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Published on: 03/01/2020
Ionic Equilibrium
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Take MCQ Chemistry Test
1.
The conjugate base of HCIO is ________.
ClO-
Cl-
H2ClO+
CIO
2.
pH of buffer depends upon concentration of _______.
acid (H+)
Conjugate base (OH-)
Salt
acid (H+) and Conjugate base (OH-)
3.
Kw represents _______.
ionic product constant of water
Solubility product of water
Equilibrium constant of water
Buffer index
4.
Pick out the incorrect statement regarding Lewis acids and bases
A Lewis acid is a electron deficient molecule
Lewis bases is one which donates an electron pair
Lewis base is a cation
Lewis acid is a electron deficient molecule and Lewis base is a cation
5.
If the solubility product of lead iodide is 3.2 × 10-8, its solubility will be _______.
2 × 10-3M
4 × 10-4M
1.6 × 10-5M
1.8 × 10-5M
6.
Conjugate base for Bronsted acids H2O and HF are _______.
OH- and H2FH+, respectively
H3O+ and F-, respectively
OH- and F-, respectively
H3O+ and H2F+, respectively
7.
Calculate the ionisation constant for the conjugate base of HF. Ionisation constant of HF at 298 K is 6.8 x 10-4
8.
Ksp of AgCl is \(1.8\times10^{-10}\). Calculate molar solubility in 1 M AgNO3
9.
When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetra ammine copper (II) complex,\({ [Cu({ H }_{ 2 }O)_4] }_{ (aq) }^{ 2+ }+ 4{ NH }_{ 3 }(aq)\rightleftharpoons { [Cu{ ({ NH }_{ 3 }) }_{ 4 }] }_{ (aq) }^{ 2+ }\) among H2O and NH3 Which is stronger Lewis base.
10.
For an aqueous solution of NH4CI, prove that [H+] = \(\sqrt { { K }_{ n }.C } \)
11.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
12.
Derive the hydrolysis constant for the hydrolysis of salt of strong base and weak acid. Deduce its pH.
13.
Calculate the pH of solution with HO+ concentrations in mol dm-3.
(i) 10-4
(ii) 10-7
(iii) 6.8 x 10-3
(iv) 3.2 x 10-5
(v) 0.035
(vi) 0.25
(vii) 5.4 x 10-9
(viii) 7.1 x 10-7
14.
Assertion: The dissociation of acetic acid decreases on addition of sodium acetate.
Reason: It is due to common ion effect.
Codes:
a) (A) and (R) are true and (R) is the correct explanation of (A)
b) Both (A) and (R) are true but (R) does not explain (A)
c) (A) is true but (R) is false
d) Both (A) and (R) are false
15.
Assertion: According to Bronsted concept. H2O is an a neutral substance.
Reason: H2O molecule can accept as well as donate a proton.
Codes:
a) (A) and (R) are true and (R) is the correct explanation of (A)
b) Both (A) and (R) are true but (R) does not explain (A)
c) (A) is true but (R) is false
d) Both (A) and (R) are false
1.
(a)
ClO-
2.
(d)
acid (H+) and Conjugate base (OH-)
3.
(a)
ionic product constant of water
4.
(c)
Lewis base is a cation
5.
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
Ksp = (s) (2s)2
3.2 × 10-8 = 4s3
s = (3.2 × 10-8/4)1/3
= ( 8 x 10-9)1/3
= 2 x 10-3 M
6.
H2O + H2O ⇌ H3O+ + OH-
acid 1 base 1 acid 2 base 2
HF + H2O ⇌ H3O+ + F-
acid 1 base 1 acid 2 base 2
∴ Conjugate bases are OH- and F- respectively.
7.
The conjugate base of HF is F-
For F-, \({ K }_{ b }=\frac { { K }_{ w } }{ { K }_{ a } } =\frac { { 10 }^{ -14 } }{ { 6.8\times 10 }^{ -14 } } \)
= 1.47 x 10-11.
8.
Ksp = 1.8 \(\times\)10-10, [AgNO3]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
s s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)
9.
(i) According to Lewis theory a species that donates a pair of electron is called Lewis base.
(ii) Nitrogen more in NH3 is less electro negative than oxygen in water. So the non - bonded electron pair on nitrogen is more available for sharing than a non - bonded electron pair on oxygen atom. So NH3 is a stronger lewis base than H2O.
10.
NH4CI is a salt of a strong acid HCI and weak base NH4OH.
\({ HCl }_{ (aq) }+{ NH }_{ 4 }OH_{(aq)}\rightleftharpoons { { NH }_{ 4 }Cl }_{ (aq) }+{ H }_{ 2 }O(I)\)
\({ NH }_{ 4 }^{ + }\) is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH- from water to produce unionised NH4OH shown below.
\({ NH }_{ 4 }^{ + }+{ H }_{ 2 }O(1)\rightleftharpoons { NH }_{ 4 }{ { { OH }_{ (aq) }+ }H }_{ (aq) }^{ + }\)
There is no such tendency shown by Cl- and therefore [H+] > [OH-]; the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the Kh and Kb as
Kh·Kb = Kw
Let us calculate the Kh value in terms of degree of hydrolysis (h) and the concentration of salt
Kb = h2C and \([{ H }^{ + }]=\sqrt { { K }_{ h }.C } \)
= \({ [H }^{ + }]=\sqrt { \frac { { K }_{ w } }{ { K }_{ b } } .C } \)
pH = - log [H+]
\(={ \left( \frac { { K }_{ w }.C }{ { K }_{ b } } \right) }^{ \frac { 1 }{ 2 } }\)
= \(-\frac { 1 }{ 2 } \log { K }_{ w }-\frac { 1 }{ 2 } \log C+\frac { 1 }{ 2 } \log{ K }_{ b }\)
\(pH=7-\frac { 1 }{ 2 } p{ K }_{ b }-\frac { 1 }{ 2 } \log C\)
11.
(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH4OH, to produce a salt, NH4CI, and water.
HCI(aq) + NH4OH(aq) ⇌ NH4CI(aq) + H2O (I)
NH4CI(aq) ⟶ NH4+ +CI-(aq)
(ii) NH4+ is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH- from water to produce unionised NH4OH
NH4+ (aq) + H2O(I) ⇌ NH4OH(aq) + H+(aq)
(iii) There is no such tendency shown by Cl- and therefore [H+] > [OH-]; the solution is acidic and the pH is less than 7.
(iv) The Kh and Kb are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H+]
12.
Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }][{ H }_{ 2 }O] } \)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }] } \) ...(1)
\({ CH }_{ 3 }{ COONH }_{ (aq) }\rightleftharpoons { C }{ H }_{ 3 }COO_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }CO{ O }^{ - }][{ H }^{ + }] }{ [{ CH }_{ 3 }{ COO }H] } \) ...(2)
(1) x (2)
⇒ Kb . Ka = [H+][OH-]
we know that [H+] [OH-] = Kw
Kh· Ka = Kw
Kh value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of Ostwald's dilution law. Kh = h2C. and i.e [OH-] = \(\sqrt { { K }_{ h }.C } \)
pH of salt solution in terms of Ka and the concentration of the electrolyte
pH + pOH = 14
pH = 14 - pOH = 14 - {-log [OH-]}
= 14 + log [OH-]
∴ pH = 14 + log (KhC)\(\frac12\)
pH =14 + log \({ \left( \frac { { K }_{ w }C }{ { K }_{ a } } \right) }^{ \frac { 1 }{ 2 } }\)
pH = 14 + (\(\frac12\) log Kw + \(\frac12\) log C - \(\frac12\) log Ka)
[∴ Kw = 10-14]
\(pH=14-7+\frac { 1 }{ 2 } \log \ C+\frac { 1 }{ 2 } p{ K }_{ a }\frac { 1 }{ 2 } \log{ K }_{ w }=\frac { 1 }{ 2 } \times { \log10 }^{ -14 }=\frac { -14 }{ 2 } (1)=-7\)
\(pH=7+\frac { 1 }{ 2 } { pK }_{ a }+\frac { 1 }{ 2 } \log \ C\) [-log Ka = pKa]
13.
pH = - log [H3O+]
(i) pH - log [104]
pH = log 1 - log 104
pH = - 4
(ii) pH = - log [10-7]
pH = log 1 - log 10-7.
pH = 7
(iii) pH = - log [H3O+]
pH = - log [6.8 x 10-3]
pH = log 1 - log 6.8 - log 10-3
= 3 - 0.8325
pH = 2.17 (or) 2.2
(iv) pH = - log [3.2 x 10-5]
pH = log 1 - log 3.2 - log 10-5
= 5 - 0.5051
pH = 4.49 (or) 4.5
(v) pH = - log [0.035]
pH = log 1 - log 0.035
= 2 - 0.5441
pH = 1.46 (or) 1.5
(vi) pH = - log [0.25]
pH = log 1 - log 0.25
= 1 - 0.3979
pH = 0.602 (or) 0.60
(vii) pH = -log [H3O+]
pH = log 1 - log 5.4 - log 10-9
= 7 - 0.7324
pH = 8.267 (or) 8.3
(viii) pH = - log [7.1 x 10-7]
pH = log 1 - log 7.1 - log 10-7
= 7 - 0.8513
pH = 6.2.
14.
a) (A) and (R) are true and (R) is the correct explanation of (A)
15.
a) (A) and (R) are true and (R) is the correct explanation of (A)
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