.
All of the following are conjugate acid_base pair except
- (a)
NH4+,NH3
- (b)
H3O+,OH-
- (c)
NH3,NH2
- (d)
HF,F-
Conjugate acid-base pair has a difference of H+
The pH of 6 * 10-7 M NaOH solution would be
- (a)
9
- (b)
7.84
- (c)
8.84
- (d)
6.84
6 X 10-7 M NaOH solution is very dilute, hence the contribution of OH due to self ionisation of water cannot be neglected.
[OH] total = \(\bar{H}\) [ from NaOH + water ]
= 6 X 10-7 + 1 X 10-7
= 7 X 10-7 mol t-1
H+ = \({ { 1.0\times{10}^{-14} }\over{7\times{10}^{-7} } }=1.43\times{10}^{-8}\) mol t-1
pH = -log[ 1.43 X 10-8 ] = - [ 0.155 - 8 ] = 7.845
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 stongest Bronsted base in the following anion is
- (a)
CIO2-
- (b)
CI-O
- (c)
CIO4-
- (d)
CIO3-
An aqueous solution is twice as alkline as water. The pH of the solution is near
- (a)
7.3
- (b)
12
- (c)
10
- (d)
13
The hydroxyl ion in this case is \(2O\bar{H}\) or 2 X 10-7
H+ = \({ { 1\times{10}^{-14} }\over{ 2\times{10}^{-7} } }={ { 1 }\over{ 2 } }\times{10}^{-7}\)
or pH = 7 log 10 - log 1/2 = 7 + 0.3010 = 7.3
The pH of 1.0 * 10-8 N HCI at room temperature is
- (a)
8
- (b)
7
- (c)
very close to 7 but less than 7
- (d)
very close to 7 but greater than 7
pH of 1 X 10-8 HCl
H+ = 1 X 10-8 due to HCl
Total H+ = 10-8 + 10-7 ( due to H2O ) = 11 X 10-8
pH = +8 log 10 -log 11
= 8 - 1.05 = 6.95
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.
A solution was prepared by mixing 50 ml of 0.2M HCI and 50 ml of 0.1M NaOH. The pH of the solution is
- (a)
2.0
- (b)
3.0
- (c)
7.0
- (d)
1.3
meq. of HCl \(\Rightarrow\) 50 ml X 0.2 M HCl
= 1.0 = 0.1 equ. L-1
meq. of NaOH \(\Rightarrow\) 50 ml X 0.1 M HCl
= 5 = 0.05 equ. L-1
Acid left after nuutralisation = 0.1 - 0.05
= 0.05 eq. L-1
pH = -log H+ = - log 0.05
= - log 5 + log 100
= - 6960 + 2 = 1.3
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
21.4 g of NH4CI and 0.4 molal solution of NH4OH are disolved in water and the content is diluted to 250 ml.Given : pKb=4.7. The pH of the solution would be
- (a)
9.3
- (b)
4.7
- (c)
7.0
- (d)
2.35
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 \(\sqrt{{K}_{sp}/4}\)
pH of a solution is defined by the exxpression
- (a)
log [H+]
- (b)
\(log\frac { 1 }{ \left[ { H }^{ + } \right] } \)
- (c)
\(\frac { 1 }{ log { H }^{ + } } \)
- (d)
\(-log\frac { 1 }{ { H }^{ + } } \)
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.
Which one of the following cannot act as a Lewis or Bronsted acid?
- (a)
BF3
- (b)
AICI3
- (c)
SnCI4
- (d)
CCI4
CCI4 is neither a Lewis or Bronsted acid because it does not contain H+ ions and has complete octet.
The pH of neutral water at 25oC is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H+ and OH- ions are equal. What will be the pH of pure water at 60oC?
- (a)
Equal to 7.0
- (b)
Greater than 7.0
- (c)
Less than 7.0
- (d)
Equal to zero
Increasing temperature, it shifts the equilibrium of water ionisation reaction to forward direction.
The pH of a 10-9 M solution of HCI in water is
- (a)
9
- (b)
- 9
- (c)
between 7 and 8
- (d)
between 6 and 7
As the solution is acidic, therefore, pH < 7. This is because [H+] from H2O[10-7] M cannot be neglected in comparison to 10-9 M. Thus,
Total [H+]=10-9+10-7=10-9(1+102)=101X10-9
pH = -log[H+]
=-log[101X10-9]
=-2.004+9=6.996
The hydrogen ion concentration of 0.1 M hydrocyanic acid is \(6.3\times { 10 }^{ -6 }M\), the K a is
- (a)
\(3.96\times { 10 }^{ -10 }\)
- (b)
\(3.78\times { 10 }^{ -8 }\)
- (c)
\(3.86\times { 10 }^{ -7 }\)
- (d)
\(3.70\times { 10 }^{ -11 }\)
[H+]=6.3X10-6 M
\(\alpha\) = Degree of dissociation
= \({ { [{H}^{+}] }\over{ C } }={ { 6.3\times{10}^{-6} }\over{ 0.1 } }=6.3\times{10}^{-5}\)
We know, that,
Ka = \({\alpha}^{2}\) C = ( 6.3 X 10-5)2 X 0.1 = 3.96 X 10-10
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 \({ log }_{ 10 }\frac { 1 }{ { K }_{ a } } \)
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 \(\rightarrow Base+{ H }^{ + }\)
The \(\triangle { G }^{ o }\) for the reaction = 57 kJ
\(MX(s)\rightleftharpoons { M }^{ + }(aq)+{ X }^{ - }(aq)\)
Hence, KSP for the reaction is
- (a)
10-15
- (b)
10-6
- (c)
10-10
- (d)
10-16
\(\triangle{G}°\) = - 2.303 RT log Ksp
57 X 1000 = - 2.303 X 8.31 X 298 X log Ksp
log Ksp = \({ { 57 \times 1000 }\over{ -2.303 \times8.31 \times 298 } }=-9.99\approx-10\)
Ksp = 10-10
In aqueous solution, The ionisation constants for carbonic acid are \(K_{1} = 4.2\times 10^{-7}\) and \(K_{2} = 4.8\times 10^{-11}\) select the correct statement for a saturated 0.0034 M solution of the carbonic acid.
- (a)
The concentration of \(CO^{2-}_{3}\) is 0.034 M
- (b)
The concentration of \(CO^{2-}_{3}\) is greater than that of \({HCO}^{-}_{3}\)
- (c)
The concentration of \(H^{+}\) and \({HCO}^{-}_{3}\) are approximately equal
- (d)
The concentration of \(H^{+}\) is double that of \(CO^{2\ -}_{3}\)
\({ H }_{ 2 }{ CO }_{ 3 }\rightleftharpoons { H }^{ + }+{ HCO }_{ 3 }^{ - };{ K }_{ 1 }=4.2\times { 10 }^{ -7 }\)
\({ HCO }_{ 3 }^{ - }\rightleftharpoons { H }^{ + }+{ CO }_{ 3 }^{ 2- };{ K }_{ 2 }=4.8\times { 10 }^{ -11 }\)
K1>>K2
∴ [H+]=[\({ HCO }_{ 3 }^{ - }\)]
\({ K }_{ 2 }=\frac { [{ H }^{ + }][{ CO }_{ 3 }^{ 2- }] }{ [{ HCO }_{ 3 }^{ - }] } \)
So, \([{ CO }_{ 3 }^{ 2- }]={ K }_{ 2 }=4.8\times { 10 }^{ -11 }\)
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 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)
\(1\times10^{-10}\)
\(AB_{2}\rightleftharpoons A^{2+}_{5} +{2B}^{-}_{25}\)
\(K_{sp} = [A^{2+}][B^{-}]^{2}= (s) (25)^{2} = 4S^{3}\)
\(= 4(1\times10^{-5}) = 4\times10^{-15}\)
The solubility of AgCl(s) with solubility product 1.6 x 10-10 in 0.1 M NaCl solution would be
- (a)
1.26 x 10-5M
- (b)
1.6 x 10-9 M
- (c)
1.6 x 10-11 M
- (d)
zero
Key Idea As solubility of AgCl(s) is asked in 0.1 M NaCl solution, so in the calculation, solubility of Cl-(from NaCl) must be added to the solubility of Cl- (from AgCl).
Let s be the solubility of Ag+ and Cl- in AgCl before the addition of NaCl.
\(\underset { \underset { 0 }{ 0.1M } }{ NaCl\left( aq \right) } \longrightarrow \underset { \underset { 0.1\quad M }{ 0 } }{ { Na }^{ + }\left( aq \right) } +\underset { \underset { 0.1+S }{ 0 } }{ { Cl }^{ - }\left( aq \right) } \)
\(AgCl\left( s \right) \rightleftharpoons \underset { s }{ { Ag }^{ + }\left( aq \right) } +\underset { s+0.1 }{ { Cl }^{ - }\left( aq \right) } \)
Given, Ksp = 1.6 x 10-10 = [Ag+] [Cl-]
or 1.6 x 10-10 = s(0.1 + s) = 0.1 s + s2
Ksp is small, so s is very less in comparison with 0.1. Hence, s2 can be neglected.
Thus, 1.6 x 10-10 = 0.1 s
or s = 1.6 x 10-9 M
Which of the following is electron deficient?
- (a)
(CH3)2
- (b)
(SiH3)2
- (c)
(BH3)2
- (d)
PH3
Boron is an element of 13 group and contains three electrons in its valence shell. When its compound BH3 dimerises, each boron atom carry only 6 electrons that is their octet is incomplete. Hence, (BH3)2 is an electron deficient compound.
What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2 ?
- (a)
0.10 M
- (b)
0.40 M
- (c)
0.0050 M
- (d)
0.12 M
Number of milliequivalents of HCl
= 20 x 0.050 x 1 = 1
Number of milliequivalents of Ba(OH)2
= 2 x 30 x 0.10 = 6
[OH-] of final solution
Milliequivalents of Ba(OH)2
\(=\frac { -milliequivalents\quad of\quad HCl }{ Total\quad volume } =\frac { 6-1 }{ 50 } \)
= 0.1 M
The hydrogen ion concentration of a 10-8 M HCl aqueous solution at 298 K (Kw = 10-14) is
- (a)
1.0 x 10-6 M
- (b)
1.0525 x 10-7 M
- (c)
9.525 x 10-8 M
- (d)
1.0 x 10-18 M
In aqueous solution of 10-8 M HCl, [H+] ion concentration is based upon the concentration of H+ ion of 10-8 M HCl and concentration of H+ ion of water.
Kw of H2O = 10-14 = [H+][OH-]
or [H+] = 10-7 M (due to its neutral behaviour)
So, in aqueous solution of 10-8 M HCl,
[H+] = [H+] of HCl + [H+] of water.
= 10-8 + 10-7
= 11 x 10-8 M = 1.10 x 10-7 M
H2S gas when passed through a solution of cations containing HCl precipitates the cations of second group in qualitative analysis but not those belonging to the fourth group. It is because
- (a)
presence of HCl decreases the sulphide ion concentration
- (b)
presence of HCl increases the sulphide ion concentration
- (c)
solubility product of group II sulphides is more than that of group IV sulphides
- (d)
Sulphides of group IV cations are unstable in HCl
In qualitative analysis of cations of second group H2S gas is passed in presence of HCl, therefore due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in the form of their sulphides due to lower value of their solubility product (Ksp). Here, fourth group cations are not precipitated because it require more sulphide ions for exceeding their ionic product to their solubility products which is not obtained here due to common ion effect.
Which has highest pH?
- (a)
CH3CO-OK+
- (b)
Na2CO3
- (c)
NH4Cl
- (d)
NaNO3
\(pH=\log { \frac { 1 }{ \left[ { H }^{ + } \right] } } \)
pH is inversely proportional to hydrogen ion concentration. As concentration of H+ decreases pH increases and vice-versa.
Ammonium chloride (NH4Cl) is a salt of weak base and strong acid. So its aqueous solution will be acidic as
\({ NH }_{ 4 }Cl+{ H }_{ 2 }O\longrightarrow \underset { Weak\quad base }{ { NH }_{ 4 }OH } +\underset { Strong\quad acid }{ HCl } \)
So, pH of NH4Cl is less than 7.
Sodium nitrate (NaNO3 ) is the salt of strong acid and strong base. So, its aqueous solution is neutral as
\(Na{ NO }_{ 3 }+{ H }_{ 2 }O\longrightarrow \underset { Strong\quad base }{ NaOH } +\underset { Strong\quad acid }{ { HNO }_{ 3 } } \)
So, pH of NaNO3 is 7.
Potassium acetate (CH3COOK) is a salt of strong base and weak acid. Its aqueous solution will be basic and pH value will be greater than 7 = 8.8
\({ CH }_{ 3 }{ COO }^{ - }{ K }^{ + }+{ H }_{ 2 }O\longrightarrow \underset { Weak\quad acid }{ { CH }_{ 3 }COOH } +\underset { Strong\quad base }{ KOH } \)
Sodium carbonate (Na2CO3 ) is a salt of strong base and weak acid. Its aqueous solution is also basic and its pH value will be more than 10.
i.e. highest among them
\({ Na }_{ 2 }{ CO }_{ 3 }+{ H }_{ 2 }O\longrightarrow \underset { Strong\quad base }{ 2NaOH } +\underset { Weak\quad acid }{ { H }_{ 2 }{ CO }_{ 3 } } \)