Step 1: Calculating the Molarity of the KOH Solution

To determine the concentration of the hydroxide ions, we first calculate the molarity of the $KOH$ solution. The molar mass of $KOH$ is calculated as: $39.10 \, g/mol (K) + 16.00 \, g/mol (O) + 1.008 \, g/mol (H) = 56.108 \, g/mol$.

Number of moles of $KOH$ ($n$):

Given the volume ($V$) = $200 \, mL = 0.200 \, L$, the molarity ($M$) is:

[Per the definition of Molarity, $M = \frac{n}{V(L)}$]

Step 2: Determining Ion Concentrations

Since $KOH$ is a strong electrolyte, it dissociates completely in water: $KOH(aq) \rightarrow K^+(aq) + OH^-(aq)$.

• $[K^+] = 0.05 \, M$
• $[OH^-] = 0.05 \, M$

To find $[H^+]$, we utilize the ionic product of water ($K_w$) at $298 \, K$:

[By the Auto-protolysis Constant of Water]

Step 3: Calculating pH

The pH of the solution is calculated using the negative logarithm of the hydrogen ion concentration:

[By the Sorensen pH scale definition]

Final Solution: The concentrations are $[K^+] = 0.05 \, M$, $[OH^-] = 0.05 \, M$, $[H^+] = 2.0 \times 10^{-13} \, M$. The pH of the solution is 12.70.

Step 1: Theoretical Foundation & Precipitation Condition

Precipitation of a sparingly soluble salt occurs when the Ionic Product ($Q_{sp}$) exceeds the Solubility Product constant ($K_{sp}$). For iron(II) sulphide ($FeS$), the dissolution equilibrium is:

$FeS_{(s)} \rightleftharpoons Fe^{2+}_{(aq)} + S^{2-}_{(aq)}$

The condition to prevent precipitation is defined as [Per the Solubility Product Principle]:

$Q_{sp} \leq K_{sp}$

$[Fe^{2+}][S^{2-}] \leq 6.3 \times 10^{-18}$

Step 2: Analysis of Mixing and Dilution

Let the initial concentration of both the $FeSO_4$ solution and the $Na_2S$ solution be $C$ $mol \cdot L^{-1}$.

When equal volumes ($V$) of these two solutions are mixed, the total volume becomes $V + V = 2V$. [Per the Dilution Law: $M_1V_1 = M_2V_2$].

The new concentration of each species in the mixture will be halved:

$[Fe^{2+}] = \frac{C \cdot V}{2V} = \frac{C}{2}$

$[S^{2-}] = \frac{C \cdot V}{2V} = \frac{C}{2}$

Step 3: Calculating Maximum Initial Concentration

To find the threshold concentration where no precipitation occurs, set $Q_{sp} = K_{sp}$:

$(\frac{C}{2})(\frac{C}{2}) = 6.3 \times 10^{-18}$

$\frac{C^2}{4} = 6.3 \times 10^{-18}$

$C^2 = 4 \times 6.3 \times 10^{-18}$

$C^2 = 25.2 \times 10^{-18}$

$C = \sqrt{25.2 \times 10^{-18}}$

$C = 5.02 \times 10^{-9} \, M$

Step 4: Graphical Representation of Equilibrium

The intersection point represents the saturation limit. At any value of $C$ such that $[Fe^{2+}][S^{2-}] < 6.3 \times 10^{-18}$, the solution remains unsaturated and no $FeS$ solid forms.

Final Solution: The maximum initial concentration of the equimolar solutions is $5.02 \times 10^{-9} \, M$.

Step 1: Analyzing the Initial Equilibrium State

At a constant temperature $T$, a liquid and its vapor exist in dynamic equilibrium within a sealed container. The equilibrium is defined by the equality of the rate of evaporation ($R_{evap}$) and the rate of condensation ($R_{cond}$):

$Liquid \rightleftharpoons Vapour$

The vapor pressure ($P_{vap}$) is a function solely of temperature for a pure substance in the presence of its liquid phase [Per the Clausius-Clapeyron Relation: $\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$]. Since $T$ is fixed, the equilibrium vapor pressure is constant regardless of the volume of the container, provided liquid remains present.

Step 2: Effect of Sudden Volume Expansion

When the volume of the container is suddenly increased ($V_{final} > V_{initial}$), the following sequence occurs:

• The number of moles of vapor ($n_v$) initially remains constant.
• According to the Ideal Gas Law ($P = \frac{n_vRT}{V}$), increasing $V$ at constant $n_v$ and $T$ results in an instantaneous drop in the pressure of the vapor.
• The system is now in a non-equilibrium state where $P_{current} < P_{equilibrium}$.

Step 3: Restoration of Equilibrium

To restore the equilibrium pressure, the system undergoes a process of net evaporation:

1. The rate of condensation ($R_{cond}$), which is proportional to the concentration of vapor molecules near the surface, decreases due to the lowered vapor density.
2. The rate of evaporation ($R_{evap}$) remains constant as it depends only on the temperature and surface area of the liquid.
3. Consequently, $R_{evap} > R_{cond}$.
4. More liquid molecules transition into the vapor phase until the concentration of molecules in the gas phase reaches the density required to exert the original equilibrium vapor pressure at temperature $T$.

Visual Representation of Vapor Pressure vs. Volume

Step 4: Final State Analysis

Once equilibrium is restored:

Final Solution: Upon the restoration of equilibrium, the system returns to its original equilibrium vapor pressure because the vapor pressure of a liquid is an intrinsic property dependent only on temperature, not on the container volume. More liquid will have evaporated to compensate for the increased volume, resulting in a decrease in the mass of the liquid phase.

Step 1: Establishing Given Parameters and Initial Concentrations

The reaction provided is the Haber process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.

To determine the state of the reaction, we must calculate the Reaction Quotient ($Q_c$) using initial molar amounts and the volume of the vessel ($V = 20\text{ L}$), then compare it to the given equilibrium constant ($K_c = 1.7 \times 10^2$).

Initial Molar Concentrations ($[X] = \frac{n}{V}$):

• $[N_2]_0 = \frac{1.57\text{ mol}}{20\text{ L}} = 0.0785\text{ M}$
• $[H_2]_0 = \frac{1.92\text{ mol}}{20\text{ L}} = 0.096\text{ M}$
• $[NH_3]_0 = \frac{8.13\text{ mol}}{20\text{ L}} = 0.4065\text{ M}$

Step 2: Calculation of the Reaction Quotient ($Q_c$)

The expression for the reaction quotient $Q_c$ is derived from the law of mass action applied at non-equilibrium states:

$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Substituting the calculated concentrations:

$Q_c = \frac{(0.4065)^2}{(0.0785)(0.096)^3}$

Performing the arithmetic:

$Q_c = \frac{0.16524225}{(0.0785)(0.000884736)}$

$Q_c = \frac{0.16524225}{0.000069451776} \approx 2379.24$

Step 3: Comparison of $Q_c$ and $K_c$

Comparing the values:

[Per the thermodynamic criterion for chemical equilibrium: If $Q_c > K_c$, the system is not at equilibrium and will proceed in the reverse direction to reach equilibrium].

Visual Representation of Reaction Directionality

Conclusion on Reaction State

Since $Q_c > K_c$, the numerator (product concentration) is too high relative to the denominator (reactant concentrations) for the system to be at equilibrium. Consequently, the system will shift toward the reactants to decrease the value of $Q_c$ until it equals $K_c$.

Final Solution: The reaction mixture is NOT at equilibrium. Because $Q_c (2379.24) > K_c (170)$, the net reaction will proceed in the reverse direction (to the left).

Step 1: Theoretical Foundation of Catalytic Action

A catalyst is a chemical substance that provides an alternative reaction pathway with a lower activation energy ($E_a$) for both the forward and backward reactions. [Per the Transition State Theory]. By lowering the energy barrier, the catalyst increases the rate at which equilibrium is attained, but it does not alter the thermodynamic stability of the reactants or products.

Step 2: Impact on Equilibrium Constant ($K_p$)

The equilibrium constant $K_p$ for a reversible reaction is defined as a function of the change in standard Gibbs free energy ($\Delta G^\circ$) and temperature ($T$), expressed by the equation:
$\Delta G^\circ = -RT \ln K_p$
Since a catalyst does not change the internal energy or entropy of the chemical species involved, $\Delta G^\circ$ remains constant. Furthermore, $K_p$ is strictly dependent on temperature. Because the catalyst does not affect the temperature, the value of $K_p$ remains unchanged.

Step 3: Impact on Equilibrium Composition

The equilibrium composition refers to the partial pressures or concentrations of reactants ($CH_4, H_2O$) and products ($CO, H_2$) once the forward rate ($r_f$) equals the backward rate ($r_b$).

• The catalyst increases the rate constant of the forward reaction ($k_f$) and the reverse reaction ($k_b$) by the same factor.
• Mathematically, $K_p = \frac{k_f}{k_b}$. If both $k_f$ and $k_b$ are scaled by the same factor due to the lowered activation energy, the ratio remains constant.
• Consequently, the position of the equilibrium (the relative concentration of species) does not shift. The system reaches the state of minimum Gibbs free energy more rapidly, but the final state itself is identical to the uncatalyzed system.

Step 4: Graphical Representation of Catalytic Effect

Final Solution: The use of a catalyst has no effect on the value of $K_p$ or the composition of the equilibrium mixture. The catalyst only serves to reduce the time required to reach the equilibrium state by lowering the activation energy of the forward and backward reactions equally.

Step 1: Chemical Equation and Phase Analysis

The given equilibrium reaction is:

$CaO(s) + CO_2(g) \rightleftharpoons CaCO_3(s)$

To determine the effect of pressure change on the equilibrium, we must identify the physical states of all reactants and products. [Per the law of mass action and Le Chatelier's Principle, only gaseous species contribute to the total pressure of the system].

• $CaO(s)$: Solid phase (does not contribute to partial pressure).
• $CO_2(g)$: Gaseous phase (contributes to total pressure).
• $CaCO_3(s)$: Solid phase (does not contribute to partial pressure).

Step 2: Quantitative Analysis of Gaseous Moles

We analyze the stoichiometric coefficients of the gaseous species on both sides of the reaction equation:

The change in the number of gaseous moles for the forward reaction is $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 0 - 1 = -1$.

Step 3: Application of Le Chatelier’s Principle

The problem states that the volume of the container is increased, which results in a decrease in the total pressure of the system [Per Boyle's Law: $P \propto 1/V$].

According to Le Chatelier’s Principle, if the pressure of a system at equilibrium is decreased, the system will shift in the direction that increases the number of gaseous moles to counteract the pressure drop.

• Forward direction: $\Delta n_g = -1$ (Pressure decreases).
• Reverse direction: $\Delta n_g = +1$ (Pressure increases).

Since the system undergoes a decrease in pressure, the equilibrium position shifts toward the side with the higher number of moles of gas, which is the reactant side ($CO_2$).

Step 4: SVG Visualization of the Shift

Conclusion on Product Moles

As the equilibrium shifts to the left (toward the reactants), the consumption of the product ($CaCO_3(s)$) increases in the reverse reaction to restore the pressure of $CO_2(g)$. Consequently, the quantity of the reaction product decreases.

Final Solution: The number of moles of reaction product decreases because the equilibrium shifts to the left to increase the partial pressure of $CO_2$.

1. Theoretical Foundation: Lewis Acid-Base Theory

According to the Lewis theory of acids and bases, a Lewis acid is defined as any species (molecule or ion) that is an electron-pair acceptor. Conversely, a Lewis base is defined as any species that is an electron-pair donor. [Per G.N. Lewis's Electron Theory of Valency].

2. Analysis of the Species: $H^{+}$

The hydrogen ion ($H^{+}$) is the nucleus of a hydrogen atom (a lone proton) stripped of its single electron. Its electronic configuration is represented as follows:

• Atomic number of Hydrogen ($Z$) = $1$
• Ground state configuration of $H$: $1s^{1}$
• Configuration of $H^{+}$: $1s^{0}$ (Empty orbital)

3. Mechanism of Action

Because the $1s$ orbital in $H^{+}$ is completely empty, the ion possesses a high affinity for electrons to complete its valence shell (achieving a duplet configuration). Therefore, it acts as an electron-pair acceptor. Any species that functions as an electron-pair acceptor is categorized as a Lewis Acid.

4. Graphical Representation of Electron Donation

5. Chemical Equation Example

The acidic nature of $H^{+}$ can be demonstrated through its reaction with ammonia ($NH_3$), where ammonia acts as the donor and $H^{+}$ acts as the acceptor to form the ammonium ion:

$:NH_{3} + H^{+} \rightarrow [NH_{3} \rightarrow H]^{+}$

[By the Principle of Coordinate Covalent Bond Formation]

Final Solution: The species $H^{+}$ is classified as a Lewis acid because it possesses an empty $1s$ orbital and acts as an electron-pair acceptor.

Step 1: Theoretical Foundation of Equilibrium Constant

The equilibrium constant, $K_c$, for a reversible chemical reaction is defined by the Law of Mass Action. It represents the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, where each concentration is raised to the power of its stoichiometric coefficient in the balanced chemical equation. [Per the Law of Chemical Equilibrium]

Step 2: Analysis of the Balanced Chemical Equation

The decomposition reaction is given as:

$PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g)$

In this balanced equation:

• Reactant: $PCl_5$ (Stoichiometric coefficient = 1)
• Products: $PCl_3$ (Stoichiometric coefficient = 1) and $Cl_2$ (Stoichiometric coefficient = 1)

Step 3: Derivation of the Expression

Applying the Law of Mass Action, we define the concentration of a species $X$ as $[X]$. The expression for $K_c$ is constructed as follows:

$K_c = \frac{[\text{Products}]^{\text{stoichiometric coefficients}}}{[\text{Reactants}]^{\text{stoichiometric coefficients}}}$

Substituting the specific chemical species from the reaction:

$K_c = \frac{[PCl_3]^1 [Cl_2]^1}{[PCl_5]^1}$

Step 4: Visualizing the Equilibrium State

Step 5: Final Formulation

The expression for the equilibrium constant $K_c$ for the decomposition of phosphorus pentachloride is strictly defined by the activities (approximated here by molar concentrations) of the gaseous species at temperature $T = 473$ K.

Final Solution: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$

Part 1: Calculation of Degree of Ionization ($\alpha$) in 0.02 M Solution

Dimethylamine is a weak base, denoted as $B$. The ionization equilibrium in water is represented as:

$B + H_2O \rightleftharpoons BH^+ + OH^-$

Step 1.1: Setting up the ICE Table

Step 1.2: Applying the Equilibrium Constant Expression

Given $K_b = 5.4 \times 10^{-4}$. [Per the Law of Mass Action]:

$K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{(0.02\alpha)(0.02\alpha)}{0.02(1-\alpha)} = \frac{0.02\alpha^2}{1-\alpha}$

Since $\alpha$ is typically small for weak bases ($1-\alpha \approx 1$), we simplify:

$5.4 \times 10^{-4} \approx 0.02\alpha^2$

$\alpha^2 = \frac{5.4 \times 10^{-4}}{0.02} = 270 \times 10^{-4} = 0.027$

$\alpha = \sqrt{0.027} \approx 0.1643$

Part 2: Calculation of Ionization in the Presence of 0.1 M NaOH

NaOH is a strong electrolyte and dissociates completely: $NaOH \rightarrow Na^+ + OH^-$. The concentration of $OH^-$ from NaOH is $0.1 \, M$. [Per the Common Ion Effect, the equilibrium shifts to the left].

Step 2.1: Defining the new equilibrium concentrations

• $[B] \approx 0.02 \, M$ (assuming $\alpha'$ is very small)
• $[BH^+] = 0.02\alpha' \, M$
• $[OH^-] = 0.1 + 0.02\alpha' \approx 0.1 \, M$

Step 2.2: Solving for $\alpha'$

$K_b = \frac{[BH^+][OH^-]}{[B]}$

$5.4 \times 10^{-4} = \frac{(0.02\alpha')(0.1)}{0.02}$

$5.4 \times 10^{-4} = \alpha'(0.1)$

$\alpha' = \frac{5.4 \times 10^{-4}}{0.1} = 5.4 \times 10^{-3}$

Visual Representation of Common Ion Effect

Step 2.3: Calculating Percentage

Percentage Ionization $= \alpha' \times 100 = 5.4 \times 10^{-3} \times 100 = 0.54\%$

Final Solution: The degree of ionization in 0.02 M solution is 0.1643; the percentage of dimethylamine ionized in the presence of 0.1 M NaOH is 0.54%.

Step 1: Identifying Variables and Initial Setup

The dissociation of acetic acid ($CH_3COOH$) in water is represented by the equilibrium equation:

$CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$

Given parameters:

• Dissociation constant ($K_a$) = $1.74 \times 10^{-5}$
• Initial concentration ($C$) = $0.05\ M$

Step 2: Calculating Degree of Dissociation ($\alpha$)

For a weak electrolyte where $\alpha \ll 1$, we apply the Ostwald's Dilution Law approximation [Per the law of chemical equilibrium]:

$K_a = C\alpha^2$ [Since $(1 - \alpha) \approx 1$]

Rearranging for $\alpha$:

$\alpha = \sqrt{\frac{K_a}{C}}$

$\alpha = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}}$

$\alpha = 0.01866$ (or $1.866\%$ dissociation)

Step 3: Calculating Concentration of Acetate Ions $[CH_3COO^-]$

From the stoichiometric equilibrium, $[CH_3COO^-] = C\alpha$:

$[CH_3COO^-] = 0.05 \times 0.01866$

$[CH_3COO^-] = 9.33 \times 10^{-4}\ M$

Step 4: Calculating pH of the Solution

The concentration of hydrogen ions $[H^+]$ is equal to the concentration of acetate ions in a simple weak acid dissociation:

$[H^+] = 9.33 \times 10^{-4}\ M$

The pH is defined by the negative logarithm of the hydronium ion concentration [By Sorensen's Definition]:

$pH = -\log[H^+]$

$pH = -\log(9.33 \times 10^{-4})$

$pH = -(\log 9.33 + \log 10^{-4})$

$pH = -(0.9699 - 4) = 3.0301$

Visual Representation of Equilibrium

Final Solution: The degree of dissociation is 0.01866, the concentration of acetate ion is $9.33 \times 10^{-4}\ M$, and the pH of the solution is 3.03.

Step 1: Analyzing the Balanced Chemical Equation

The given reaction is:

$COCl_2 (g) \rightleftharpoons CO (g) + Cl_2 (g)$

To determine the effect of pressure, we must first calculate the change in the number of moles of gaseous species ($\Delta n_g$) between the product side and the reactant side [Per Le Chatelier's Principle].

Step 2: Calculating Moles of Gaseous Species

The stoichiometric coefficients for the gaseous species are as follows:

• Total moles of gaseous products ($n_p$): $1 \text{ (from } CO) + 1 \text{ (from } Cl_2) = 2 \text{ moles}$
• Total moles of gaseous reactants ($n_r$): $1 \text{ (from } COCl_2) = 1 \text{ mole}$

The change in gaseous moles is calculated as:

$\Delta n_g = \sum n_p - \sum n_r = 2 - 1 = 1$

Since $\Delta n_g \neq 0$, the reaction is sensitive to pressure changes.

Step 3: Applying Le Chatelier's Principle

According to Le Chatelier’s Principle, if the pressure of a system at equilibrium is increased, the equilibrium will shift in the direction that reduces the total number of moles of gas to minimize the effect of the applied stress.

Step 4: Predicting the Direction of Shift

In this reaction, the reactant side has $1$ mole and the product side has $2$ moles of gas.

• Increasing the pressure increases the concentration of gaseous particles per unit volume.
• The system will attempt to counteract this increase by favoring the side with the lower number of moles [Le Chatelier's Principle].
• Since $1 < 2$, the equilibrium will shift towards the reactant side.

Visual Representation of Pressure Effect

Final Solution: The reaction will be affected by increasing the pressure. Because the number of moles of gaseous products (2) is greater than the number of moles of gaseous reactants (1), an increase in pressure will cause the equilibrium to shift in the backward direction (towards the formation of $COCl_2$).

Step 1: Dissociation of Chloroacetic Acid ($ClCH_{2}COOH$)

Chloroacetic acid is a weak acid that dissociates in water according to the equilibrium:

$ClCH_{2}COOH \rightleftharpoons ClCH_{2}COO^{-} + H^{+}$

Given:

• Acid concentration ($C$) = $0.1\,M$
• Dissociation constant ($K_{a}$) = $1.35 \times 10^{-3}$

Step 2: Calculating Hydrogen Ion Concentration ($[H^{+}]$)

Since $K_{a} = 1.35 \times 10^{-3}$, the degree of dissociation $\alpha = \sqrt{\frac{K_{a}}{C}}$:

$\alpha = \sqrt{\frac{1.35 \times 10^{-3}}{0.1}} = \sqrt{1.35 \times 10^{-2}} \approx 0.1162$

The concentration of $H^{+}$ ions is calculated as $[H^{+}] = C \times \alpha$:

$[H^{+}] = 0.1 \times 0.1162 = 0.01162\,M$

Step 3: Calculating pH of the Acid Solution

The pH is defined as $-\log[H^{+}]$ [By Sörensen's pH definition]:

$pH = -\log(0.01162) \approx 1.935$

Step 4: Hydrolysis of Sodium Chloroacetate ($ClCH_{2}COONa$)

Sodium chloroacetate is a salt of a weak acid and a strong base. It undergoes anionic hydrolysis:

$ClCH_{2}COO^{-} + H_{2}O \rightleftharpoons ClCH_{2}COOH + OH^{-}$

The hydrolysis constant ($K_{h}$) is given by $K_{h} = \frac{K_{w}}{K_{a}}$:

$K_{h} = \frac{10^{-14}}{1.35 \times 10^{-3}} = 7.407 \times 10^{-12}$

Step 5: Calculating pH of the Salt Solution

For the salt of a weak acid and strong base, the concentration of $[OH^{-}]$ is given by:

$[OH^{-}] = \sqrt{K_{h} \times C} = \sqrt{\frac{K_{w} \times C}{K_{a}}}$

$[OH^{-}] = \sqrt{\frac{10^{-14} \times 0.1}{1.35 \times 10^{-3}}} = \sqrt{7.407 \times 10^{-13}} \approx 8.606 \times 10^{-7}\,M$

Calculate $pOH = -\log(8.606 \times 10^{-7}) \approx 6.065$

Finally, $pH = 14 - pOH = 14 - 6.065 = 7.935$

Final Solution: The pH of the 0.1M chloroacetic acid is 1.94, and the pH of the 0.1M sodium chloroacetate salt solution is 7.94.

Step 1: Identifying the Given Equilibrium Parameters

The forward reaction is defined as the decomposition of phosphorus pentachloride:

$PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g)$

Given equilibrium constant for the forward reaction ($K_{c,f}$):

$K_{c,f} = 8.3 \times 10^{-3}$

Temperature ($T$) = $473\ K$

Enthalpy change ($\Delta_r H^{\circ}$) = $124.0\ kJ\ mol^{-1}$ [Note: This value indicates an endothermic process, per the positive sign].

Step 2: Defining the Reverse Reaction

The reverse reaction is the synthesis of phosphorus pentachloride from its products:

$PCl_3 (g) + Cl_2 (g) \rightleftharpoons PCl_5 (g)$

The equilibrium constant for a reverse reaction ($K_{c,r}$) is mathematically related to the equilibrium constant of the forward reaction ($K_{c,f}$) by the following thermodynamic identity [Per the Law of Mass Action for reversible reactions]:

$K_{c,r} = \frac{1}{K_{c,f}}$

Step 3: Calculating the Value of $K_{c,r}$

Substituting the provided value of $K_{c,f}$ into the reciprocal equation:

$K_{c,r} = \frac{1}{8.3 \times 10^{-3}}$

Perform the arithmetic division:

$K_{c,r} = \frac{1}{0.0083}$

$K_{c,r} \approx 120.4819$

Step 4: Significant Figure Adjustment

Given that $K_{c,f} = 8.3 \times 10^{-3}$ possesses two significant figures, the result should ideally be expressed to a similar degree of precision [Per scientific convention for experimental data].

$K_{c,r} \approx 1.2 \times 10^{2}$

Visual Representation of Equilibrium Reciprocity

Final Solution: The value of $K_c$ for the reverse reaction is $120.48$ (or $1.2 \times 10^2$ to two significant figures).

Step 1: Theoretical Foundation

The hydrogen ion concentration $[H^+]$ is defined as the molar concentration of hydronium ions in an aqueous solution. The relationship between pH and the hydrogen ion concentration is governed by the logarithmic definition established by S.P.L. Sørensen:

$pH = -\log[H^+]$

[Per the definition of the pH scale for dilute aqueous solutions]

Step 2: Algebraic Rearrangement

To isolate $[H^+]$, we apply the properties of logarithms:

$-pH = \log[H^+]$

Applying the antilogarithm (base 10) to both sides:

$10^{-pH} = 10^{\log[H^+]}$

$[H^+] = 10^{-pH}$

Step 3: Substitution of Given Data

Given the pH of the human muscle fluid is $6.83$:

$[H^+] = 10^{-6.83}$

Step 4: Calculation of the Exponential Value

To compute $10^{-6.83}$, we express the exponent as a sum of an integer and a decimal fraction to facilitate manual calculation or precision checking:

$10^{-6.83} = 10^{-7 + 0.17}$

$[H^+] = 10^{-7} \times 10^{0.17}$

Utilizing the value $10^{0.17} \approx 1.479$ [By logarithmic identity $10^x = \text{antilog}(x)$]:

$[H^+] = 1.479 \times 10^{-7} \, M$

Step 5: Significant Figures Analysis

In logarithmic expressions, the number of digits after the decimal point in the pH value determines the number of significant figures in the concentration result. Since $6.83$ has two decimal places, the final result must be reported with two significant figures.

$[H^+] = 1.479 \times 10^{-7} \approx 1.5 \times 10^{-7} \, M$

Final Solution: The hydrogen ion concentration of the human muscle fluid is $1.5 \times 10^{-7} \, M$.

Given Reaction and Thermodynamic Principles

The chemical reaction provided is:

$Fe^{3+} (aq) + 3OH^{-} (aq) \rightleftharpoons Fe(OH)_3 (s)$

To determine the expression for the equilibrium constant ($K_c$), we apply the Law of Mass Action, which states that for a reversible reaction at equilibrium, the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients, is a constant at a given temperature [Per the Law of Chemical Equilibrium].

Step 1: Application of Heterogeneous Equilibrium Rules

In this system, we encounter a heterogeneous equilibrium involving aqueous species and a solid phase. According to the conventions of chemical thermodynamics:

• The concentration (or activity) of a pure solid or pure liquid is defined as unity ($1$) because its density and composition remain constant regardless of the amount present. [By Convention of Activity for Pure Condensed Phases]
• Only species in the aqueous ($aq$) or gaseous ($g$) state are included in the $K_c$ expression.

Step 2: Constructing the Expression

The general expression for $K_c$ is defined as:

$K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}$

Substituting the species from the reaction into the template:

$K_c = \frac{[Fe(OH)_3(s)]}{[Fe^{3+}(aq)]^1 [OH^-(aq)]^3}$

Step 3: Simplification based on Phase Conventions

Since $[Fe(OH)_3(s)] = 1$, the expression simplifies as follows:

$K_c = \frac{1}{[Fe^{3+}(aq)] [OH^-(aq)]^3}$

Visual Representation of the Equilibrium System

Final Solution: The equilibrium constant expression for the reaction is $K_c = \frac{1}{[Fe^{3+}(aq)] [OH^-(aq)]^3}$

Initial Chemical Equilibrium Setup

The given reversible reaction at equilibrium is represented as:

$2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g)$

The reaction quotient ($Q_c$) for this system is defined by the ratio of the molar concentrations of products to reactants, raised to the power of their respective stoichiometric coefficients [Per the Law of Mass Action]:

$Q_c = \frac{[CH_3OH]}{[H_2]^2 [CO]}$

At equilibrium, $Q_c = K_c$, where $K_c$ is the equilibrium constant at a constant temperature.

Step 1: Analyzing the Effect of Adding $H_2$

Upon the addition of $H_2(g)$ at constant temperature and volume, the concentration of the reactant $[H_2]$ increases instantaneously. This perturbation shifts the value of the reaction quotient $Q_c$ such that the denominator of the expression increases [Per Le Chatelier's Principle].

Mathematically, if $[H_2]_{new} > [H_2]_{initial}$, then:

$Q_c = \frac{[CH_3OH]}{([H_2] + \Delta[H_2])^2 [CO]} < K_c$

Step 2: Determining the Direction of Equilibrium Shift

Since the system is no longer at equilibrium ($Q_c < K_c$), the system must undergo a net reaction to restore the state of equilibrium [Per the Reaction Quotient Criterion]. To increase the value of $Q_c$ back to $K_c$, the numerator ($[CH_3OH]$) must increase, and the denominator ($[H_2]^2[CO]$) must decrease.

This is achieved by the consumption of reactants ($H_2$ and $CO$) to produce more product ($CH_3OH$). Consequently, the equilibrium position shifts toward the right (the forward direction).

Visual Representation of the Equilibrium Shift

Step 3: Justification based on Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as concentration, pressure, or temperature), the position of equilibrium moves to counteract the change. By adding $H_2$, the system acts to decrease the concentration of $H_2$ by converting it into $CH_3OH$ [Per Le Chatelier's Principle].

Final Solution: The addition of $H_2$ increases the concentration of reactants, causing the reaction quotient $Q_c$ to become less than the equilibrium constant $K_c$. To re-establish equilibrium, the system shifts to the right, favoring the forward reaction and resulting in an increased yield of $CH_3OH$.

Step 1: Analyzing the Balanced Chemical Equation

The given equilibrium reaction is:

$3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)$

To determine the effect of pressure change on the equilibrium, we must identify the state of matter for each participant. In gaseous equilibria, the shift is dictated by the total number of moles of gaseous species on each side of the equation [Per Le Chatelier's Principle].

Step 2: Counting Gaseous Moles ($n_g$)

We classify the species by their phase ($s = solid$, $g = gas$):

• Reactants: $3Fe(s)$ and $4H_2O(g)$. Total moles of gas ($n_{g, reactants}$) = 4.
• Products: $Fe_3O_4(s)$ and $4H_2(g)$. Total moles of gas ($n_{g, products}$) = 4.

Step 3: Calculating Change in Gaseous Moles ($\Delta n_g$)

The change in the number of gaseous moles is calculated as follows:

$\Delta n_g = \sum n_{g, products} - \sum n_{g, reactants}$

$\Delta n_g = 4 - 4 = 0$

Step 4: Evaluating the Effect of Pressure/Volume Change

Le Chatelier's Principle states that when pressure is decreased (by increasing volume) in a system at equilibrium, the system will shift to the side with a greater number of gaseous moles to counteract the pressure drop. Conversely, if pressure is increased, it shifts toward the side with fewer gaseous moles.

In this specific reaction, since $\Delta n_g = 0$:

• The number of moles of gaseous products does not depend on the total pressure of the system.
• A change in volume/pressure does not create a concentration gradient that would favor the forward or backward reaction [Per the Law of Mass Action].

Step 5: Conclusion

Because the number of moles of gaseous reactants is equal to the number of moles of gaseous products, the equilibrium position remains unaffected by changes in pressure or volume.

Final Solution: The number of moles of reaction products remains the same.

Step 1: Identifying the General Expression for Reaction Quotient ($Q_c$)

For a reversible chemical reaction represented by the general equation:
$aA + bB \rightleftharpoons cC + dD$
The reaction quotient ($Q_c$) is defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, with each concentration term raised to the power of its respective stoichiometric coefficient [Law of Mass Action].

The mathematical form is expressed as:
$Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

Step 2: Analyzing the Specific Chemical Equilibrium

The given reaction is the esterification of acetic acid with ethanol:

$CH_3COOH (l) + C_2H_5OH (l) \rightleftharpoons CH_3COOC_2H_5 (l) + H_2O (l)$

In this specific case, the stoichiometric coefficients are as follows:

• Coefficient of $CH_3COOH = 1$
• Coefficient of $C_2H_5OH = 1$
• Coefficient of $CH_3COOC_2H_5 = 1$
• Coefficient of $H_2O = 1$

Step 3: Determining the Inclusion of Species

Typically, in equilibrium expressions for dilute aqueous solutions, water ($H_2O$) is treated as a solvent, and its concentration is considered constant, thus omitted from the expression. However, the problem explicitly states that water is not in excess and is not a solvent in this reaction. Therefore, the concentration of water must be explicitly included in the reaction quotient [By the principle that all components in a homogeneous liquid-phase reaction where no single component acts as an invariant solvent must be represented].

Step 4: Formulating the $Q_c$ Expression

Substituting the specific chemical species into the general formula:

$Q_c = \frac{[CH_3COOC_2H_5] \times [H_2O]}{[CH_3COOH] \times [C_2H_5OH]}$

Final Solution: The concentration ratio (reaction quotient), $Q_c$, for the given equilibrium is: $Q_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$

Step 1: Calculating $[HS^-]$ in 0.1 M $H_2S$ Solution

The dissociation of $H_2S$ is given by: $H_2S(aq) \rightleftharpoons H^+(aq) + HS^-(aq)$.

Given: $C = 0.1\,M$, $K_{a1} = 9.1 \times 10^{-8}$.

Since $K_{a1}$ is very small, we use the approximation $\alpha = \sqrt{\frac{K_{a1}}{C}}$ [Per the Ostwald Dilution Law].

$\alpha = \sqrt{\frac{9.1 \times 10^{-8}}{0.1}} = \sqrt{9.1 \times 10^{-7}} = 9.54 \times 10^{-4}$.

The concentration $[HS^-] = C \times \alpha = 0.1 \times 9.54 \times 10^{-4} = 9.54 \times 10^{-5}\,M$.

Step 2: Calculating $[HS^-]$ in the presence of 0.1 M $HCl$

$HCl$ is a strong electrolyte and dissociates completely: $HCl \to H^+ + Cl^-$. Thus, $[H^+] = 0.1\,M$ from $HCl$.

Using the equilibrium expression: $K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}$.

Assuming $[H_2S] \approx 0.1\,M$ (initial concentration) and $[H^+] \approx 0.1\,M$ (common ion effect):

$9.1 \times 10^{-8} = \frac{(0.1)[HS^-]}{0.1} \implies [HS^-] = 9.1 \times 10^{-8}\,M$.

The concentration of $HS^-$ is significantly suppressed due to the Common Ion Effect.

Step 3: Calculating $[S^{2-}]$ in 0.1 M $H_2S$ Solution

The second dissociation is: $HS^- \rightleftharpoons H^+ + S^{2-}$, with $K_{a2} = 1.2 \times 10^{-13}$.

From Step 1, $[H^+] \approx [HS^-] = 9.54 \times 10^{-5}\,M$.

$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$.

Since $[H^+] = [HS^-]$, the expression simplifies to $[S^{2-}] = K_{a2} = 1.2 \times 10^{-13}\,M$.

Step 4: Calculating $[S^{2-}]$ in the presence of 0.1 M $HCl$

From Step 2, $[H^+] = 0.1\,M$ and $[HS^-] = 9.1 \times 10^{-8}\,M$.

$1.2 \times 10^{-13} = \frac{(0.1)[S^{2-}]}{9.1 \times 10^{-8}}$.

$[S^{2-}] = \frac{(1.2 \times 10^{-13})(9.1 \times 10^{-8})}{0.1} = 1.09 \times 10^{-19}\,M$.

Summary Table

Final Solution: The concentration of $[HS^-]$ is $9.54 \times 10^{-5}\,M$ and $[S^{2-}]$ is $1.2 \times 10^{-13}\,M$ in pure solution. In the presence of $0.1\,M$ $HCl$, $[HS^-] = 9.1 \times 10^{-8}\,M$ and $[S^{2-}] = 1.09 \times 10^{-19}\,M$.

Theoretical Foundation: The Brönsted-Lowry Theory

According to the Brönsted-Lowry concept, an acid is defined as a proton ($H^+$) donor, while a base is defined as a proton acceptor. A conjugate base is the species remaining after an acid has donated a proton. The relationship is governed by the following general equilibrium expression:

$Acid \rightleftharpoons Conjugate\ Base + H^+$

Step 1: Analyzing Hydrofluoric Acid (HF)

To determine the conjugate base of $HF$, we must remove one proton ($H^+$) from the molecule:

• Chemical Equation: $HF \rightarrow H^+ + F^-$
• Justification: By removing $H^+$, the neutral molecule $HF$ loses one unit of positive charge, resulting in the fluoride ion ($F^-$) [Per the Law of Conservation of Charge].
• Conjugate Base: $F^-$

Step 2: Analyzing Sulfuric Acid ($H_2SO_4$)

For a polyprotic acid like $H_2SO_4$, the conjugate base is formed by the loss of a single proton:

• Chemical Equation: $H_2SO_4 \rightarrow H^+ + HSO_4^-$
• Justification: The loss of one $H^+$ cation from the neutral sulfuric acid molecule leaves behind the hydrogen sulfate anion ($HSO_4^-$).
• Conjugate Base: $HSO_4^-$

Step 3: Analyzing Bicarbonate Ion ($HCO_3^-$)

The bicarbonate ion acts as a Brönsted acid by donating its remaining proton:

• Chemical Equation: $HCO_3^- \rightarrow H^+ + CO_3^{2-}$
• Justification: The removal of $H^+$ from an ion with a $-1$ charge results in the carbonate anion ($CO_3^{2-}$) with a $-2$ charge [By the Principle of Electroneutrality].
• Conjugate Base: $CO_3^{2-}$

Summary of Conjugate Base Pairs

Visual Representation of Proton Transfer

Final Solution: The conjugate bases are: (1) for $HF$ is $F^-$, (2) for $H_2SO_4$ is $HSO_4^-$, and (3) for $HCO_3^-$ is $CO_3^{2-}$.