Practice Start Test

Ionic Equilibrium

JEE Chemistry · 82 questions · Page 4 of 9 · Click an option or "Show Solution" to reveal answer

Q31
pH of water is 7 at 25∘C25^{\circ} \mathrm{C}. If water is heated to 80∘C80^{\circ} \mathrm{C}., it's pH will :
A Decrease
B Remains the same
C Increase
D H+\mathrm{H}^{+}concentration increases, OH−\mathrm{OH}^{-}concentration decreases
Correct Answer
Option A
Solution

When water is heated, its self-ionization increases. At 25°C, the water ionization constant is given by

Kw=[H+][OH−]=1.0×10−14,K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14},

so in pure water,

[H+]=[OH−]=1.0×10−7 M,[\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7} \, \text{M},

which corresponds to a pH of 7. As water is heated to 80°C, the value of

KwK_w

increases due to the endothermic nature of water’s autoionization. This means that both

[H+][\mathrm{H}^+]

and

[OH−][\mathrm{OH}^-]

increase.

However, since their concentrations remain equal (which defines neutrality at that temperature), the pH isn’t 7 anymore.

Instead, the increased concentration of

H+\mathrm{H}^+

ions leads to a pH value lower than 7. Key points: Heating water increases

KwK_w

. Higher

KwK_w

means higher

[H+][\mathrm{H}^+]

(and equally higher

[OH−][\mathrm{OH}^-]

).

Thus, the pH decreases (e.g., it might be around 6.5–6.6 at 80°C) even though the water is still neutral.

So, the correct answer is: Option A – Decrease.

Q32
Ksp\mathrm{K}_{\mathrm{sp}} for Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3 is 1.6×10−301.6 \times 10^{-30}. What is the molar solubility of this salt in water?
A 1.8×10−305\sqrt[5]{1.8 \times 10^{-30}}
B 1.8×10−3027\dfrac{1.8 \times 10^{-30}}{27}
C 1.6×10−30274\sqrt[4]{\dfrac{1.6 \times 10^{-30}}{27}}
D 1.6×10−302\sqrt[2]{1.6 \times 10^{-30}}
Correct Answer
Option C
Solution

To find the molar solubility of Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3 in water, we start with the dissolution equilibrium: Cr(OH)3( s)⇌Cr(aq)3+s+3OH(aq)−3s \mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathop {\mathrm{Cr^{3+}_{(aq)}}}\limits_s + \mathop {\mathrm{3OH^-_{(aq)}}}\limits_{3s} At equilibrium, the solubility product Ksp\mathrm{K}_{\mathrm{sp}} is given by: Ksp=(s)⋅(3 s)3=27 s4 \mathrm{K}_{\mathrm{sp}} = (\mathrm{s}) \cdot (3 \mathrm{~s})^3 = 27 \mathrm{~s}^4 Substituting the given Ksp\mathrm{K}_{\mathrm{sp}} value: 27 s4=1.6×10−30 27 \mathrm{~s}^4 = 1.6 \times 10^{-30} Solving for s\mathrm{s}, we divide both sides by 27: s=(1.6×10−3027)1/4 \mathrm{s} = \left(\dfrac{1.6 \times 10^{-30}}{27}\right)^{1/4} Thus, the molar solubility of Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3 is: (1.627×10−30)1/4 \left(\dfrac{1.6}{27} \times 10^{-30}\right)^{1/4}

Q33
An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would (( Given log⁡2=0.30)\log 2=0.30)
A increase to 1.3
B reduce to 0.5
C increase to 2
D remain same
Correct Answer
Option A
Solution

An aqueous solution of HCl initially has a pH of 1.0, meaning the hydrogen ion concentration is [H+]=10−1 M[\text{H}^+]=10^{-1} \, \text{M}.

When we dilute the solution by adding an equal volume of water, the concentration of hydrogen ions will be halved: [H+]new=10−12=5×10−2 M [\text{H}^+]_{\text{new}} = \dfrac{10^{-1}}{2} = 5 \times 10^{-2} \, \text{M} To find the new pH, we use the pH formula: pH=−log⁡[H+] \text{pH} = -\log[\text{H}^+] Substituting the new hydrogen ion concentration gives: pH=−log⁡(5×10−2) \text{pH} = -\log(5 \times 10^{-2}) Using the logarithm property log⁡(a×b)=log⁡a+log⁡b\log(a \times b) = \log a + \log b: pH=−(log⁡5+log⁡10−2)=−log⁡5+2 \text{pH} = -(\log 5 + \log 10^{-2}) = -\log 5 + 2 Given log⁡2=0.30\log 2 = 0.30, we need log⁡5\log 5, which can be calculated as: log⁡10=log⁡(2×5)=log⁡2+log⁡5⇒log⁡5=log⁡10−log⁡2=1−0.30=0.70 \log 10 = \log(2 \times 5) = \log 2 + \log 5 \Rightarrow \log 5 = \log 10 - \log 2 = 1 - 0.30 = 0.70 Thus: pH=−(0.70)+2=1.3 \text{pH} = -(0.70) + 2 = 1.3 Therefore, the pH of the solution increases to 1.3.

Q34
The pH of a 0.02 M NH4Cl solution will be : [given Kb (NH4OH) = 10–5 and log 2 = 0.301]
A 2.56
B 5.35
C 4.35
D 4.65
Correct Answer
Option B
Solution

NH4+ + H2O ⇋ NH4OH + H+ [H+] = cα\alpha =

ka(NH4+)×c\sqrt {{k_a}\left( {NH_4^ + } \right) \times c}

=

kwkb(NH4OH)×c\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c}

=

10−1410−5×0.02\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02}

=

20×10−6\sqrt {20} \times {10^{ - 6}}

∴\therefore pH =

−log⁡(20×10−6)- \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)

= 5.35

Q35
The incorrect statement for the use of indicators in acid-base titration is :
A Phenolphthalein is a suitable indicator for a weak acid vs strong base titration.
B Methyl orange may be used for a weak acid vs weak base titration.
C Methyl orange is a suitable indicator for a strong acid vs weak base titration.
D Phenolphthalein may be used for a strong acid vs strong base titration.
Correct Answer
Option B
Solution

.tg .tg Indicator pH range Methyl orange 3.2 - 4.5 Phenolpthalein 8.3 - 10.5 Option B is the incorrect statement for the use of indicators in acid-base titration.

Methyl orange is a suitable indicator for a strong acid vs strong base titration, not for a weak acid vs weak base titration.

In a weak acid vs weak base titration, the pH at the equivalence point is typically around 7, which is within the pH range where methyl orange undergoes a color change.

Therefore, methyl orange is not suitable for use in weak acid vs weak base titrations.

The other options are correct: Phenolphthalein is a suitable indicator for a weak acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.

Methyl orange is a suitable indicator for a strong acid vs weak base titration because the pH at the equivalence point is acidic, which causes methyl orange to undergo a color change.

Phenolphthalein may be used for a strong acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.

Q36
Solid Ba(NO3)2 is gradually dissolved in a 1.0 ×\times 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ? (Ksp for BaCO3 = 5.1 ×\times 10−9 )
A 5.1 ×\times 10-5 M
B 8.1 ×\times 10-8 M
C 8.1 ×\times 10-7 M
D 4.1 ×\times 10-5 M
Correct Answer
Option A
Solution
Na2CO31×10−4M→2Na+1×10−4M  +  CO32−1×10−4M\mathop {N{a_2}C{O_3}}\limits_{1 \times {{10}^{ - 4}}M} \to \mathop {2N{a^ + }}\limits_{1 \times {{10}^{ - 4}}M} \,\, + \,\,\mathop {C{O_3}^{2 - }}\limits_{1 \times {{10}^{ - 4}}M}
KSP(BaCO3)=[Ba2+][CO32−]{K_{SP\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]
[Ba2+]=5.1×10−91×10−4\left[ {B{a^{2 + }}} \right] = {{5.1 \times {{10}^{ - 9}}} \over {1 \times {{10}^{ - 4}}}}
=5.1×10−5M= 5.1 \times {10^{ - 5}}M
Q37
The molar solubility (in ol L-1) of a sparingly soluble salt MX4 is "s". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :
A s = (256 Ksp)1/5
B s = (128 Ksp)1/4
C s = ( Ksp / 128)1/4
D s = (Ksp / 256)1/5
Correct Answer
Option D
Solution
MX4⇌ M4+S       +4X−4SM{X_4}\rightleftharpoons\,\mathop {{M^{4 + }}}\limits_{S\,\,\,\,\,\,\,} + \mathop {4{X^ - }}\limits_{4S}
Ksp=[s][4s]4=256 s5{K_{sp}} = \left[ s \right]{\left[ {4s} \right]^4} = 256\,{s^5}

∴\therefore

     s=(Ksp256)1/5\,\,\,\,\,s = {\left( {{{{K_{sp}}} \over {256}}} \right)^{1/5}}
Q38
Which one of the following statements is not true?
A pH + pOH = 14 for all aqueous solutions
B The pH of 1 ×\times 10-8 M HCl is 8
C 96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of copper at the cathode
D The conjugate base of H2PO4−H_2PO_4^- is HPO42−HPO_4^{2-}
Correct Answer
Option B
Solution
pHpH

of an acidic solution should be less than

7.7.

The reason is that from

H2O.[H+]=10−7M{H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M

which cannot be neglected in comparison to

10−8M.{10^{ - 8}}M.

The

pHpH

can be calculated as. from acid,

   [H+]=10−8M.\,\,\,\left[ {{H^ + }} \right] = {10^{ - 8}}M.

from

H2O,  [H+]=10−7M{H_2}O,\,\,\left[ {{H^ + }} \right] = {10^{ - 7}}M

∴\therefore

   \,\,\,

Total

[H+]=10−8+10−7\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}}
=10−8(1+10)=11×10−8= {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}}

∴\therefore

   pH=−log⁡[H+]\,\,\,pH = - \log \left[ {{H^ + }} \right]
=−log⁡11×10−8= - \log 11 \times {10^{ - 8}}
=−[log⁡11+8log⁡ 10]= - \left[ {\log 11 + 8\log \,10} \right]
=−[1.0414−8]=6.9586= - \left[ {1.0414 - 8} \right] = 6.9586
Q39
Four species are listed below i. HCO3−HCO_3^− ii. H3O+H_3O^+ iii. HSO4−HSO_4^− iv. HSO3FHSO_3F Which one of the following is the correct sequence of their acid strength?
A iv < ii < iii < I
B ii < iii < i < iv
C i < iii < ii < iv
D iii < i < iv < ii
Correct Answer
Option C
Solution

The correct order of acidic strength of the given species in

HSO3F(iv)>H3O+(ii)>HSO4−(iii)>HCO3−(i)\mathop {HS{O_3}F}\limits_{\left( {iv} \right)} > \mathop {{H_3}{O^ + }}\limits_{\left( {ii} \right)} > \mathop {HS{O_4}^ - }\limits_{\left( {iii} \right)} > \mathop {HC{O_3}^ - }\limits_{\left( i \right)}

or

   (i)<(iii)<(ii)<(iv)\,\,\,\left( i \right) < \left( {iii} \right) < \left( {ii} \right) < \left( {iv} \right)

It corresponds to choice

(c)(c)

which is correct answer.

Q40
For the following Assertion and Reason, the correct option is Assertion (A): When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid. Reason (R): The equilibrium constant of Cu2+(aq) + S2–(aq) ⇌ CuS(s) is high because the solubility product is low.
A (A) is false and (R) is true.
B Both (A) and (R) are true but (R) is not the explanation for (A).
C Both (A) and (R) are true and (R) is the explanation for (A).
D Both (A) and (R) are false.
Correct Answer
Option C
Solution

Ksp value of CuS is very low 10–36 (3.6 × 10–36) due to low Ksp value Cu+2 ion gets precipitated very quickly even with very low concentration of S–2 ion.

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