Chemistry 5 Marks Questions

1. To calculate the concentration of HS^– ion:

Case I (in the absence of HCl):

Let the concentration of HS^– be xM.

Then, $\text{K}_{\text{a}_1}=\frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}$

$9.1\times10^{-8}=\frac{\text{(x)(x)}}{0.1-\text{x}}$

$(9.1\times10^{-8})(0.1-\text{x})=\text{x}^2$

$\text{Taking 0.1}-\text{xM};0.1\text{M, we have }(9.1\times10^{-8})(0.1)=\text{x}^2.$

$9.1\times10^{-9}=\text{x}^2$

$\text{x}=\sqrt{9.1\times10^{-9}}$

$=9.54\times10^{-5}\text{M}$

$\Rightarrow[\text{HS}^-]=9.54\times10^{-5}\text{M}$

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let [HS^-] be yM.

Then,

Also,

Now, $\text{K}_{\text{a}_1}=\frac{[\text{HS}^-][\text{H}^+]}{[\text{H}_2\text{S}]}$

$\text{K}_{\text{c}_1}=\frac{[\text{y}(0.1+\text{y})]}{0.1-\text{y}}$ $(\because\ 0.1-\text{y};0.1\text{M})$

$9.1\times10^{-8}=\frac{\text{y}\times0.1}{0.1}$

$9.1\times10^{-8}=\text{y}$ $(\text{and}\ 0.1+\text{y};0.1\text{M})$

$\Rightarrow[\text{HS}^-]=9.1\times10^{-8}$

2. To calculate the concentration of [S^2-]:

Case I (in the absence of 0.1 M HCl):

$\text{HS}^-\leftrightarrow\text{H}^++\text{S}^{2-}$

$[\text{HS}^-]=9.54\times10^{-5}\text{M}$ (From first ionization, case I)

Let $[\text{S}^{2-}]$ be x.

Also, $[\text{H}^+]=9.54\times10^{-5}\text{M}$ (From first ionization, case I)
$\text{K}_{\text{a}_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]}$

$\text{K}_{\text{a}_2}=\frac{(9.54\times10^{-5})(\text{x})}{9.54\times10^{-5}}$

$1.2\times10^{-13}=\text{x}=[\text{S}^{2-}]$

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^– be X'M.

$[\text{HS}^-]=9.1\times10^{-8}\text{M}$ (From first ionization, case II)

$[\text{H}^+]=0.1\text{M}$ (From HCl, case II)

$[\text{S}^{2-}]=\text{x}'$

Then, $\text{K}_{\text{a}_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]}$

$1.2\times10^{-13}=\frac{(0.1)(\text{x}')}{9.1\times10^{-8}}$

$10.92\times10^{-21}=0.1\text{x}'$

$\frac{10.92\times10^{-21}}{0.1}=\text{x}'$

$\text{x}'=\frac{1.092\times10^{-20}}{0.1}$

$=1.092\times10^{-19}\text{M}$

$\Rightarrow\text{K}_{\text{a}_1}=1.74\times10^{-5}$

1. $\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+\ \text{K}_\text{a}=1.74\times10^{-5}$

2. $\text{H}_2\text{O}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{OH}^-\ \text{K}_\text{w}=1.0\times10^{-14}$

Since Ka >> K_w:

$\text{K}_\text{a}=\frac{(.05\alpha)(.05\alpha)}{(.05-0.05\alpha)}$

$=\frac{(.05\alpha)(0.05\alpha)}{.05(1-\alpha)}$

$=\frac{.05\alpha^2}{1-\alpha}$

$1.74\times10^{-5}=\frac{0.05\alpha^2}{1-\alpha}$

$1.74\times10^{-5}-1.74\times10^{-5}\alpha=0.05\alpha^2$

$0.05\alpha^2+1.74\times10^{-5}\alpha-1.74\times10^{-5}$

$\text{D}=\text{b}^2-4\text{ac}$

$=(1.74\times10^{-5})^2-4(.05)(1.74\times10^{-5})$

$=3.02\times10^{-25}+.348\times10^{-5}$

$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$

$\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$

$=\sqrt{\frac{34.8\times10^{-5}\times10}{10}}$

$=\sqrt{3.48\times10^{-6}}$

$=\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$

$\alpha=1.86\times10^{-3}$

$[\text{CH}_3\text{COO}^-]=0.05\times1.86\times10^{-3}$

$=\frac{0.93\times10^{-3}}{1000}$

$=.000093$

Method 2

Degree of dissociation,

$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$

c = 0.05M

K_a = 1.74 × 10^–5

^{Then, $\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$}

$\alpha=\sqrt{34.8\times10^{-5}}$

$\alpha=\sqrt{3.48}\times10^{-4}$

$\alpha=1.8610^{-2}$

$\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$

Thus, concentration of CH₃COO– = c.α

$=.05\times1.86\times10^{-2}$

$=.093\times10^{-2}$

$=.00093\text{M}$

$\text{Since}[\text{oAc}^-]=[\text{H}^+],$

$[\text{H}^+]=.00093=.093\times10^{-2}.$

$\text{pH}=-\log[\text{H}^+]$

$=-\log(.093\times10^{-2})$

$\therefore\ \text{pH}=3.03$

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

1. $\text{K}_\text{p}=\frac{\text{p}_\text{CO}\times\text{p}^3_{\text{H}_2}}{\text{p}_{\text{CH}_4}\times\text{p}_{\text{H}_2\text{O}}}$
2.  

1. By Le Chatelier’s principle, on increasing pressure, equilibrium will shift in the backward direction where decreases number of moles.
2. As the given reaction is endothermic, by Le Chatelier ‘s principle, equilibrium will shift in the forward direction with increasing temperature.
3. Equilibrium composition will not be disturbed by the presence of catalyst but equilibrium will be attained quickly.

$\text{K}_{\text{c}}=\frac{(\text{x/2})(\text{x/2})}{(3.30\times10^{-3}-\text{x})^2}=32\text{ (Given)}$

$\therefore\ \frac{\text{x}^2}{4(3.30\times10^{-3}-\text{x})^2}=32$

$\text{or, }\frac{\text{x}}{2(3.30\times10^{-3}-\text{x})^2}=\sqrt{32}=5.66$

$\text{or, }\text{x=11.32(3.30}\times10^{-3}-\text{x})$

$\text{or, }12.32\text{x}=11.32\times3.30\times10^{-3}$

$\text{or, }\text{x}=3.0\times10^{-3}$

$\therefore\ \text{At eqm., [BrCl]}=(3.30\times10^{-3}-3.0\times10^{-3})$

$=0.30\times10^{-3}$

$=3.0\times10^{-4}\text{mol L}^{-1}$

It is given that $\text{K}_\text{p}=10.1.$

Now,

$\frac{\text{p}_{\text{CO}_2}\times\text{p}_{\text{H}_2}}{\text{p}_\text{COH}\times\text{p}_{\text{H}_2\text{O}}}=\text{K}\text{p}$

$\Rightarrow\frac{\text{p}\times\text{p}}{(4.0-\text{p})(4.0-\text{p})}=10.1$

$\Rightarrow\frac{\text{p}}{4.0-\text{p}}=3.178$

$\Rightarrow\text{p}=12.712-3.178\text{p}$

$\Rightarrow4.178\text{p}=12.712$

$\Rightarrow\text{p}=3.04$

Hence, at equilibrium, the partial pressure of H₂ will be 3.04 bar.

Applying law of chemical equilibrium,

$\text{K}_{\text{c}}=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}\Rightarrow0.14=\frac{\text{x.x}}{(0.78-2\text{x})^2}$

$\text{x}^2=0.14(0.78-2\text{x})^2$

$>$

$\text{or }\frac{\text{x}}{0.78-2\text{x}}=\sqrt{0.14}=0.374$

$\text{or }\text{x}=0.292-0.748\text{x}$

$\text{or }1.748\text{x}=0.292$

$\text{or }\text{x}=0.167$

Hence at equilibrium, $[\text{I}_2]=[\text{Cl}_2]=0.167\text{ M}$

$[\text{ICl}]=0.78-2\times0.167=0.446\text{ M}$

$\therefore\ [\text{H}_2\text{O}]=\frac{0.6}{10}\text{mol L}^{-1}=0.06\text{mol L}^{-1}$

$[\text{CO}]=\frac{0.6}{10}\text{mol L}^{-1}=0.06\text{mol L}^{-1}$

$[\text{H}_2]=\frac{0.4}{10}=0.04\text{mol L}^{-1}$ and $\text{[CO}_2]=\frac{0.4}{10}$

$=0.04\text{mol L}^{-1}$

$\text{K}_{\text{c}}=\frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO]}}=\frac{\frac{0.4}{10}\times\frac{0.4}{10}}{\frac{0.6}{10}\times\frac{0.6}{10}}=0.44$

$1.0\times10^{-38}=27\text{s}^4$

$.037\times10^{-38}=\text{s}^4$

$.00037\times10^{-36}=\text{s}^4\ \Rightarrow1039\times10^{-10}\text{M=S}$

$\Rightarrow0.4\times10^{-5}=\text{s}^3$

$4\times10^{-6}=\text{s}^3\Rightarrow1.58\times10^{-2}\text{M}=\text{S.1}$

$\Rightarrow\text{s}=2.24\times10^{-10}\text{M}$

Case I: When 0.01M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05-x and 0.01 + xcan be taken as 0.05 and 0.01 respectively.

$\text{K}_\text{a}=\frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$

$\therefore\ \text{K}_\text{a}=\frac{(0.01)\text{x}}{0.05}$

$\text{x}=\frac{1.82\times10^{-5}\times0.05}{0.01}$

$\text{x}=1.82\times10^{-3}\times0.05\text{M}$

Now,

$\alpha=\frac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}$

$=\frac{1.82\times10^{-3}\times0.05}{0.05}$

$=1.82\times10^{-3}$

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

$[\text{CH}_3\text{COOH}]=0.05-\text{x ; }0.05\text{M}$

$[\text{CH}_3\text{COO}^-]=\text{x}$

$[\text{H}^+]=0.1+\text{x ; }0.1\text{M}$

$\text{K}_\text{a}=\frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$

$\therefore\ \text{K}_\text{a}=\frac{(0.01)\text{x}}{0.05}$

$\text{x}=\frac{1.82\times10^{-5}\times0.05}{0.01}$

$\text{x}=1.82\times10^{-4}\times0.05\text{M}$

Now,

$\alpha=\frac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}$

$=\frac{1.82\times10^{-4}\times0.05}{0.05}$

$=1.82\times10^{-4}$

Then, 

$\text{Ag}_2\text{CrO}_4\rightarrow2\text{Ag}^++\text{CrO}_4^{-}$

$\text{K}_\text{sp}=(2\text{s})^2.\text{s}=4\text{s}^3$

$1.1\times10^{-12}=4\text{s}^3$

$\text{s}=6.5\times10^{-5}\text{M}$

$.275\times10^{-12}=\text{s}^3$

$\text{s}=0.65\times10^{-4}\text{M}$

Let s' be the solubility of $\text{AgBr}.$

$\text{AgBr}_\text{(s)}\leftrightarrow\text{Ag}^++\text{Br}^-$

$\text{K}_\text{sp}=\text{s}'^2=5.0\times10^{-13}$

$\therefore\ \text{s}'=7.07\times10^{-7}\text{M}$

Therefore, the ratio of the molarities of their saturated solution is

$\frac{\text{s}}{\text{s}'}=\frac{6.5\times10^{-5}\text{M}}{7.07\times10^{-7}\text{M}}=91.9.$

Also,

$\text{K}_\text{p}$ for the reaction i.e., $\text{H}_{2\text{ (g)}}\text{ + Br }_{2\text{(g)}}\leftrightarrow2\text{HBr}_\text{(g)}\text{ is }1.6\times10^5.$

Therefore, for the reaction $$$2\text{HBr}_\text{(g)}\leftrightarrow\text{H}_{2\text{ (g)}}\text{ + Br }_{2\text{(g)}},$ the equilibrium constant will be,

$\text{K}'_\text{p}=\frac{1}{\text{K}_\text{p}}$

$=\frac{1}{1.6\times10^5}$

$=6.25\times10^{-6}$

$\text{p}=2.49\times10^{-2}\text{ bar }=2.5\times10^{-2}(\text{approximately})$

$\text{Q}_{\text{p}}=\frac{\text{p}_{\text{co}_2}}{\text{p}_{\text{co}}}$

$\Rightarrow\text{pH}=-\log(10.75\times10^{-9})$

$\text{ Q}_{\text{c}}=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O]}}{[\text{CH}_3\text{COOH][}\text{C}_2\text{H}_5\text{OH}] }$

$=0204$ (approximately)

After mixing:

Ionic product of copper iodate:

Since the ionic product (1 × 10^–9) is less than K_sp (7.4 × 10^–8), precipitation will not occur.

Now,

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.

If the concentrations of both solutions are equal to or less than 5.02 × 10^–9M, then there will be no precipitation of iron sulphide.

$\text{K}_{\text{c}}=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O]}}{[\text{CH}_3\text{COOH][}\text{C}_2\text{H}_5\text{OH}] }$

$=3.92$ (approximately)

1. pH of milk = 6.8

Since, $\text{pH}=-\log[\text{H}^+]$

$6.8=-\log[\text{H}^+]$

$\log[\text{H}^+]=-6.8$

$[\text{H}^+]=\text{anitlog}(-6.8)$

$=1.5\times10^{-7}\text{M}$

2. pH of black coffee = 5.0

Since, $\text{pH}=-\log[\text{H}^+]$

$5.0=-\log[\text{H}^+]$

$\log[\text{H}^+]=-5.0$

$[\text{H}^+]=\text{anitlog}(-5.0)$

$=10^{-5}\text{M}$

3. pH of tomato juice = 4.2

Since, $\text{pH}=-\log[\text{H}^+]$

$4.2=-\log[\text{H}^+]$

$\log[\text{H}^+]=-4.2$

$[\text{H}^+]=\text{anitlog}(-4.2)$

$=6.31\times10^{-5}\text{M}$

4. pH of lemon juice = 2.2

Since, $\text{pH}=-\log[\text{H}^+]$

$2.2=-\log[\text{H}^+]$

$\log[\text{H}^+]=-2.2$

$[\text{H}^+]=\text{anitlog}(-2.2)$

$=6.31\times10^{-3}\text{M}$

5. pH of egg white = 7.8

Since, $\text{pH}=-\log[\text{H}^+]$

$7.8=-\log[\text{H}^+]$

$\log[\text{H}^+]=-7.8$

$[\text{H}^+]=\text{anitlog}(-7.8)$

$=1.58\times10^{-8}\text{M}$

$\text{p}_1=\frac{40}{100}\times\text{p}_{\text{total}}$

$=\frac{40}{100}\times10^5$

$=4\times10^4\text{ Pa}$

Partial pressure of I₂ molecules,

$\text{p}_{\text{I}_2}=\frac{60}{100}\times\text{p}_{\text{total}}$

$=\frac{60}{100}\times10^5$

$=6\times10^4\text{ Pa}$

Now, for the given reaction,

$\text{K}_{\text{p}}=\frac{(\text{pI})^2}{\text{p}_{\text{I}_2}}$

$=\frac{(4\times10^4)^2\text{ Pa}^2}{6\times10^4\text{ Pa}}$

$=2.67\times10^4\text{ Pa}$

1. Dissociation Constant of an Acid K_a: It measures the ability of an acid to lose [H⁺] experimentally.

$\text{CH}_3\text{COOH}(\text{l})+\text{H}_2\text{O}(\text{l})\\\rightleftharpoons\text{CH}_3\text{COO}^-(\text{aq})+\text{H}_3\text{O}^+(\text{aq})$

$\text{K}_{\text{a}}=\frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}$

2. Buffer Solution: The solution whose pH does not change by adding small amount of H+ or OH⁺ is called buffer solution. e.g. mixture of CH₃COOH and CH₃COONa is a buffer solution.

3. Solubility Product: It is a product of molar concentration of ions formed in a saturated solution at a given temperature raised to the power equal to the number of each ions formed by 1 mole of sparingly soluble compound.

$\text{e.g. AgCl}(\text{s})\rightleftharpoons\text{Ag}^+(\text{aq})+\text{Cl}^-\text{(aq)};$

$\text{K}_{\text{sq}}=[\text{Ag}^+][\text{Cl}^-]$

Let S be the solubility of Al(OH)₃

$\text{K}_\text{sp}=[\text{Al}^{3+}][\text{OH}^-]^3=\text{(S)(3S)}^3=27\text{S}^4$

$\text{S}^4=\frac{\text{K}_\text{sp}}{27}=\frac{27\times10^{-11}}{27\times10}=1\times10^{-12}$

$\text{S}=1\times10^{-3}\text{mol L}^{-1}$

Solubility of Al(OH)₃

Molar mass of Al(OH)₃ is 78g. Therefore,

Solubility of Al(OH)₃ in g L^-1 = 1 × 10^-3 × 78g L^-1

= 78 × 10^-3g L^-1

= 7.8 × 10^-2g L^-1

pH of the solution

S = 1 × 10^-3mol L^-1

[OH^-] = 3S = 3 × 1 × 10^-3 = 3 × 10^-3

pOH = 3 -log 3

pH = 14 -pOH = 11 + log 3 = 11.4771

Applying law of chemical equilibrium,

$\text{K}_{\text{c}}=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}\Rightarrow0.14=\frac{\text{x.x}}{(0.78-2\text{x})^2}$

$\text{x}^2=0.14(0.78-2\text{x})^2$

$>$

$\text{or }\frac{\text{x}}{0.78-2\text{x}}=\sqrt{0.14}=0.374$

$\text{or }\text{x}=0.292-0.748\text{x}$

$\text{or }1.748\text{x}=0.292$

$\text{or }\text{x}=0.167$

Hence at equilibrium, $[\text{I}_2]=[\text{Cl}_2]=0.167\text{ M}$

$[\text{ICl}]=0.78-2\times0.167=0.446\text{ M}$

$\text{K}_\text{sp}=\text{[Pb}^{2+}][\text{Cl}^-]^2=\text{(S)(2S)}^2=4\text{S}^3$

$\text{S}^3=\frac{\text{K}_\text{sp}}{4}=\frac{3.2\times10^{-8}}{4}8\times10^{-9}\text{mol L}^{-1}$

$\text{S}=\sqrt[3]{8\times10^{-9}}=2\times10^{-3}\text{mol L}^{-1}$

Solubility of $\text{PbCl}_4=2\times10^{-3}\times278$ (molar mass of PbCl₂) = 556 × 10^-3g L^-1

= 0.556 g L^-1

To get saturated solution, 0.556g PbCl₂ is dissolved in 1L water.

0.1g of PbCl₂, i disobed in $\frac{0.1}{0.556}=0.1798\text{L}$ water.

To make a saturated solution, 0.1g PbCl₂ is dissolved in $0.1798\approx0.2\text{L}$ water.

1. $\text{K}_{\text{p}}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}=41$

$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$

$\text{K}'_{\text{p}}=\frac{\text{[N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$

$=\frac{1}{\text{K}_{\text{p}}}=\frac{1}{41}=0.024$

2. $\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2\rightleftharpoons\text{NH}_3(\text{g})$

$\text{K}''_{\text{p}}=\frac{[\text{NH}_3]}{[\text{NH}_2]^{\frac{1}{2}}[\text{H}_2]^{\frac{3}{4}}}$

$=\sqrt{\text{K}_{\text{p}}}=\sqrt{41}=\sqrt{6.4}$

3. $2\text{N}_2\text{(g)}+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$

$\text{K}'''_{\text{p}}=\frac{[\text{NH}_3]^4}{[\text{N}_2]^2[\text{H}_2]^6}$

$=(\text{K}_{\text{p}})^2=(41)^2=1681$

$=1.68\times10^3$

$\text{K}_{\text{p}}=\frac{\text{p}_{\text{C}_2\text{H}_4}.\text{p}_{\text{H}_2}}{\text{p}_{\text{C}_2\text{H}_6}}$

$=\frac{\text{p}\times\text{p}}{4.0-\text{p}}$

$0.04=\frac{\text{p}^2}{4.0-\text{p}}$

$\text{or }0.16-0.04\text{p}=\text{p}^2$

$\text{p}=-0.04\pm\frac{\sqrt{0.0016-4(-0.16)}}{2}$

$\text{p}=\frac{-0.04\pm0.80}{2}$

$\Rightarrow\text{p}=0.38$

(by taking positive value)

Hence, $\text{p}_{\text{C}_2\text{H}_6}=4.0-0.38=3.62\text{atm}.$

1. $\begin{matrix}&\text{H}_2(\text{g})&+&\text{Cl}_2(\text{g})\rightleftharpoons&\text{2Hcl}(\text{g})\\\text{Initial Conc. }&0.25&&0.25\text{M}&0\\\text{Final Conc.at equilibrium}&0.0314\text{M}&&0.0314\text{M}&0\end{matrix}$

$2(0.25-0.0314)=0.2186\text{M}\times2$

$=0.219\text{M}\times2=0.438$

$\text{K}_{\text{c}}=\frac{[\text{HCl}]^2}{[\text{H}_2][\text{Cl}_2]}$

$=\frac{(0.438)^2}{0.0314\times0.314}=195$

2. $\text{K}_{\text{p}}=\text{K}_{\text{c}}(\text{RT})^{\Delta\text{n}}$

$\Delta\text{n}=0$

$\text{K}_{\text{p}}=\text{K}_{\text{c}}=195$

$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]}$

$=-\log(1.75\times10^{-5})+\log\frac{0.15}{0.10}$

$= (5 - 0.2430) + 0.1761$

$= 4.757 + 0.1761$

$= 4.933$

1. 1cc of 1M NaOH contains NaOH = 10^-3mol.

This will convent 10^-3mol of acetic acid into the salt so that salt formed = 10^-3mol.

Now, [Acid] = 0.10 - 0.001 = 0.099M

[Salt] = 0.15 + 0.001 = 0.151

$\text{pH}=4.757+\log\frac{0.151}{0.099}$

$\therefore$ Increase in pH = 4.940 - 4.933 = 0.007 which is negligible.

2. 1cc of 1M HCl contains HCl= 10^-3mol.

This will convert 10^-3mol CH₃COONa into CH₃COOH

$\therefore$ Now, [Acid] = 0.10 + 0.001 = 0.101M

[Salt] = 0.15 - 0.001 = 0.149M

$\text{pH}=4.757+\log\frac{0.149}{0.101}$

$= 4.757 + 0.169 = 4.925$

$\therefore$ Decrease in pH = 4.933 = 0.007 which is again negligible.

3. Caculation of buffer index

No. of moles of HCl or NaOH added = 0.001mol Change in pH = 0.007

Hence, buffer index $=\frac{\text{dn}}{\text{dpH}}=\frac{0.001}{0.007}$

$=\frac{1}{7}=0.143$

1. $\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+\ \text{K}_\text{a}=1.74\times10^{-5}$

2. $\text{H}_2\text{O}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{OH}^-\ \text{K}_\text{w}=1.0\times10^{-14}$

Since Ka >> K_w:

$\text{K}_\text{a}=\frac{(.05\alpha)(.05\alpha)}{(.05-0.05\alpha)}$

$=\frac{(.05\alpha)(0.05\alpha)}{.05(1-\alpha)}$

$=\frac{.05\alpha^2}{1-\alpha}$

$1.74\times10^{-5}=\frac{0.05\alpha^2}{1-\alpha}$

$1.74\times10^{-5}-1.74\times10^{-5}\alpha=0.05\alpha^2$

$0.05\alpha^2+1.74\times10^{-5}\alpha-1.74\times10^{-5}$

$\text{D}=\text{b}^2-4\text{ac}$

$=(1.74\times10^{-5})^2-4(.05)(1.74\times10^{-5})$

$=3.02\times10^{-25}+.348\times10^{-5}$

$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$

$\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$

$=\sqrt{\frac{34.8\times10^{-5}\times10}{10}}$

$=\sqrt{3.48\times10^{-6}}$

$=\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$

$\alpha=1.86\times10^{-3}$

$[\text{CH}_3\text{COO}^-]=0.05\times1.86\times10^{-3}$

$=\frac{0.93\times10^{-3}}{1000}$

$=.000093$

Method 2

Degree of dissociation,

$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$

c = 0.05M

K_a = 1.74 × 10^–5

^{Then, $\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$}

$\alpha=\sqrt{34.8\times10^{-5}}$

$\alpha=\sqrt{3.48}\times10^{-4}$

$\alpha=1.8610^{-2}$

$\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$

Thus, concentration of CH₃COO– = c.α

$=.05\times1.86\times10^{-2}$

$=.093\times10^{-2}$

$=.00093\text{M}$

$\text{Since}[\text{oAc}^-]=[\text{H}^+],$

$[\text{H}^+]=.00093=.093\times10^{-2}.$

$\text{pH}=-\log[\text{H}^+]$

$=-\log(.093\times10^{-2})$

$\therefore\ \text{pH}=3.03$

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

1. $\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4(\text{aq});$

(mol. mass of CaSO₄ = 136g/ mol)

Let the solubility of CaSO₄ in mol/L is x.

$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]=\text{x}^2$

$\text{x}=\sqrt{\text{K}_\text{sp}}=\sqrt{9.1\times10^{-6}}$

$=3.02\times10^{-3}\text{mol/L}$

$=3.02\times10^{-3}\times136\text{g/L}$

$=0.411\text{g/L};$

For dissolving 0.411g of CaSO₄, water required is 1L

For dissolving 1g of CaSO₄, water required $=\frac{1}{0.411}\text{L}=2.43\text{L}$

2. For the reaction:

$2\text{SO}_2(\text{g})+\text{O}_2\text{(g)}\rightleftharpoons2\text{SO}_3(\text{g})$

$\Delta\text{n}_\text{g}=2-3=-1$

$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$

$\text{K}_\text{c}=\text{K}_\text{p}(\text{RT})^{-\Delta\text{n}}$

For this reaction

$\text{K}_\text{c}=\text{K}_\text{p}\text{RT}=(2\times10^{10}\text{bar}^{-1}) $

$(0.083\text{L bar K}^{-1}\text{mol}^{-1})\times450\text{K}$

$=7.48\times10^{11}\text{L mol}^{-1}$

1. The reaction is endothermic because equilibrium constant is increasing with increase in temperature.
2. $\text{s}=1.7\times10^{-3}\text{g/ 100ml}$

$=\frac{1.7\times10^{-3}}{100}\times1000$

$=1.7\times10^{-2}\text{g L}^{-1}$

$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{(40+38)\text{g mol}^{-1}}$

$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{78\text{g mol}^{-1}}$

$=2.18\times10^{-4}\text{mol L}^{-1}$

$\text{CaF}_2\rightleftharpoons\text{Ca}^{2+}+\text{2F}^-$

$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{F}^-]^2$

$=(2.18\times10^{-4})(2\times2.18\times10^{-4})^2$

$=41.44\times10^{-12}$

$=4.144\times10^{-11}$

1. HCOOH is conjugate acid of HCOO^-.

2. $\text{H}_2\text{O}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{OH}^-$

$\text{K}_\text{w}=1.0\times10^{-14}=[\text{H}_3\text{O}^+][\text{OH}^-]$ $(\because[\text{H}_3\text{O}^+]=[\text{OH}^-])$

$=1.0\times10^{-14}=[\text{H}_3\text{O}^+]^2$

$[\text{H}_3\text{O}^+]=\sqrt{1.0\times10^{-14}}=10^{-7}\text{mol L}^{-1}$

$\text{H}_2\text{O}+\text{HCl}\xrightarrow{\ \ \ \ }\text{H}_3\text{O}s^++\text{Cl}^-$

$10^{-8}$

Total $[\text{H}_3\text{O}^+]=10^{-8}+10^{-7}=10^{-7}(1+10^{-1})$

$=1.1\times10^{-7}\text{mol L}^{-1}$

$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log1.1\times10^{-7}$

$=-0.04139+7.00=6.958$

3. $\text{A}_2\text{X}_3\rightleftharpoons2\text{A}^{3+}+3\text{X}^{2-}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{s}\ \ \ \ \ \ \ \ \ \ \ 3\text{s}$

where solubility is 's' mol L^-1

$\text{K}_\text{sp}=[\text{A}^{3+}]^2[\text{X}^{2-}]^3$

$1.1\times10^{-23}=(2\text{s})^2(3\text{s})^3=108\text{s}^5$

$\Rightarrow \sqrt{\frac{110}{108}\times10^{-25}}=1\times10^{-5}\text{mol L}^{-1}$

$\text{s}\simeq1\times10^{-5}\text{mol L}^{-1}$

1. $\text{PCl}_5(\text{s})\rightleftharpoons\text{PCl}_3(\text{s})+\text{Cl}_2(\text{g});$

$\Delta _\text{r}\text{H}^\text{o}=124.0\text{kJ mol}^{-1}$

1. $\text{K}_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=8.3\times10^{-3}$

2. K_c for the reverse reaction $=\frac{1}{\text{K}_\text{c}\text{for the forward reaction}}$

$=\frac{1}{8.3\times10^{-3}}=120.48$

1. When pressure increases K_c reamains unchanged.
2. K_c increses with increases in temprature because the reaction is endothermic.

1. $\text{K}=\frac{[\text{BaO(s)}][\text{CO}_2(\text{g})]}{[\text{BaCO}_3(\text{g})]}$

Since $[\text{BaCO}_3(\text{s})]=[\text{BaO(s)}]=1$

$\text{K}=[\text{CO}_2(\text{g})]$

2. ​​​​​​​$\text{K}=\frac{[\text{CO}_2(\text{g})][\text{H}_2\text{O(g)}]^2}{[\text{CH}_2(\text{g})][\text{O}_2(\text{g})]^2}$

1. Lewis acids are electron deficient species. So, Lewis acids are $\text{BF}_3,\text{H}^+$ and $\text{NH}^+_4.$

2. $\text{pH}=-\log[\text{H}^+]$

$\Rightarrow\log[\text{H}^+]=-\text{pH}=-3.76$

$\Rightarrow \log[\text{H}^+]=-3.76+1-1$

$[\text{H}^+]=\bar{4}.24\text{ Antilog}$

$[\text{H}^+]=1.738\times10^{-4}$

$=1.74\times10^{-4}\text{M}$

3. Common ion effect is the suppression of the dissociation of weak electrolyte in the presence of a strong electrolyte having a common ion. Sulphides of II group radicals are precipitated by passing H₂S gas in presence of HCl. H₂S being weak electrolyte ionises slightly and HCl which is a strong electrolyte is completely ionised.

$\text{H}_2\text{S}\rightleftharpoons2\text{H}^++\text{S}^{2-}$

$\text{HCl}\xrightarrow{\ \ \ \ \ \ \ }\text{H}^++\text{Cl}^-$

Due to common ion effect the degree of dissociation of H₂S decreases and concentration of S^2- ions in solution becomes small enough to precipitate only II group cations and not group IV cations. The second group cations have lower solubility product than IV group cations.

1. The reaction is

$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$

Given: $\text{K}_\text{p}=3.6\times10^{-2}$ at 500K

The reation between K_p and K_c is

$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$

For the above reaction

$\Delta \text{n}=2-4=-2$

$\text{K}_\text{c}=\frac{\text{K}_\text{p}}{(\text{RT})^{\Delta\text{n}}}$ (R = 0.083bar L K^-1mol^-1)

$=\frac{3.6\times10^{-2}}{(0.083\times500)^{-2}}$

$=3.6\times10^{-2}\times(0.083\times500)^2$

$=62$

2.  
3. The equilibrium will shift in backward reaction because number of moles of products are more than reactants $\Delta \text{n}>0.$
4. No effect because number of moles of reactants and products are equal, i.e., $\Delta \text{n}=0.$

1. If Q_c < K_c; the reaction tends towards forward direction to attain equilibrium.
2. Le Chatelier's Principle: If a system in equilibrium is subjected to a change in concentration, temperature or pressure, the equilibrium shifts in a direction that tends to undo the effect of the change.
3. Hydroxides of group III are precipitated by adding NH₄OH in presence of NH₄Cl. The role of NH₄Cl is to produce common ion effect.

$\text{NH}_4\text{OH}\rightleftharpoons\text{NH}^+_4+\text{OH}^-$

$\text{NH}_4\text{Cl}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{NH}_4^++\text{Cl}^-$

Due to common ion effect, the degree of dissociation of NH₄OH is suppressed and less OH^‑ are formed. This less concentration of OH^- is sufficient to precipitate group III cations but not the cations of higher groups since the K_sp of group III < subsequent groups.

1. $\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$

$\text{K}_{\text{c}}2\times10^{13},\text{T}=300\text{K},\Delta\text{G}^\circ=?$

$\Delta\text{G}^\circ=-2.30\text{ RT }\log\text{K}_{\text{c}}$

$=-2.303\times8.314\times300\log2\times10^{13}$

$=-19.147\times300(\log2+\log^{13})$

$=\frac{-19.147\times300\times13.3010\text{J}}{1000}$

$=-76.402\text{kJ mol}^{-1}$

2. For A,

$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log4.0\times10^{-7}$

$=-\log4.0-\log10^{-7}$

$=-0.6021+7.000=6.3979$

3. For B,

$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log3.2\times10^{-6}$

$=-\log3.2-\log10^{-6}$

$=-0.5050+6.000=5.4950$

pH of 'A' is higher than 'B'. 

1. Cations are grouped on the basis of their solubility product (K_sp). i.e. value of K_sp is closed to each other, e.g., K_sp of sulphides of groups II cations are close to each other.
2. For the reaction, the reaction quotient Q_c is given by $\text{Q}_\text{c}=\frac{[\text{B}][\text{C}]}{[\text{A}]^2}$

as $[\text{A}]=[\text{B}]=[\text{C}]=3\times10^{-4}\text{M}$

$\text{Q}_\text{c}=\frac{(3\times10^{-4})(3\times10^{-4})}{(3\times10^{-4})^2}=1$

as $\text{Q}\text{c}>\text{K}_\text{c}$

So, the reaction will proceed in the reverse direction.

3. Molar mass of Sr(OH)₂ = 87.6 + 34

= 121.6g mol^-1

Solubility of Sr(OH)₂ in mol L^-1

$=\frac{19.23\text{g L}^{-1}}{121.6\text{g mol}^{-1}}$

$\text{Sr(OH)}_2\xrightarrow{\ \ \ \ \ }\text{Sr}^{2+}+2\text{OH}^-$

$[\text{Sr}^{2+}]=0.1581\text{M},[\text{OH}^-]$

$=2\times0.1581=0.3162\text{M}$

$\text{p}[\text{OH}]=-\log0.3162=0.5$

$\Rightarrow \text{pH}=14-0.5=13.5$

1. Solubility Product: It is defined as the product of molar concentrations of the ions (formed in the saturated solution at a given temperature) raised to the power equal to the number of times each ion occurs in the equation for solubility equilibrium,

 $\text{Zr}_3(\text{PO}_4)_4\rightleftharpoons3\text{Zr}^{4+}+4\text{PO}^{3-}_4$

$\text{K}_{\text{sp}}=[\text{Zr}^{4+}][\text{PO}^{3-}_4]^4$

2. $\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log\text{C}\alpha$

$=-\log\sqrt{\text{K}_\text{a}\times\text{C}}$

$=-\log\sqrt{1.74\times10^{-5}\times0.01}$

$=-\log\sqrt{1.74\times10^{-7}}$

$=-\log\sqrt{17.4\times10^{-8}}$

$=-\log4.17\times10^{-4}$

$=-\log4.17-\log10^{-4}$

$=-0.6217+4.000$

$=3.3783$

3. It is due to common ion, CI^- increase, therefore rate of backward reaction increases, solubility of NaCl decreases.

1. $\text{NH}^+_4$ is conjugate acid of NH₃.
2.  

1. NH₄Cl is salt of weak base NH₄OH and strong acid HCl, therefore H⁺ ions are more than OH^- ions thus, pH is less than 7.
2. It is done to decrease [OH^-] due to common ion effect, so that only group III radicals Fe³⁺ or Al³⁺ get precipitated and higher group radicals do not.

1. When temperature is increased equilibrium will shift to backward direction as reaction is exothermic.
2. When pressure is increased rate of forward reaction will increase as there is decrease in number of moles from reactants to products.

1. Solubility product is defined as the product of molar concentration of ions raised to the power the number of ions formed per formula of the compound.

$\text{FeCl}_3(\text{s})\rightleftharpoons\text{Fe}^{3+}+3\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ '\text{s}' \ \ \ \ \ \ \ \ \ \ 3\text{s}$

$\text{K}_{\text{sp}}=[\text{Fe}^{3+}][\text{Cl}^-]^3$

$\text{K}_\text{sp}=(\text{s})(3\text{s})^3,$

where 's' mol L^-1 is solubility.

$\Rightarrow\text{K}_\text{sp}= 27\text{s}^4$

$\Rightarrow\text{s}=4\sqrt{\frac{\text{K}_\text{sp}}{27}}$

2. Solubility of gases in liquids decreases with increase in temperature because force of attraction between gas and liquid decreases at high temperature.
3. K for the reaction = 4

$\therefore \text{K}'$ for reverse reaction $=\frac{1}{4}=0.25$ [$\because \text{K}'=\frac{1}{ \text{K}}$ for reverse reaction]

4. $$pH of 10^-8M HCl solution.

$\text{HCl}\xrightarrow{\ \ \ \ \ \ }\text{H}^++\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{10^{-8}\text{M}}$

$\text{H}_2\text{O}\rightleftharpoons\text{H}^++\text{OH}^-$

$\text{K}_\text{w}=1\times10^{-14}$

$\Rightarrow [\text{H}^+][\text{OH}^-]=10^{-14}$

$\because[\text{H}^+]=[\text{OH}]$

$\therefore [\text{H}^+]^2=10^{-14}$

$\Rightarrow [\text{H}^+]=10^{-7}\text{mol L}^{-1}$

Total concentration of

$[\text{H}^+]=(10^{-8}+10^{-7})$

$=10^{-7}(1+0.1)$

$=1.1\times10^{-7}$

$\therefore \text{pH}=-\log[\text{H}^+]$

$=-\log1.1\times10^{-7}$

$=-\log1.1-\log10^{-7}$

$\Rightarrow \text{pH}=-0.0454+7.000$

$=6.9546$

1. $\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$

1. $\text{K}_{\text{p}}=\frac{(\text{p}_{\text{CO}})(\text{p}_{\text{H}_2})^3}{(\text{p}_{\text{CH}_4})(\text{p}_{\text{H}_2\text{O}})}$
2.  

1. On increasing pressure, the reaction equilibria will shift in the backward direction.

2. As the given reaction is endothermic, on increasing temperature the given equilibrium will shift in forward direction.

3. There is no effect of catalyst in equilibrium composition, however the equilibrium will be attained faster.

2. 10ml of 0.1M H₂SO₄ is mixed with 10ml of 0.1M KOH.

The reaction is

$2\text{KOH}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{H}_2\text{O}$

10ml of 0.1M H₂SO₄ = 0.1 × 10 = 1 millimole

10ml of 0.1M KOH = 0.1 × 10 = 1 millimole

1 millimole of KOH will react with 0.5 millimole of H₂SO₄

$[\because$ 0.5 millimole will produce 1 millimole of H⁺]

$\therefore$ H₂SO₄ left = 1 - 0.5 = 0.5 millimole

Volume of reaction mixtrue

= 10 + 10 = 20ml

$\therefore$ Molarity of × in the mixtrue.

$=\frac{0.5}{20}=2.5\times10^{-2}\text{M}$

$[\text{H}^+]=2\times2.5 \times10^{-2}$

$=5\times10^{-2}\text{M}$

$\text{pH}=-\log(5\times10^{-2})$

$=-\log5-\log10^{-2}$

$=-0.6990+2.0000=1.30$

1. If we increase the concentration of NO, the rate of forward reaction will increase, i.e. more NO₂ will be formed.
2. Decrease in pressure will favour backward reaction, i.e. less NO₂ will be formed.

1 mole of CO₂ is obtained by decomposition of 1 mole CaCO₃. Therefore, moles of CaCO₃ decomposed is equal to the moles of CO₂ = 0.065mol.

Now NaCl(aq) = Na⁺(aq) + Cl^-(aq)

and $\text{AgNO}_3 \text{aq}= \text{Ag}+ \text{(aq}) +\text{NO}_3^- \text{(aq)}$

[Cl^-] = [NaCl] = 0.005N

[Ag⁺] = [AgNO₃] = 0.005M

$\therefore$ Ionic product of [Ag⁺][Cl^-] = 0.005 × 0.005 = 2.5 × 10^-5

Since ionic product is greater than its solubility product, precipitation will occur.

$\Delta\text{G}=\Delta\text{G}^{\ominus}+\text{RT InQ}$

$\Delta\text{G}=$ change in free energy as the reaction proceeds

$\Delta\text{G}^{\ominus}$ standard free energy

Q = reaction quotienten

R= gas constant

T = absolute temperature in Kual

1. Since, $\Delta\text{G}^{\ominus}=-\text{RT In K}$

$\therefore\Delta\text{G}=-\text{RT In K}+\text{RT In Q;}$

$\Delta\text{G}=\text{RT In}\frac{\text{Q}}{\text{K}}$

will be negative and the reaction proceeds in the forward direction. $\text{Q}=\text{K},\Delta\text{G}=0$ If reaction is in equilibrium and there is no net reaction.

2. $\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$

$\text{K}_{\text{c}}=\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$

On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,

$\text{Q}_{\text{c}}=\frac{2[\text{CH}_4].2[\text{H}_2\text{O}]}{2[\text{CO}]\{2[\text{H}_2]\}^3}$

$\frac{1}{4}\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}=\frac{1}{4}\text{K}_{\text{c}}$

Therefore, Q_c is less than K_C so Q_c will tend to increase to re-establish equilibrium and the reaction will go in forward direction.

$\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$

If the concentrations of both solutions are equal to or less than 5.02 × 10^–9M, then there will be no precipitation of iron sulphide.

$=\frac{\Big(\frac{\alpha}{1+\alpha}\text{p}\Big)\Big(\frac{\alpha}{1+\alpha}\Big)}{\Big(\frac{1-\alpha}{1+\alpha}\Big)\text{p}}$

[salt] = 10 × 0.5 = 5.0

Conc. of salt in the mixture

Also,

$\Rightarrow\text{pH}=-\log(10.75\times10^{-9})$

1. pH of milk = 6.8

Since, $\text{pH}=-\log[\text{H}^+]$

$6.8=-\log[\text{H}^+]$

$\log[\text{H}^+]=-6.8$

$[\text{H}^+]=\text{anitlog}(-6.8)$

$=1.5\times10^{-7}\text{M}$

2. pH of black coffee = 5.0

Since, $\text{pH}=-\log[\text{H}^+]$

$5.0=-\log[\text{H}^+]$

$\log[\text{H}^+]=-5.0$

$[\text{H}^+]=\text{anitlog}(-5.0)$

$=10^{-5}\text{M}$

3. pH of tomato juice = 4.2

Since, $\text{pH}=-\log[\text{H}^+]$

$4.2=-\log[\text{H}^+]$

$\log[\text{H}^+]=-4.2$

$[\text{H}^+]=\text{anitlog}(-4.2)$

$=6.31\times10^{-5}\text{M}$

4. pH of lemon juice = 2.2

Since, $\text{pH}=-\log[\text{H}^+]$

$2.2=-\log[\text{H}^+]$

$\log[\text{H}^+]=-2.2$

$[\text{H}^+]=\text{anitlog}(-2.2)$

$=6.31\times10^{-3}\text{M}$

5. pH of egg white = 7.8

Since, $\text{pH}=-\log[\text{H}^+]$

$7.8=-\log[\text{H}^+]$

$\log[\text{H}^+]=-7.8$

$[\text{H}^+]=\text{anitlog}(-7.8)$

$=1.58\times10^{-8}\text{M}$

Now,

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.