Ionic Equilibrium

NH₄⁺ + H₂O ⇋ NH₄OH + H⁺

[H⁺] = c$\alpha $

= $\sqrt {{k_a}\left( {NH_4^ + } \right) \times c} $

= $\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c} $

= $\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02} $

= $\sqrt {20} \times {10^{ - 6}}$

$ \therefore $ pH = $ - \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)$ = 5.35

(a) Equivalance of strong acid = 0.1 $ \times $ 2 $ \times $ 400 = 80

Equivalance of strong base = 0.1 $ \times $ 400 = 40

$ \therefore $ [H⁺] of mixture = ${{80 - 40} \over {800}}$ = ${1 \over {20}}$

$ \therefore $ pH = $ - \log \left[ {{H^ + }} \right]$ = $ - \log \left( {{1 \over {20}}} \right)$ = 1.3

(b) Ionic product of water increases with increase of temperature because ionisation of water is endothermic.

(c)

k_a = ${{{{10}^{ - 5}} \times c\alpha } \over {c\left( {1 - \alpha } \right)}}$

$ \Rightarrow $ 10^-5 = ${{{{10}^{ - 5}} \times \alpha } \over {\left( {1 - \alpha } \right)}}$

$ \Rightarrow $ ${\alpha \over {\left( {1 - \alpha } \right)}}$ = 1

$ \Rightarrow $ $\alpha $ = 0.5

$ \therefore $ Degree of dissociation($\alpha $) = 50%

(d) The Le Chatelier's principle is always applicable to common-ion effect.

When strong acid is added to the strong base there will be sharp in the pH. So (A) is the correct answer.

Zr3(PO4)4	⇌	3Zr+4	+	4PO4-3
S		3S		4S

K_sp = [Zr⁺⁴]³ [PO₄^-3]⁴

= [3S]³ [4S]⁴

= 27S³ $ \times $ 256S⁴

= 6912S⁷

$ \therefore $ $S = {\left( {{{{K_{sp}}} \over {6912}}} \right)^{1/7}}$

$ \therefore $   8 $ \times $ 10^$-$12 = (2s + 0.1)² s

$ \Rightarrow $   s $ \times $ 10^$-$2 = 8 $ \times $ 10^$-$12

$ \Rightarrow $   s = 8 $ \times $ 10^$-$10

Ca(OH)₂ + Na₂SO₄ $ \to $ CaSO₄ + 2NaOH

100 m mol 14 m mol      $-$     $-$      $-$

$-$    $-$      $-$  14 m mol 28 m mol

w_CasO4 = 14 $ \times $ 10^$-$3 $ \times $ 136 = 1.9 gm

[OH^$-$] = ${{28} \over {100}}$ = 0.28 M

pH of rain water is approximate 5.6

H₂SO₄ + 2NH₄OH $ \to $ (NH₄)₂SO₄ + H₂O

Initially,

H₂SO₄ present = 20 $ \times $ 0.1 $ \times $ 2 = 4 miliequivalent

NH₄OH present = 30 $ \times $ 0.2 = 6 miliequivalent

Here H₂SO₄ is the limiting reagent,

So, finally. H₂SO₄ present = 0

and NH₄OH present = (6 $-$ 4) = 2

and (NH₄)₂SO₄ produced = 4 miliequivalent.

As in the solution there is (NH₄)₂ SO₄ present so it a basic buffer.

$ \therefore $   P^OH = P^K_b + log ${{\left[ {Salt} \right]} \over {\left[ {base} \right]}}$

= 4.7 + log ${4 \over 2}$

= 4.7 + log2

= 4.7 + 0.3

= 5

$ \therefore $   P^H = 14 $-$ P^OH

= 14 $-$ 5

= 9

Methyl orange is a pH indicator frequently used in titrations because of its clear and distinct colour change. Under acidic conditions, it is red or pinkish red and under alkaline conditions, it turns yellow.

Let's consider each of the options :

Option A : When a weak base is titrated against a strong acid, the solution at the end point is slightly acidic. However, the color change from colorless to pink is not associated with methyl orange.

Option B : When a strong base is titrated against a strong acid, the solution is neutral at the end point. Methyl orange changes color from pinkish-red in acidic conditions to yellow in basic conditions, so this color change doesn't match the description.

Option C : When a weak base is titrated against a strong acid, the solution is slightly acidic at the end point. Methyl orange would change from yellow in the basic solution to red in the acidic solution at the end point. This corresponds to a "yellow to pinkish-red" color change.

Option D : When a strong base is titrated against a strong acid, the solution is neutral at the end point. The color change described here is "pink to colorless" which does not correspond with the characteristic changes of methyl orange.

Therefore, the correct answer is Option C: Weak base, Strong acid, Yellow to pinkish-red.

Note :
When a salt is made with strong base and weak acid then that salt will be basic nature in aqueous solution.

(a)   Pb(CH₃COO)₂  is a salt of weak base Pb(OH)₂ and weak acid CH₃COOH.

(b)   Al(CN)₃  is a salt of weak base Al(OH)₃ and weak acid HCN.

(c)   CH₃COOK is a salt of weak acid CH₃COOH and strong base KOH.

(d)   FeCl₃ is a salt of weak base Fe(OH)₃ and strong acid HCl.

Let initially concentration of Ba⁺² = x m.

After adding 50 ml Na₂SO₄ in Ba⁺² solution final volume becomes 500 ml.

$\therefore\,\,\,$ Initial volume of Ba⁺² solution

= (500 $-$ 50) ml = 450 ml

As at the begining of precipitation, ionic product = solubility product.

$ \Rightarrow \,\,\,$ [Ba²⁺] [SO${_4^{ - 2}}$] = K_sp of BaSO₄

$ \Rightarrow \,\,\,$ [Ba²⁺] $\left( {{{50 \times 1} \over {500}}} \right)$ = 1 $ \times $ 10^$-$10

$ \Rightarrow \,\,\,$ [Ba²⁺] = 10^$-$9 M.

So, the concentration of Ba⁺² in final solution is 10^$-$9 M.

$\therefore\,\,\,$ concentration of Ba⁺² in original solution,

M₁ V₁ = M₂ V₂

$ \Rightarrow \,\,\,$ x $ \times $ 450 = 10^$-$9 $ \times $ 500

$ \Rightarrow \,\,\,$ x = 1.1 $ \times $ 10^$-$9 M

The compound which have the ability to accepted at least are lone pair electron.

Structure of BCl₃ is



Here B is electron deficient atom, so it can accepted lone pair. So it is a lewis acid.

Structure of AlCl₃



Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair. So it is also lewis acid.



Here in PH₃ there is vacant 3d orbital but it Can't take. Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases. So PH₃ is not lewis acid.



Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital.

This can be shown by following reaction.



So, here option (A) and (C) both are correct. But as SiCl₄ is not as strong lewis acid as BCl₃ and AlCl₃, So we can say option (A) is more correct.

HCl $ \to $ H⁺ + Cl^$-$

H⁺ concentration is = 0.2 M.

H₂S $\rightleftharpoons$ H⁺ + HS^$-$; K₁ = 1.0 $ \times $ 10^$-$7

HS^$-$ $\rightleftharpoons$ H⁺ + S^2$-$; K₂ = 1.2 $ \times $ 10^$-$13

H₂S $\rightleftharpoons$ S^2$-$ + 2H⁺

K = K₁ $ \times $ K₂ = 1.0 $ \times $ 10^$-$7 $ \times $ 1.2 $ \times $ 1.0^$-$13 = 1.2 $ \times $ 10^$-$20

as K₁ and K₂ both are very low for this reaction so dissociation of H₂S and HS^$-$ will be very low so, the produced H⁺ from this reaction will also be very low.

So, we can say the concentration of H⁺ will be almost same as H⁺ in HCl.

$\therefore\,\,\,$ [ H⁺ ] = 0.2 M.

From the reaction, H₂S $\rightleftharpoons\,$ 2H⁺ + S^2$-$

We get [ H⁺ ]² [ S^2$-$] = K $ \times $ [ H₂ S ]

$ \Rightarrow \,\,\,$ [ S^2$-$ ] = $ {{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}$

$ \Rightarrow \,\,\,$ [ S^2$-$ ] = 3 $ \times $ 10^$-$20 M

(a)    100 mL  ${M \over {10}}$ NaOH will nutalise

100 mL   ${M \over {10}}$ HCl, so number extra

HCl   will remain.

This will be neutral solution

$\therefore\,\,\,$ pH = 7

(b)    Here 25 mL  ${M \over 5}$ NaOH   nutralise

25 mL  ${M \over 5}$ HCl

$\therefore\,\,\,$ Extra HCl   =   75 $-$ 25   =   50 mL

Total volume   =   75 + 25   =   100 mL

$\therefore\,\,\,$ Milimole of HCl   =   ${{50} \over 5}$ = 10

$\therefore\,\,\,$ Concentration of HCl  =  ${{10} \over {100}}$  = 0.1

$\therefore\,\,\,$ pH  =  $-$ log[H⁺]  =  $-$ log(0.1)  =  1

(c)    HCl left  =  60 $-$ 40  =  20 mL

$\therefore\,\,\,$ milimole of HCl   =   ${{20} \over {10}}$   =  2

$\therefore\,\,\,$ Concentration of HCl   =   ${2 \over {100}}$   =  0.02 M

$\therefore\,\,\,$ pH  =  $-$ log (0.02)   =   1.69

(d)    HCl left  = 55 $-$ 45  =  10 mL

$\therefore\,\,\,$ milimole of HCl   =  ${{10} \over {10}}$  = 1

$\therefore\,\,\,$ Concentration of HCl  = ${{1} \over {100}}$   =  0.01 M

$\therefore\,\,\,$ pH = $-$ log (0.01) = 2

Given,

K_sp of PbCl₂ = 3.2 $ \times $10^$-$8

	PbCl2 $\rightleftharpoons$	Pb2+	2Cl
Initially	1	0	0
At equilibrium	1-s	s	2s

K_sp = [s] [2s]²

$\therefore\,\,\,\,$ K_sp = 4s³

$\therefore\,\,\,\,$ 4s³ = 3.2 $ \times $ 10^$-$8

$ \Rightarrow $ $\,\,\,\,$ s³ = 8 $ \times $ 10^$-$9

$ \Rightarrow $ $\,\,\,\,$ s = 2 $ \times $ 10 ^$-$3 M

Solubility = ${{no.\,\,of\,moles\,\,of\,\,pbc{l_2}\,} \over {volume\,(in\,\,litre)}}$

$\therefore\,\,\,\,$ 2 $ \times $ 10^$-$3 = ${{0.1} \over {278}}$ $ \times $ ${1 \over V}$

$ \Rightarrow $ $\,\,\,\,$ V = ${{0.1} \over {278}}$ $ \times $ ${{{{10}^3}} \over 2}$ = 0.18 L

As B can accept electron pair, so it is Lewis Acid.

P. Sodium hydroxide reacts with acetic acid to form sodium acetate.



The final solution will be a buffer of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$ as $\mathrm{CH}_3 \mathrm{COOH}$ is not completely neutralised by $\mathrm{NaOH}$.

The final moles of $\mathrm{CH}_3 \mathrm{COOH}=$ moles of $\mathrm{CH}_3 \mathrm{COONa}$ = 0.001 mol

$ \mathrm{Ph}=p \mathrm{~K}_a+\log \frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} $

$\Rightarrow\left[\mathrm{H}^{+}\right]$will not change on dilution as concentration of salt and acid does not change on dilution.

$\therefore$ Correct match $\mathrm{P}-1$

Q.

$ \begin{aligned} & \left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]_{\mathrm{old}}=\frac{20 \times 0.1}{40}=\frac{2}{40} \\\\ & {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]_{\text {new }}=\frac{2}{80}} \end{aligned} $

$ \begin{aligned} {\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_{\mathrm{H}} \mathrm{C}} \\\\ & =\sqrt{\left(\frac{\mathrm{K}_w}{\mathrm{~K}_a} \mathrm{C}\right)} \\\\ {\left[\mathrm{H}^{+}\right]_1 } & =\sqrt{\frac{\mathrm{K}_w \mathrm{~K}_a}{\mathrm{C}}} \\\\ \frac{\left[\mathrm{H}^{+}\right]_2}{\left[\mathrm{H}^{+}\right]_1}=\sqrt{\frac{\mathrm{C}_1}{\mathrm{C}_2}} & =\sqrt{\frac{0.05}{0.025}}=\sqrt{2} \end{aligned} $

Correct match Q-5

R.

As per the condition given in $R$ the resultant solution before dilution contains 2 millimoles of $\mathrm{NH}_4 \mathrm{Cl}$ in $40 \mathrm{~mL}$ of solution. Hence, a salt of strong acid and weak base is formed. For this,

$ \left[\mathrm{H}^{+}\right]_{\text {initial }}=\sqrt{\frac{K_w \times C}{K_b}} $

Now on dilution upto $80 \mathrm{~mL}$ new conc. becomes $C / 2$.

So $ \left[\mathrm{H}^{+}\right]_{\text {new }}=\sqrt{\frac{K_w \times \frac{C}{2}}{K_b}} $

or $ \left[\mathrm{H}^{+}\right]_{\text {new }}=\left[\mathrm{H}^{+}\right]_{\text {initial }} \times \frac{1}{\sqrt{2}} $

(S) For a saturated solution,


$ \begin{aligned} K_{s p}=s \times(2 s)^2=4 s^3 \\\\ \Rightarrow s=\left[\mathrm{OH}^{-}\right]=\sqrt[3]{\frac{K_{s p}}{4}} \end{aligned} $

Irrespective of volume of solution, $\left[\mathrm{H}^{+}\right]$remains constant.

(S) $\rightarrow$ (1)

NH₃ + HCl $ \to $ NH₄Cl

moles of HCl = 0.2 M × 25 × 10^–3 L = 0.005 moles HCl (total consumed)
moles of NH₃ = 0.2 M × 50 × 10^–3 L = 0.01 moles HCl
excess NH₃ = 0.01 – 0.005 = 0.005 moles

From reaction, 1 mole ammonia = 1 mole NH4Cl

$ \therefore $ 0.005 NH₃ = 0.005 NH₄Cl

Total Volume = V_HCl + V_NH3 = 25 + 50 = 75 mL

[NH₃] = [NH₄Cl] = ${{0.005} \over {75 \times {{10}^{ - 3}}}}$ = 0.066 M

pOH = pK_b + log ${{\left[ {salt} \right]} \over {\left[ {base} \right]}}$

= pK_b + log${{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}$

= 4.75 + log${{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}$

$ \Rightarrow $ pOH = 4.75

$ \therefore $ pH = 14 – 4.75 = 9.25

HA   $\rightleftharpoons$    H⁺ + A^$-$

K_a = ${{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$ = 10^$-$5

pH = pK_a + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$ \Rightarrow $$\,\,\,$ 6 = $-$ log [10^$-$5] + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$ \Rightarrow $$\,\,\,$ 6 = 5 + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$ \Rightarrow $ log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$ = 1

$ \Rightarrow $$\,\,\,$ ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$ = 10

$\therefore\,\,\,$ Salt : Acid = 10 : 1

The salt (AB) given is a salt is of weak acid and weak base. Hence its pH can be calculated by the formula

$ \therefore $ pH = 7 + ${1 \over 2}p{K_a} - {1 \over 2}p{K_b}$

= 7 + ${1 \over 2}\left( {3.2} \right) - {1 \over 2}\left( {3.4} \right)$

= 6.9

Energy evolved on neutralization of HCl and NaOH is

0.1 $\times$ 57 = 5.7 kJ = 5700 J

Energy utilized to rise the temperature of the solution is

ms . $\Delta$T = 200 $\times$ 1 $\times$ 4.2 $\times$ 5.7 = 4788 J

Energy used to increase temperature of calorimeter is

= 5700 $-$ 4788 = 912 J

ms . $\Delta$T = 912

m $\times$ s $\times$ 5.7 = 912 $\rightarrow$ ms = 160 J$^\circ$C^$-$1 [Calorimeter constant]

Energy evolved by neutralization of CH₃COOH and NaOH is

= 200 $\times$ 4.2 $\times$ 5.6 + 160 $\times$ 5.6 = 5600 J

So, the energy used in dissociation of 0.1 mol CH₃COOH is

= 5700 $-$ 5600 = 100 J

Thus, enthalpy of dissociation is 1 kJ mol^$-$1.

In Expt. 2, the final solution is a buffer as it contains equimolar amounts of acid and salt.

$pH = p{K_a} + \log {{(salt)} \over {(acid)}}$ ..... (1)

$p{K_a} = - \log (2 \times {10^{ - 5}})$

$ = - 0.3010 + 5$

$ = 4.699 \approx 4.7$

$[Salt] = [C{H_3}COONa] = {{100} \over {200}} \times 2 = 1\,M$

$[Acid) = [C{H_3}COOH] = {{200 - 100} \over {200}} \times 2 = {{100} \over {200}} \times 2 = 1\,M$

Substituting the values in Eq. (1), we get

$pH = 4.7 + \log {1 \over 1} = 4.7$

As $\,\,\,pH = 1;\,\,{H^ + } = {10^{ - 1}} = 0.1\,M$

$\,\,\,pH = 2;\,\,{H^ + } = {10^{ - 2}} = 0.01\,M$

$\therefore$ $\,\,\,{M_1} = 0.1\,\,{V_1} = 1;\,\,$

${M_2} = 0.01\,\,{V_2} = ?$

From

${M_1}{V_1} = {M_2}{V_2};0.1 \times 1 = 0.01 \times {V_2}$

${V_2} = 10\,$ litre

$\therefore$ volume of water added $ = 10 - 1 = 9\,$ litre.

To determine the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution, let’s consider the solubility equilibrium and how it is affected by the common ion effect.

First, let’s establish the dissociation equation for Ag₂CrO₄ :

$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq)$

The solubility product constant, K_sp, is given by :

$K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$

Given that K_sp = 1.1 $\times$ 10^-12.

In a 0.1 M AgNO₃ solution, the concentration of Ag⁺ ions from AgNO₃ alone is already 0.1 M. Let’s denote the additional solubility of Ag₂CrO₄ in this solution as $s$.

Therefore, the total concentration of Ag⁺ ions will be (0.1 + 2s) M, and the concentration of CrO₄^2- ions will be $s$ M.

Substituting these into the K_sp expression :

$K_{sp} = (0.1 + 2s)^2 s$

Given that K_sp is a very small number, we can assume that $2s$ is much smaller than 0.1, therefore $0.1 + 2s \approx 0.1$. This simplifies our equation to :

$K_{sp} \approx (0.1)^2 s$

Substitute $K_{sp} = 1.1 \times 10^{-12}$ into the equation :

$1.1 \times 10^{-12} \approx (0.1)^2 s$

$1.1 \times 10^{-12} \approx 0.01 s$

Solving for $s$ :

$s = \frac{1.1 \times 10^{-12}}{0.01} = 1.1 \times 10^{-10}$

Hence, the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution is 1.1 $\times$ 10^-10 mol/L.

The correct option is B: 1.1 $\times$ 10^-10.

Let's break down this problem step-by-step:

Understanding the Reaction:

The hydrolysis of methyl acetate (MeCOOCH₃) is catalyzed by both strong acids (like HX) and weak acids (like HA). The reaction proceeds as follows:

$ MeCOOCH_3 + H_2O \xrightarrow[or \, HA]{H^+ \, or \, HX} MeCOOH + CH_3OH $

Rate of Reaction and Acid Strength:

The rate of acid-catalyzed hydrolysis depends on the concentration of H⁺ ions.

• Strong acids ionize completely. So, a 1M solution of HX provides 1M H⁺ ions.


• Weak acids ionize partially. So, a 1M solution of HA will provide significantly less than 1M H⁺ ions.

Information from the Problem:

We are told that the initial rate of hydrolysis with HA is 1/100^th of that with HX. Since the rate is directly proportional to [H⁺], this means the [H⁺] from HA is 1/100^th of that from HX.

Calculating K_a:

1. [H⁺] from HX: Since HX is a strong acid and fully ionizes, [H⁺] = 1M.


2. [H⁺] from HA: This is 1/100^th the [H⁺] from HX, so [H⁺] = (1/100) \* 1M = 10^-2 M.


3. Setting up the K_a expression:

$ K_a = \frac{[H^+][A^-]}{[HA]} $

Since the initial concentration of HA is 1M and it ionizes to a small extent, we can approximate:

• [HA] ≈ 1 M


• [H⁺] = [A^-] = 10^-2 M

1. Calculating K_a:

$ K_a = \frac{(10^{-2})(10^{-2})}{1} = 1 \times 10^{-4} $

Answer:

Therefore, the K_a of the weak acid HA is 1 × 10^-4 (Option A).

${H^ + } = C\alpha ;\alpha = {{\left[ {{H^ + }} \right]} \over C}$

or $\,\,\,\alpha = {{{{10}^{ - 3}}} \over {0.1}} = {10^{ - 2}}$

$Ka = C{\alpha ^2} = 0.1 \times {10^{ - 2}} \times {10^{ - 2}}$

$ = {10^{ - 5}}$

To determine in which reactions $ H_2PO_4^- $ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases. According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.

Let's analyze each reaction :

1. 1. $ H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^- $

   
   
   • In this reaction, $ H_3PO_4 $ is donating a proton to $ H_2O $, forming $ H_3O^+ $ and $ H_2PO_4^- $.
   
   
   • $ H_2PO_4^- $ is the product of this reaction and is not acting as an acid here.



2. 2. $ H_2PO_4^- + H_2O \to HPO_4^{2−} + H_3O^+ $

   
   
   • In this reaction, $ H_2PO_4^- $ donates a proton to $ H_2O $, resulting in $ HPO_4^{2−} $ and $ H_3O^+ $.
   • 
     $ H_2PO_4^- $ is acting as an acid in this reaction.
3. 
   

   3. $ H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-} $

   • 
     This reaction is not correctly balanced and does not follow standard chemical reaction rules. The product $ O^{2-} $ is highly unlikely in aqueous solutions due to its reactivity.
   • 
     A more correct reaction would be $ H_2PO_4^- + OH^- \to HPO_4^{2−} + H_2O $. In this corrected reaction, $ H_2PO_4^- $ is donating a proton to $ OH^- $, which makes it an acid.

Given the information, the correct option is :

Option A : (ii) only

This is because in reaction (ii), $ H_2PO_4^- $ is clearly acting as an acid by donating a proton to water. In reaction (i), $ H_2PO_4^- $ is not acting as an acid but rather is formed as a product. Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $ H_2PO_4^- $ acting as an acid.

$AgBr\,\rightleftharpoons\,A{g^ + } + B{r^ - }$

${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]$

For precipitation to occur

Ionic product $>$ Solubility product

$\left[ {B{r^ - }} \right] = {{{K_{sp}}} \over {\left[ {A{g^ + }} \right]}} = {{5 \times {{10}^{ - 13}}} \over {0.05}} = {10^{ - 11}}$

i.e., precipitation just starts when ${10^{ - 11}}\,$

moles of $KBr$ is added to $1\ell \,AgN{O_3}\,$ solution

$\therefore$$\,\,\,$ Number of moles of $\,\,\,B{r^ - }\,\,$ needed

from $KBr = {10^{ - 11}}$

$\therefore$ $\,\,\,$ Mass of $KBr = {10^{ - 11}} \times 120$

$ = 1.2 \times {10^{ - 9}}g$

$Mg{\left( {OH} \right)_2}\,\rightleftharpoons\,M{g^{ + + }} + 2O{H^ - }$

${K_{sp}} = \left[ {M{g^{ + + }}} \right]{\left[ {O{H^ - }} \right]^2}$

$1.0 \times {10^{ - 11}} = {10^{ - 3}} \times {\left[ {O{H^ - }} \right]^2}$

$\left[ {O{H^ - }} \right] = \sqrt {{{{{10}^{ - 11}}} \over {{{10}^{ - 3}}}}} = {10^{ - 4}}$

$\therefore$ $\,\,\,pOH = 4$

$\therefore$ $\,\,\,pH + pOH = 14$

$\therefore$ $\,\,\,pH = 10$

$\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} $

${K_{SP\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]$

$\left[ {B{a^{2 + }}} \right] = {{5.1 \times {{10}^{ - 9}}} \over {1 \times {{10}^{ - 4}}}}$

$ = 5.1 \times {10^{ - 5}}M$

In aqueous solution $BA$ (salt) hydrolyses to give

$BA\,\, + \,\,{H_2}O\,\rightleftharpoons\,BOH\,\, + \,\,HA$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ Base acid

Now $pH$ is given by

$pH = {1 \over 2}p{K_w} + {1 \over 2}pKa - {1 \over 2}p{K_b}$

Substituting given values, we get

$pH = {1 \over 2}\left( {14 + 4.80 - 4.78} \right) = 7.01$

The correct order of acidic strength of the given species in

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

It corresponds to choice $(c)$ which is correct answer.

MX > M$_3$X > MX$_2$

$\Rightarrow~~\mathrm{MX} \to$ $\mathrm{\mathop {{M^ + }}\limits_S + \mathop {{M^ - }}\limits_S}$

Solubiliity, (S$_1$) of

$MX = \sqrt {{K_{sp}}} = \sqrt {4 \times {{10}^{ - 8}}} $

$ = 2 \times {10^{ - 4}}\,M$

$ \Rightarrow M{X_2} \to \mathop {{M^ + }}\limits_S + \mathop {2{X^ - }}\limits_S $

${K_{sp}} = {(s)^1}{(2s)^2} = 4{s^3}$

$S = {\left( {{{3.2 \times {{10}^{ - 14}}} \over 4}} \right)^{{1 \over 3}}}$

${K_{sp}} = 2 \times {10^{ - 5}}\,M$

$ \Rightarrow {M_3}X \to \mathop {3{M^ + }}\limits_{3S} + \mathop {{X^{3 - }}}\limits_S $

${K_{sp}} = {(3s)^3}{s^1} = 27{s^4}$

$S = {\left( {{{2.7 \times {{10}^{ - 15}}} \over {27}}} \right)^{{1 \over 9}}}$

$ = {({10^{ - 16}})^{{1 \over 2}}}$

$S = {10^{ - 4}}\,M = 1 \times {10^{ - 14}}\,M$

Hence, option (D) is correct.

MX > M$_3$X > MX$_2$

Weak monoacidic base, e.g., BOH is neutralised as follows:

BOH + HCl $\to$ BCl + H$_2$O

At the equivalence point, all BOH get converted into the salt. The concentration of H$^+$ (or pH of solution) is due to hydrolysis of the resultant salt (BCl, cationic hydrolysis here.)

Volume of HCl used up,

${V_a} = {{{N_b}{V_b}} \over {{N_a}}}$

${V_a} = {{2.5 \times 2 \times 15} \over {2 \times 5}}$

${V_a} = 7.5$

Concentration of salt,

$[BCl] = {{conc.\,of\,base} \over {total\,volume}}$

$ = {{2 \times 2.5} \over {5(7.5 + 2.5)}} = {1 \over {10}} = 0.1$

${K_h} = {{C{h^2}} \over {1 - h}} = {{{K_w}} \over {{K_b}}} = {{{{10}^{ - 14}}} \over {{{10}^{ - 12}}}}$

$ = {10^{ - 2}} = {{0.1 \times {h^2}} \over {1 - h}}$ ..... (i)

(h should be estimated whether that can be neglected or not.)

On calculating, $h = 0.27$ (significant, not negligible)

$[{H^ + }] = Ch = 0.1 \times 0.27 = 2.7 \times {10^{ - 2}}$ M

For acidic buffer $pH = p{K_a} + \log \left[ {{{salt} \over {acid}}} \right]$

or $\,\,\,pH = p{K_a} + \log {{\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$

Given $p{K_a} = 4.5\,\,$ and acid is $50\% $ ionised.

$\left[ {HA} \right] = \left[ {{A^ - }} \right]\,\,\,$ (when acid is $50\% $ ionised)

$\therefore$ $\,\,\,pH = p{K_a} + \log \,1$

$\therefore$ $\,\,\,pH = p{K_a} = 4.5$

$pOH = 14 - pH$

$ = 14 - 4.5$

$ = 9.5$

${H_2}A\,\rightleftharpoons\,{H^ + }\,\, + \,\,H{A^ - }$

$\therefore$ $\,\,\,{K_1} = 1.0 \times {10^{ - 5}} = {{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]} \over {\left[ {{H_2}A} \right]}}$ (Given)

$H{A^ - } \to {H^ + } + {A^ - }$

$\therefore$ $\,\,\,{K_2} = 5.0 \times {10^{ - 10}}$

$ = {{\left[ {{H^ + }} \right]\left[ {{A^{ - - }}} \right]} \over {\left[ {H{A^ - }} \right]}}$ (Given)

$K = {{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]} \over {\left[ {{H_2}A} \right]}}$

$ = {K_1} \times {K_2}$

$ = \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right)$

$ = 5 \times {10^{ - 15}}$

Let $\,\,\,s = \,\,\,$ solubility

$Ag{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s $

${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]$

$ = s \times s = {s^2}$

Given $\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}$

$\therefore$ $\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} $

$ = 1.0 \times {10^4}\,\,mol/lit$

$ = 1.0 \times {10^{ - 4}} \times 283\,g/lit$

(as Molecular mass of $Ag\,{\rm I}{O_3} = 283$ )

$ = {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml$

$ = 2.83 \times {10^{ - 3}}\,gm/100\,ml$

Conjugate acid-base pair differ by only one proton.

$O{H^ - } \to {H^ + } + {O^2}^ - \,\,$ Conjugate base of $O{H^ - }$ is ${O^{2 - }}$

$M{X_2}\,\rightleftharpoons\,\,\mathop {{M^{ 2 + }}}\limits_{s\,\,\,\,\,\,\,\,} \,\, + \,\,\mathop {2{X^ - }}\limits_{2s} $

Where $s$ is the solubility of $M{X_2}$

then ${K_{sp}} = $ $s \times {\left( {2s} \right)^2}$ = $4{s^3}$;

$ \Rightarrow $$ 4 \times {10^{ - 12}} = 4{s^3};$

$ \Rightarrow $ $s = 1 \times {10^{ - 4}}$

$\left[ {{M^{ 2+ }}} \right] = s $ $ = 1 \times 10^ {- 4}$

$pH = - \log \left[ {{H^ + }} \right] = \log {1 \over {\left[ {{H^ + }} \right]}};$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5.4 = \log {1 \over {\left[ {{H^ + }} \right]}}$

On solving, $\left[ {{H^ + }} \right] = 3.98 \times {10^{ - 6}}$

$M{X_4}\rightleftharpoons\,\mathop {{M^{4 + }}}\limits_{S\,\,\,\,\,\,\,} + \mathop {4{X^ - }}\limits_{4S} $

${K_{sp}} = \left[ s \right]{\left[ {4s} \right]^4} = 256\,{s^5}$

$\therefore$ $\,\,\,\,\,s = {\left( {{{{K_{sp}}} \over {256}}} \right)^{1/5}}$

NOTE : Conjugate acid-base differ by ${H^ + }$

$\mathop {{H_2}PO_4^ - }\limits_{Acid} \mathop \to \limits^{ - H} \,\mathop {\,HPO{{_4^ - }^ - }}\limits_{conjugate\,\,base} $

$pH$ of an acidic solution should be less than $7.$ The reason is that from ${H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M$ which cannot be neglected in comparison to ${10^{ - 8}}M.$ The $pH$ can be calculated as.

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

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

$\therefore$ $\,\,\,$ Total $\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}}$

$ = {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}}$

$\therefore$ $\,\,\,pH = - \log \left[ {{H^ + }} \right]$

$ = - \log 11 \times {10^{ - 8}}$

$ = - \left[ {\log 11 + 8\log \,10} \right]$

$ = - \left[ {1.0414 - 8} \right] = 6.9586$

$A{B_2}\rightleftharpoons\,{A^{ + 2}} + 2{B^ - }$

$\left[ A \right] = 1.0 \times {10^{ - 5}},\,\,$

$\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right],$

${K_{sp}} = {\left[ B \right]^2}\left[ A \right]$

$ = {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right]$

$ = 4 \times {10^{ - 15}}$

The rain water after thunderstorm contains dissolved acid and therefore the $pH$ is less than rain water without thunderstorm.