Ionic Equilibrium

Given that,

Mass of $\mathrm{PbCl}_2$ to dissolve $=0.1 \mathrm{~g}$

$ K_{\mathrm{sp}}\left(\mathrm{PbCl}_2\right)=3.2 \times 10^{-8} $

Atomic mass of $\mathrm{Pb}=207 \mathrm{u}$

Atomic mass of $\mathrm{Cl}=35.5 \mathrm{u}$

For $\mathrm{PbCl}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$, the $K_{\mathrm{sp}}$ expression is

$ \begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2=(S)(2 S)^2=4 S^3 \\ 3.2 \times 10^{-8} & =4 S^3 \\ S^3 & =8 \times 10^{-9} \\ S & =2 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \\ = & \left(2 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\right) \times(278.0 \mathrm{~g} / \mathrm{mol}) \\ = & 0.556 \mathrm{~g} / \mathrm{L} \\ \text { Volume }(\mathrm{L}) & =\frac{\text { Mass of } \mathrm{PbCl}_2}{\text { Solubility }(\mathrm{g} / \mathrm{L})} \end{aligned} $

$ \begin{aligned} &=\frac{0.1 \mathrm{~g}}{0.556 \mathrm{~g} / \mathrm{L}} \approx 0.17985 \mathrm{~L}\\ &\text { So, volume } \approx 180 \mathrm{~mL} \end{aligned} $

In the saturated solution, both salts are in equilibrium with their ions. The concentration of the common ion, $\mathrm{CO}_3^{2-}$, links the two equilibria.

Step 1 Find $\left[\mathrm{CO}_3^{2-}\right]$ from the $\mathrm{MgCO}_3$ equilibrium.

$ \begin{aligned} K_{s p}\left(\mathrm{MgCO}_3\right) & =\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right] \\ {\left[\mathrm{CO}_3^{2-}\right] } & =\frac{1.6 \times 10^{-6}}{3.2 \times 10^{-5}}=0.05 \mathrm{M} \end{aligned} $

Step 2 Find $\left[\mathrm{Ag}^{+}\right]$from the $\mathrm{Ag}_2 \mathrm{CO}_3$ equilibrium,

$ \begin{aligned} & K_{s p}\left(\mathrm{Ag}_2 \mathrm{CO}_3\right)=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & {\left[\mathrm{Ag}^{+}\right]^2 }=\frac{K_{s p}\left(\mathrm{Ag}_2 \mathrm{CO}_3\right)}{\left[\mathrm{CO}_3^{2-}\right]} \\ &=\frac{8.0 \times 10^{-12}}{0.05}=1.6 \times 10^{-10} \\ & {\left[\mathrm{Ag}^{+}\right]=\sqrt{1.6 \times 10^{-10}}=\sqrt{1.6} \times 10^{-5} \mathrm{M} } \end{aligned} $

For $\mathrm{HCl},\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}$,

$ \mathrm{HCl}=10^{-2} \mathrm{M} $

For NaOH ,

$ \left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}, \mathrm{NaOH}=10^{-2} \mathrm{M} $

$ \begin{aligned} \text { Moles of } \mathrm{H}^{+} & =\left[\mathrm{H}^{+}\right] \times V_{\mathrm{HCl}}=10^{-2} \times 0.2 \\ & =0.002 \mathrm{~mol} \\ \text { Moles of } \mathrm{OH}^{-} & =\left[\mathrm{OH}^{-}\right] \times V_{\mathrm{NaOH}} \\ & =10^{-2} \times 0.300=0.003 \mathrm{~mol} \end{aligned} $

Moles of $\mathrm{OH}^{-}$remaining

$ \begin{aligned} &=\text { Moles of } \mathrm{OH}^{-}-\text {Moles of } \mathrm{H}^{+} \\ &=0.003-0.002=0.001 \mathrm{~mol} \\ & {\left[\mathrm{OH}^{-}\right]_{\text {final }} }=\frac{\text { Moles of } \mathrm{OH}^{-} \text {remaining }}{V_{\text {final }}} \\ &=\frac{0.001}{1}=0.001 \mathrm{M} \\ & \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]_{\text {final }}=-\log (0.001)=3 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-3=11 \end{aligned} $

Given, $K_{\mathrm{sp}}$ of $M X_2=5 \times 10^{-13}$

$K_{\mathrm{sp}}$ of $M X=1.6 \times 10^{-11}$

Ratio of molar solubility of $\frac{M X_2}{M X}=$ ?

From $M X_2=S=3 \sqrt{\frac{K_{\text {sp }}}{4}}=3 \sqrt{\frac{5 \times 10^{-13}}{4}}$

For $M X, S=\sqrt{K_{\mathrm{sp}}}=\sqrt{1.6 \times 10^{-11}}$

So, molar solubility of

$ \frac{M X_2}{M X}=\frac{3 \sqrt{1.25 \times 10^{-13}}}{\sqrt{1.6 \times 10^{-11}}}=12.5 $

The dissociation equation of $\mathrm{PbI}_2$ is,

$ \mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $

Let $s$ be molar solubility of $\mathrm{PbI}_2$

Initial $\left[\mathrm{Pb}^{2+}\right]=0.2 \mathrm{M}$

At equilibrium $\left[\mathrm{Pb}^{2+}\right]=(0.2+s) \mathrm{M}$

At equilibrium $\left[\mathrm{I}^{-}\right]=2 \mathrm{~s} \mathrm{M}$

$ K_{s p}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^2 $

$ =(0.2+s)(2 s)^2=(0.2)\left(4 s^2\right) $

Since $s$ is very small compared to 0.2 there, neglecting it

$ \begin{aligned} & =0.8 s^2 \\ s & =\left(\frac{K_{s p}}{0.8}\right)^{1 / 2} \end{aligned} $

Convert the percentage ionisation to decimal

$ \begin{aligned} \alpha_A & =\frac{4.242}{100}=0.04242 \\ \alpha & =\sqrt{\frac{K_\alpha}{C}} \\ \Rightarrow \quad C_A & =\frac{K_\alpha}{\alpha_A^2} \\ C_A & =\frac{1.8 \times 10^{-5}}{\left(0.04242^2\right.}, C_A=0.01 \mathrm{M} \\ \alpha_B & =\frac{3}{100}=0.03 \\ C_B & =\frac{K_\alpha}{\alpha_B^2}=\frac{1.8 \times 10^{-5}}{(0.03)^2}=0.02 \mathrm{M} \end{aligned} $

Moles from solution $A$,

$ \begin{gathered} n_A=C_A \times V_A=0.01 \mathrm{M} \times 1 \mathrm{~L}=0.01 \mathrm{~mol} \\ n_B=c_B \times V_B=0.02 \mathrm{M} \times 1 \mathrm{~L}=0.02 \mathrm{~mol} \\ n_{\text {Total }}=n_A+n_B=0.03 \mathrm{~mol} \\ V_{\text {Total }}=V_A+V_B=1+1=2 \mathrm{~L} \\ c_{\text {Final }}=\frac{n_{\text {Total }}}{V_{\text {Total }}}=\frac{0.03}{2 \mathrm{~L}}=0.015 \mathrm{M} \end{gathered} $

pOH of buffer solution

$ p \mathrm{OH}=14-\mathrm{pH} \Rightarrow \mathrm{pH}=14-8.2=5.8 $

Moles of $\mathrm{NH}_4 \mathrm{OH}=0.2 \times 0.030$

$ =0.006 \mathrm{~mol} $

Moles of $\mathrm{NH}_4 \mathrm{Cl}=2 \times 0.030=0.060 \mathrm{~mol}$

Total volume $=30 \mathrm{~mL}+30 \mathrm{~mL}=0.060 \mathrm{I}$

Concentration of conjugate base and acid in buffer

$ \begin{aligned} & {\left[\mathrm{NH}_4 \mathrm{OH}\right]=\frac{0.006 \mathrm{~mol}}{0.060 \mathrm{~L}}=0.1 \mathrm{M}} \\ & {\left[\mathrm{NH}_4 \mathrm{Cl}\right]=\frac{0.060 \mathrm{~mol}}{0.060 \mathrm{~L}}=1 \mathrm{M}} \end{aligned} $

Using Henderson equation

$ \begin{aligned} \mathrm{pOH} & =\mathrm{p} K_b+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]} \\ 5.8 & =\mathrm{p} K_b+\log \left(\frac{1}{01}\right) \\ \mathrm{p} K_b & =4.8 \end{aligned} $

$ \begin{aligned} &\text{The ion product of water}~(K_w)~\text{is the product of the acid dissociation constant}~(K_a)~\text{and the base dissociation constant}~(K_b): \\[4pt] &K_a \times K_b = K_w \\[4pt] &\text{So, you can solve for}~K_b~\text{using:} \\[4pt] &K_b = \frac{K_w}{K_a} \end{aligned} $

$ \begin{aligned} &\text{We know:} \\[4pt] &K_w = 1.0 \times 10^{-14}~\text{at}~25^\circ \mathrm{C} \\[4pt] &K_a = 1.8 \times 10^{-4} \\[8pt] &\text{Plug these values into the formula:} \\[8pt] &K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \\[8pt] &K_b = 5.55 \times 10^{-11} \end{aligned} $

$x$ be the initial concentration of acetic acid.

Let $\alpha$ be degree of ionisation $=0.04242$

Concentration of ionised acid $=\alpha x$

$ \begin{aligned} k_a & =\frac{\left[\mathrm{H}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}=\frac{\alpha x(\alpha x)}{x-\alpha x} \\ k_a & =\frac{\alpha^2 x}{1-\alpha} \\ \Rightarrow \quad x & =\frac{k_a(1-\alpha)}{\alpha^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Substituting the values

$ x=\frac{1.8 \times 10^{-5}(1-0.04242)}{(0.04242)^2} \simeq 0.01 $

(a) Percentage ionisation

$ =\frac{\left[\mathrm{H}^{+}\right]}{\text {Initial concentration of acid }} \times 100 \% $

$ \begin{aligned} 4.242 & =\frac{\left[\mathrm{H}^{+}\right]}{x} \times 100 \\ {\left[\mathrm{H}^{+}\right] } & =\frac{4.242 \times x}{100}=0.04242 x \end{aligned} $

$ \left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=0.04242 x $

                                                   [ $\because$ % ionisation is small]

$ \begin{aligned} {\left[\mathrm{CH}_3 \mathrm{COOH}\right] } & =x-0.04242 x \\ & =0.95758 x \end{aligned} $

For equilibrium reaction,

$ \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} $

$ \begin{aligned} K_a & =\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ 1.8 \times 10^{-5} & =\frac{0.04242 x \times 0.04242 x}{0.95758 \mathrm{x}} \\ x & =0.009579 \\ {\left[\mathrm{H}^{+}\right] } & =0.04242 \times 0.009579 \\ & =0.00040693 \\ \mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\ & =-\log [0.00040693]=3.37 \end{aligned} $

Dissociation equation of AgBr

$ \mathrm{AgBr}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Br}^{-}(a q) $

Let solubility of $\mathrm{AgBr}=s$

$ \begin{aligned} \mathrm{Ag}^{+} & =s(\text { concentration }), \mathrm{Br}^{-}=0.1+s \\ K_{\mathrm{sp}} & =\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] \\ 4 \times 10^{-13} & =s \times(0.1+s) \\ & s=4 \times 10^{-12} \mathrm{M} \end{aligned} $

To find the pH of the final solution, follow these steps:

Calculate the moles of HCl and NaOH before mixing:

HCl: Moles = $ 0.5 \text{ M} \times 0.1 \text{ L} = 0.05 \text{ mol} $

NaOH: Moles = $ 0.4 \text{ M} \times 0.1 \text{ L} = 0.04 \text{ mol} $

Determine the reaction outcome:

HCl and NaOH react in a 1:1 molar ratio as shown in the equation:

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Thus, 0.04 mol of HCl will completely react with 0.04 mol of NaOH. The remaining moles of HCl are:

$0.05 \text{ mol} - 0.04 \text{ mol} = 0.01 \text{ mol}$

Calculate the final volume after dilution:

By adding 800 mL of distilled water to 200 mL of the resultant solution, the total volume becomes:

$200 \text{ mL} + 800 \text{ mL} = 1000 \text{ mL} = 1 \text{ L}$

Determine the concentration of HCl after dilution:

$\text{Concentration of HCl} = \frac{0.01 \text{ mol}}{1 \text{ L}} = 0.01 \text{ M}$

Since HCl fully dissociates:

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

Therefore, the concentration of $\text{H}^+$ ions is also 0.01 M.

Calculate the pH of the solution:

$ \text{pH} = -\log [\text{H}^+] \\ \text{pH} = -\log [0.01] = 2.0 $

Thus, the pH of the final solution is 2.0.

To determine the concentrations of $\mathrm{H}_3 \mathrm{O}^{+}$, $A^{-}$, and $\mathrm{H} A$ at equilibrium, we analyze the dissociation of the weak acid $\mathrm{H} A$ in a 0.5 M solution where its degree of dissociation, $\alpha$, is 1%.

The dissociation reaction is as follows:

$ \mathrm{H} A + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+} + A^{-} $

At equilibrium:

The concentration of undissociated $\mathrm{H} A$ is $(1-\alpha)C$.

The concentrations of $\mathrm{H}_3 \mathrm{O}^{+}$ and $A^{-}$, which are produced in equal amounts due to the stoichiometry of the reaction, are both $\alpha C$.

Given that:

$C = 0.5 \, \mathrm{M}$

$\alpha = \frac{1}{100} = 0.01$

We calculate:

$ \left[\mathrm{H}_3 \mathrm{O}^{+}\right] = \left[A^{-}\right] = \alpha C = 0.01 \times 0.5 = 0.005 \, \mathrm{M} $

The concentration of $\mathrm{H} A$ at equilibrium is:

$ [\mathrm{H} A] = (1-\alpha)C = (1-0.01) \times 0.5 = 0.495 \, \mathrm{M} $

Thus, the concentrations at equilibrium are:

$\left[\mathrm{H}_3 \mathrm{O}^{+}\right] = 0.005 \, \mathrm{M}$

$\left[A^{-}\right] = 0.005 \, \mathrm{M}$

$[\mathrm{H} A] = 0.495 \, \mathrm{M}$

When 100 mL of 0.1 M HA (a weak acid) is mixed with 100 mL of 0.2 M NaA (its conjugate base), we can determine the pH of the solution using the Henderson-Hasselbalch equation. This equation is suitable for calculating the pH of a buffer solution and is given by:

$ \text{pH} = \text{p}K_a + \log \left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) $

Given:

The acid dissociation constant, $ K_a = 10^{-5} $

The concentration of acid HA is $ 0.1 \, \text{M} $

The concentration of salt NaA is $ 0.2 \, \text{M} $

First, calculate the $\text{p}K_a$:

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

Now, use the Henderson-Hasselbalch equation:

$ \text{pH} = 5 + \log\left(\frac{0.2}{0.1}\right) $

$ \text{pH} = 5 + \log(2) $

Given that $\log(2) = 0.3$:

$ \text{pH} = 5 + 0.3 = 5.3 $

Therefore, the pH of the resultant solution is 5.3.

Original solution,

$ \left[\mathrm{H}^{+}\right]=0.01 \mathrm{M}=10 \times 10^{-3} \mathrm{M} $

and $\quad \mathrm{pH}=2$

$\Rightarrow\left[\mathrm{H}^{+}\right]$and pH of the mixture as in,

$ \text { } \begin{aligned} (a)\,\,{\left[\mathrm{H}^{+}\right]=} & \frac{20 \times 0.01+20 \times 0.02}{40} \\ = & 15 \times 10^{-3} \mathrm{M}, \mathrm{pH}=1.82 \\ & {\left[\because[\mathrm{HCl}]=\left[\mathrm{H}^{+}\right]\right] } \end{aligned} $

$ \begin{aligned} \,\,(b)\,\,\, {\left[\mathrm{H}^{+}\right] } & =\frac{20 \times 0.01+20 \times 0.005}{40} \\ & =7.5 \times 10^{-3} \mathrm{M} ; \mathrm{pH}=2.12 \end{aligned} $

$ \begin{aligned}\,\,(c)\,\,\,\,\, {\left[\mathrm{H}^{+}\right] } & =\frac{20 \times 0.01+20 \times 0.01}{40} \\ & =10 \times 10^{-3} \mathrm{M} ; \mathrm{pH}=2 \end{aligned} $

$ \begin{aligned}\,(d) \,\,\,\,\,\,{\left[\mathrm{H}^{+}\right] } & =\frac{20 \times 0.01+40 \times 0.005}{60} \\ & =6.67 \times 10^{-3} ; \mathrm{pH}=2.17 \end{aligned} $

So, decrease in pH , i.e., $\mathrm{pH}<2$ takes place in option (a) only.

The solubility of barium phosphate, with molar mass $ M $, in water is $ x \, \text{g} $ per 100 mL at 298 K. To convert this to moles per liter, the solubility is expressed as $\frac{10 \times x}{M} \, \text{mol/L}$.

The solubility product ($ K_{\text{sp}} $) expression for $\text{Ba}_3(\text{PO}_4)_2$ is given by:

$ K_{\text{sp}} = [\text{Ba}^{2+}]^3[\text{PO}_4^{3-}]^2 $

Substituting the molar solubility into this expression, we have:

$ K_{\text{sp}} = \left(3 \times \frac{10 \times x}{M}\right)^3 \left(\frac{2 \times 10 \times x}{M}\right)^2 $

Simplifying this:

$ K_{\text{sp}} = 108 \left(\frac{10 \times x}{M}\right)^5 = 1.08 \times 10^7 \left(\frac{x}{M}\right)^5 $

From this, the values of $ a $ and $ b $ in the solubility product expression $ 1.08 \times \left(\frac{x}{M}\right)^a \times 10^b $ are 5 and 7, respectively.

$ \text { Initial concentration at equilibrium } $

$ K_{\text {sp }}=\left[A^{+}\right]\left[X^{-}\right]=10^{-10} $

$ \begin{gathered} 10^{-10}=S[S \times 0.1] \\ \text { As } \left.K_{s p} \lll 1 \text { then, } 10^{-10}=S \times 0\right] \\ S=10^{-9} \mathrm{M} \end{gathered} $

A basic buffer solution consists of a weak base and the salt of its conjugate strong acid.

In this specific reaction, we have:

100 mL of 0.1 M HCl and 200 mL of 0.1 M NH₄OH.

First, calculate the millimoles:

HCl: $ n = 100 \, \text{mL} \times 0.1 \, \text{M} = 10 \, \text{millimoles} $

NH₄OH: $ n = 200 \, \text{mL} \times 0.1 \, \text{M} = 20 \, \text{millimoles} $

The balanced chemical reaction is:

$ \mathrm{HCl} (10 \, \text{millimoles}) + \mathrm{NH_4OH} (20 \, \text{millimoles}) \rightarrow \mathrm{NH_4Cl} (10 \, \text{millimoles}) + \mathrm{H_2O} $

After the reaction, 10 millimoles of NH₄OH remain, along with 10 millimoles of NH₄Cl. As NH₄OH is a weak base and NH₄Cl is its conjugate acid's salt, this mixture forms a basic buffer solution. Therefore, the solution will effectively resist changes in pH upon the addition of small amounts of acids or bases. Thus, the formation of a basic buffer makes the scenario correct.

First step is to determine the number of moles of HCl and NaOH before mixing.

$ \begin{aligned} & \mathrm{HCl}=0.4 \mathrm{M} \times 0.1 \mathrm{~L}=0.04 \mathrm{~mol} \\ & \mathrm{NaOH}=0.5 \mathrm{M} \times 0.1 \mathrm{~L}=0.05 \mathrm{~mol} \end{aligned} $

Since, HCl and NaOH react in a $1: 1$ molar ratio.

$ \mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} $

0.04 moles of HCl will react with 0.04 moles of NaOH

$ \begin{aligned} \text { Remaining } \mathrm{NaOH} & =0.05-0.04 \\ & =0.01 \mathrm{~mol} \end{aligned} $

Final volume after adding 800 mL of distilled water

$ 200 \mathrm{~mL}+800 \mathrm{~mL}=1 \mathrm{~L} $

Concentration of NaOH after dilution

$ \begin{aligned} & =\frac{0.01}{1}=0.01 \mathrm{M} \\ \mathrm{NaOH} & \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \end{aligned} $

Concentration of $\mathrm{OH}^{-}$is equal to concentration of $\mathrm{NaOH}=0.01 \mathrm{M}$

$ \begin{aligned} & \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\ & \mathrm{pOH}=-\log [0.01] \\ & \mathrm{pOH}=2 ; \mathrm{pH}=14-2=12 \end{aligned} $

The conjugate base of chloric acid is formed when chloric acid donates a proton (H⁺).

$ \mathrm{HClO}_3 \rightleftharpoons \mathrm{H}^{+} + \mathrm{ClO}_3^{-} $

Thus, the conjugate base of chloric acid is $\mathrm{ClO}_3^{-}$.

We know that,

$ \begin{aligned} & \mathrm{Cd}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Cd}^{2+}+2 \mathrm{OH}^{-} \\ & \mathrm{KOH}=0.1 \mathrm{M}=\left[\mathrm{OH}^{-}\right] \end{aligned} $

Let solubility of $\mathrm{Cd}^{2+}$ ions be $S$.

$ \begin{aligned} & K_{\mathrm{sp}}=\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & 2.5 \times 10^{-14}=S \times[0.1]^2 \\ & 2.5 \times 10^{-12} \mathrm{M}=S \\ & \text { or } 25 \times 10^{-13} \mathrm{M}=S \end{aligned} $

On comparing the answer with $x \times 10^{-y}$.

We get $x=25$ and $y=13$

Buffer is a solution of weak acid and its conjugate base or weak base and its conjugate acid in which pH cannot be changed much after, adding a certain amount of acid and base.

Hence, $1: 1$ mole ratio of ( $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}$ )

is not a buffer solution, since both acid and base has same ratio.

(A) is false but (R) is true. $\mathrm{pK}_a$ of phenol is 10 and that of benzoic acid is 4.19 . This is because in the resonating structures of benzoate lon, there will be equivalent resonating structures, with negative sign at oxygen. However, in case of phenoxide ion, it has non-equivalent resonance structure as negative charge appears on less electronegative carbon which leads to less stability.

The solution prepared by mixing the given quantities of acetic acid and sodium acetate is an acidic buffer.

To find the pH of an acidic buffer, we use the Henderson-Hasselbalch equation:

$ \mathrm{pH} = \mathrm{p}K_a + \log \frac{[\text{Salt}]}{[\text{Acid}]} $

Given:

$\mathrm{p}K_a = 4.76$ (provided)

Concentration of the salt, $[\text{Salt}]$, is calculated as:

$ \frac{20 \, \text{mL} \times 0.5 \, \text{M}}{100 \, \text{mL}} = 0.1 \, \text{M} $

Concentration of the acid, $[\text{Acid}]$, is calculated as:

$ \frac{10 \, \text{mL} \times 1.0 \, \text{M}}{100 \, \text{mL}} = 0.1 \, \text{M} $

Substituting these values into the Henderson-Hasselbalch equation:

$ \mathrm{pH} = 4.76 + \log \frac{0.1}{0.1} = 4.76 + \log 1 = 4.76 $

Therefore, the pH of the solution is 4.76.

Ammonia is Lewis base because of presence of lone pair of electron on nitrogen. Lewis bases are the electron rich species which have lone pair of electron for donation to electron deficient species. $\ddot{\mathrm{N}} \mathrm{H}_3$ electron donor.

Diprotic acids are those acid which consist of two protons. Out of citric acid, chromic acid, oxalic acid, pyrosulphuric acid and sulphurous acid, only citric acid is not a diprotic acid.

Citric acid $-\mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_7$

Chromic acid $-\mathrm{H}_2 \mathrm{CrO}_4$

Oxalic acid $-\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$

Pyrosulphuric acid $-\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$

Sulphurous acid $-\mathrm{H}_2 \mathrm{SO}_3$

$ \begin{aligned} &\text { The correct match is }\\ &\text { A-(II), B-(I), C-(IV), D-(III) } \end{aligned} $

The value of $K_w$ increases with rise in temperature because dissociation of water is endothermic process. At $37^{\circ} \mathrm{C}$, $K_w$ is $10^{-14}$ and pH of water is 7 . So, at $80^{\circ} \mathrm{C}, K_w$ will be more than $10^{-14}$ and pH of water will be less than 7 .

Lime water is $\mathrm{Ca}(\mathrm{OH})_2$. Acidity of lime water is 2.

$\begin{aligned} & \text { So, Normality }=\text { Acidity } \times \text { Molarity } \\ & \Rightarrow \text { Molarity }=\frac{0.01}{2} \mathrm{M} \end{aligned}$

Concentration of $\left[\mathrm{OH}^{-}\right]$ ions, $\left[\mathrm{OH}^{-}\right]=2 \times \frac{0.01}{2}=0.01 \mathrm{M}$

Now, $\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]$

$\begin{aligned} \mathrm{pOH} & =-\log 10^{-2} \\ \Rightarrow \quad \mathrm{pOH} & =2 \log 10=2 \\ \mathrm{pH}+\mathrm{pOH} & =14 \\ \Rightarrow \quad \mathrm{pH} & =14-\mathrm{pOH}=14-2 \end{aligned}$

Thus, $\mathrm{pH}=12$

Given, concentration of acetic acid,

$\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.5 \mathrm{~N}$

Concentration of sodium acetate,

$\left[\mathrm{CH}_3 \mathrm{COONa}\right]=0.5 \mathrm{~N}$

$\mathrm{p} K_a$ of $\mathrm{CH}_3 \mathrm{COOH}=4.75$

Using Henderson-Hasselbalch equation,

$\begin{aligned} & \mathrm{pH}=\mathrm{pK}{ }_a+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & \Rightarrow \mathrm{pH}=4.75+\log \frac{0.5}{0.5} \\ \end{aligned}$

So, $\mathrm{pH}=4.75+\log 1=4.75+0 =4.75$

Conjugate acid is a chemical compound formed when an acid donates a proton [H$^+$] to a base.

• Concentration of acetic acid = 0.1 M

• Degree of dissociation (α) = 0.0132

We need to find the pH.

Step 1: Find the hydronium ion concentration

$ [\text{H}^+] = C \times \alpha = 0.1 \times 0.0132 = 0.00132\ \text{M} $

Step 2: Calculate pH

$ \text{pH} = -\log_{10} ( [\text{H}^+] ) = -\log_{10} (0.00132) $

$ \text{pH} = 2.88 $

✅ Final Answer:

Option D — 2.88

Given,

$\begin{aligned} {\left[K_{\mathrm{sp}}\right]_{\mathrm{AgBr}} } & =5 \times 10^{-10} \\ {[\mathrm{NaBr}] } & =0.2 \mathrm{M} \\ {\left[\mathrm{Na}^{+}\right] } & =\left[\mathrm{Br}^{-}\right]=0.2 \mathrm{M} \\ {\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] } & =\left[\mathrm{K}_{\mathrm{sp}}\right]_{\mathrm{AgBr}} \\ {\left[\mathrm{Ag}^{+}\right](0.2) } & =5 \times 10^{-10} \\ {\left[\mathrm{Ag}^{+}\right] } & =\frac{5 \times 10^{-10}}{0.2}=25 \times 10^{-10} \mathrm{M}\end{aligned}$

HCl is the strongest acid, it dissociate into ions,

$\begin{aligned} & \text {We know that } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\\\ & \mathrm{pH}=-\log [0.10] \Rightarrow \mathrm{pH}=1 \end{aligned}$

Using formula of $\mathrm{pH}+\mathrm{pOH}=14$ at $273 \mathrm{~K}$,

$\begin{aligned} 1+\mathrm{pOH} & =14 \\\\ \mathrm{pOH} & =14-1=13 \end{aligned}$

Acidic buffer $\Rightarrow$ weak acid + salt with strong base. An acidic buffer contains weak acid and its salt with a strong base.

$\mathrm{HClO}_4$ and $\mathrm{NaClO}_4$ contains strong acid $\mathrm{HClO}_4$ and its salt with strong base $\mathrm{NaOH} . \mathrm{HClO}_4$ and $\mathrm{NaClO}_4$ are not an acidic buffer because strong acid with its salt cannot form buffer solution.

So, $\mathrm{HClO}_4$ and $\mathrm{NaClO}_4$ cannot form acidic buffer.

∵ (i) Sodium acetate, i.e. $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ in water gives $\mathrm{CH}_3 \mathrm{COOH}$ and NaOH as follows :

$ \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} $

Hence, for (i) $\mathrm{pH}>7$.

[ ∵ solution is basic]

Also, (ii) when, ammonium chloride, i.e. $\left(\mathrm{NH}_4 \mathrm{Cl}\right)$ is added in water, it gives $\mathrm{NH}_4^{+}$(aqueous) and $\mathrm{Cl}^{-}$(aqueous) ions.

$\mathrm{NH}_4^{+}$(aqueous)

$ +\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_3+\mathrm{H}^{+}+\mathrm{OH}^{-} $

Hence, $\mathrm{pH}<7$

Hence (a) is the correct answer, i.e. for (i) $\mathrm{pH}>7$

and for (ii) $\mathrm{pH}<7$.

Given, Ionic product of $\mathrm{Ca}(\mathrm{OH})_2=5.5 \times 10^{-6}$

Let, solubility $=S$

$ \begin{gathered} \mathrm{Ca}^{2+}+2\left[\mathrm{OH}^{-}\right] \\ S \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2 S)^2 \end{gathered} $

Also, $K_{\mathrm{sp}}=4 S^3$

where, $K_{\mathrm{sp}}=$ solubility product ∴ Due to common ion effect, Molar solubility of

$ \mathrm{Ca}(\mathrm{OH})_2=\frac{5.5 \times 10^{-6}}{0.1 \times 0.1}=5.5 \times 10^{-4} $

Hence, option (c) is the correct answer.

∵ Various factors with which ionisation energy (IE) varies are :

(A) Atomic size $\propto \frac{1}{\text { IE }}$

(B) Screening effect $\propto \frac{1}{\text { IE }}$

(C) Nuclear charge $\propto \mathrm{IE}$

Thus, option (a) is the correct answer, i.e.