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} $

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.

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 .

∵ (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.