Ionic Equilibrium

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 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 $\mathrm{ClO}_3^{-}$.

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.