(a) CH₃COONa is a salt of weak acid
(CH₃COOH) and a strong base (NaOH).
Hence, the solutions is alkaline due to hydrolysis.
\(CH_{3}COO^{-}_{(aq)}+H_{2}O_{(aq)}\rightleftharpoons CH_{3}COOH_{(aq)}+OH^{-}_{(aq)}\)
(i)\(h=\sqrt{\frac{K_{w}}{K_{a}\times C}}\)
Given that pK_a =4.74
pK_a = -log K_a
ie., K_a = antilog of (-pK_a)
= antilog of (-4.74)
= antilog of (-5 + 0.26)
= 10^-5 \(\times\) 1.8 = 1.8 \(\times\) 10^-5
[antilog of 0.26 = 1.82 \( \simeq\) 1.8]
\(\therefore\) h=\(\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times0.1}}\)
h = 7.5 x 10^-5
(ii) \(K_{h}=\frac{K_{w}}{K_{a}}=\frac{1\times10^{-14}}{1.8\times10^{-5}}\)
\(=5.56\times10^{-10}\)
iii) \(pH=7+\frac{pK_{a}}{2}+\frac{logC}{2}\)
= \(7+\frac{4.74}{2}+\frac{log0.1}{2}\)
= 7 + 2.37 - 0.5
= 8.87

\(\mathrm{M} =\frac{\mathrm{V}_{1} \mathrm{M}_{1}-\mathrm{V}_{2} \mathrm{M}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}}  \)
\(=\frac{(50 \times 0.05)-(50 \times 0.025)}{50+50}  \)
\(\text { Molarity }=\frac{\text { Number of millimoles }}{\mathrm{V}_{\mathrm{m} l}}\)
\(=\frac{2.5-1.25}{100}=\frac{1.25}{100}=0.0125 \mathrm{M}  \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10} 0.0125  \)
Normality = Molarity \(\times\) basicity
\(=-\log _{10}\left(1.25 \times 10^{-2}\right)  \)
\(=-\left[\log _{10} 1.25-2 \log _{10} 10\right]  \)
\(=2-\log _{10} 1.25=2-0.0969 \)
pH = 1.9031

For a first order reaction
\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ 1 } } \)
V_∞= 58.3 ml.

t(min)	Vt	V∞=Vt	\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } \)
6	19.3	58.3-19.3=39.0	\(k=\frac { 2.303 }{ 6 } \log\left( \frac { 58.3 }{ 39 } \right) =0.0670\) min-1
12	32.6	58.3-32.6=25.7	\(k=\frac { 2.303 }{ 12 } \log\left( \frac { 58.3 }{ 25.7 } \right) =0.06838\) min-1
18	41.3	58.3-41.3=17.0	\(k=\frac { 2.303 }{ 18 } \log\left( \frac { 58.3 }{ 17 } \right) =0.06838\) min-1
24	46.5	58.3-46.5=11.8	\(k=\frac { 2.303 }{ 24 } \log\left( \frac { 58.3 }{ 11.8 } \right) =0.0666\) min-1
30	50.4	58.3 - 50.4 = 7.9	\(k=\frac{2.303}{30} \log \left(\frac{58.3}{7.9}\right)=0.067\) min-1
			Mean value of k = 0.0674 min-1

As the rate constants are constant through out it is a first order reaction. 

A Reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. For a first order reaction,
A ⟶ product
Rate law can be expressed as
Rate = k[A]^-1
\(\frac { -d\left[ A \right] }{ \left[ A \right] } =k dt\)    ...(1)
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial: concentration [A_o] to [A] at the later time.
\(\int _{ { [A }_{ 0 }] }^{ [A] }{ \frac { -d\left[ A \right] }{ \left[ A \right] } } =k\int _{ 0 }^{ t }{ dt } \)
\({ \left( -In\left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =k\left( t-0 \right) \)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =kt\)
\(In\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\) ...(2)
This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303.

\(2.303\quad log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\)
\(k=\frac { 2.303 }{ t } log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) \)    ...(3)
Equation (2) can be written in the form
y = mx+c as below
In [A₀]-In[A] = kt
In[A] = In[A₀]-kt
⇒ y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval 't' a plot of In[A] against 't' yields a straight line with a negative slope. From this, the rate constant is calculated.

Given that H₃O⁺ = \(2\times10^{-3}M\)
\(K_{w}=[H_{3}O^{+}][OH^{-}]\)
\(\therefore [OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1\times10^{-14}}{2\times10^{-3}}=0.5\times10^{-11}M\)
\(2\times10^{-3} >>0.5\times10^{-11}\)
i.e., [H₃O⁺]>>[OH^-], hence the juice is acidic in nature

(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.

\(\log K=\log A-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \)
y = c + mx
\(m=-\frac { { E }_{ a } }{ 2.303R } \)
E_a = -2.303 Rm
E_a = -2.303 x 8.314 x (-4000)
E_a = 76,589J  mol^-1
E_a = 76.589 KJ mol^-1

K = 1.54 \(\times\) 10^-3s^-1; t_1/2 = ?
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 1.54\times { 10 }^{ -3 } } \)
= 450 s

(i) First order reaction
(ii) First order reaction
(iii) \(\frac{1}{2}+2=2 \frac{1}{2}\); Pseudo first order reaction

(i) Ostwald's dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).
where  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
(ii) The dissociation of acetic acid can be represented as
\(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
The dissociation constant of acetic acid is,
\({ K }_{ a }=\frac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }COO^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)  ........(1) 

	CH3COOH	H+	CH3COO-
Initial number of moles	1	-	-
Degree of dissociation of CH3COOH	α	-	-
Number of moles at equilibrium	1-α	α	α
Equilibrium concentration	(1-α)C	αC	αC

Substituting the equilibrium concentration in equation (1)
\({ K }_{ a }=\cfrac { \left( \alpha C \right) \left( \alpha C \right) }{ \left( 1-\alpha \right) C } \) 
\({ K }_{ a }=\cfrac { { \alpha }^{ 2 }C }{ 1-\alpha } \) .......(2) 
(iii) We know that weak acid dissociates only to a very small extent compared to one, a is so small and hence in the denominator (1 - α) ⋍1. The above expression (2) now becomes,
k_a =a²C \(\Rightarrow { \alpha }^{ 2 }=\cfrac { { k }_{ a } }{ C } \) ; \(\alpha =\sqrt { \cfrac { { K }_{ a } }{ C } } \) 
(iv) When dilution increases, the degree of dissociation of weak electrolyte also increases. This is called Ostwald's dilution law
Also \(;\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}\) and \(\left[\mathrm{H}^{+}\right]=\left(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}\right) \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \mathrm{C}^{2}}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
Similarly for a weak base
\(\begin{aligned} & \mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} ; \quad \therefore \alpha=\sqrt{\frac{\mathrm{k}_{\mathrm{b}}}{\mathrm{C}}}, \\ \end{aligned}\)
\(\begin{aligned} & {\left[\mathrm{OH}^{-}\right] \alpha \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}} \times \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}} \mathrm{C}^2}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}} \end{aligned}\)

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %