Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

Mixing a strong base with a weak acid in excess creates an acidic buffer solution, where the concentration ratio of the weak acid and its conjugate base dictates the acidity of the mixture. Tracking the millimoles of the reactants shows that 20 mL of 0.1 M NaOH reacts completely with the 50 mL of 0.1 M acetic acid. This reaction produces 2 mmol of the salt, sodium acetate, and consumes 2 mmol of the acetic acid, leaving 3 mmol of unreacted acid behind.

Since the total volume is consistent for both species in the mixture, the concentration ratio is equivalent to the ratio of their millimoles. Calculating the acidity using the Henderson-Hasselbalch equation and the provided constants yields:
$\text{pH} = 4.76 + \log_{10} \left( \frac{2}{3} \right)$
$\text{pH} = 4.76 + \log_{10} 2 - \log_{10} 3 = 4.76 + 0.30 - 0.48 = 4.58$
This result is $458 \times 10^{-2}$.

The solubility of a sparingly soluble salt like $\mathrm{BaSO}_4$ is reduced by the common ion effect, as the presence of sulfate ions from the strong electrolyte $\mathrm{K}_2\mathrm{SO}_4$ suppresses the dissolution of the barium salt. In a 0.1 M solution of $\mathrm{K}_2\mathrm{SO}_4$, the sulfate ion concentration is essentially 0.1 M. Using the solubility product expression K_{\mathrm{sp}} = \[\mathrm{Ba}^{2+}]$$[\mathrm{SO}_4^{2-}]$ = 1 \times 10^{-10}$, the concentration of barium ions is calculated as$[\mathrm{Ba}^{2+}] = (1 \times 10^{-10}) / 0.1 = 10^{-9}\ \text{M}$. To find the solubility in grams per liter, multiply this molar solubility by the molar mass of 233 g/mol. This calculation yields$10^{-9} \times 233 = 233 \times 10^{-9}\ \text{g\ L}^{-1}$, so the required integer value is 233.

Acid-base indicators function by changing color within a specific pH interval dictated by their dissociation constant, $K_{\text{In}}$. For an indicator, the effective color change range is empirically determined to span $pK_{\text{In}} \pm 1$. Using the provided $K_{\text{In}} = 4 \times 10^{-10}$, the calculation for $pK_{\text{In}}$ is $pK_{\text{In}} = -\log(4 \times 10^{-10}) = -(\log 4 - 10) = -(0.6 - 10) = 9.4$. With a $pK_{\text{In}}$ of 9.4, the operational transition range extends from 9.4 - 1 to 9.4 + 1, resulting in an interval of 8.4 to 10.4, which confirms that the visual transition initiates at pH 8.4.

The dilution process relies on the principle that the total amount of solute remains constant before and after adding solvent, which is described by the equation $M_1V_1 = M_2V_2$. Because the molarity of an acid corresponds to its hydronium ion concentration $[H^+] = 10^{-\mathrm{pH}}$, the initial solution with pH 1 has a concentration of 0.1 M, while the desired solution with pH 2 requires a concentration of 0.01 M. Substituting these values into the relationship $0.1 \times 1 = 0.01 \times V_2$ leads to a final volume of $V_2 = 10$ L. Subtracting the initial volume of 1 L from this total volume indicates that 9 L of water must be added, which is equal to 9000 mL.

The solubility product constant $K_{\text{sp}}$ represents the equilibrium between a solid salt and its dissolved ions in a saturated solution, where the ion concentration is determined by the molar solubility $S$. First, find the molar mass of silver chloride by adding the atomic masses of silver (107.9 g/mol) and chlorine (35.5 g/mol) to obtain 143.4 g/mol. Converting the provided mass solubility of 1.434 x 10^-3 g L^-1 to molar solubility requires dividing the mass value by the molar mass, which yields $S = 10^{-5}$ mol L^-1. The dissociation follows the equation $\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)$, meaning the solubility product constant is given by the expression $K_{\text{sp}} = S^2$. Substituting the calculated molar solubility results in $K_{\text{sp}} = (10^{-5})^2 = 10^{-10}$. Consequently, the value of $-\log K_{\text{sp}}$ is found by evaluating $-\log(10^{-10})$, which equals 10.

Mixing 25 mL of 1 M $\mathrm{AgNO}_3$ with 25 mL of 1.05 M $\mathrm{KI}$ initiates a precipitation reaction that forms solid $\mathrm{AgI}$. Because silver iodide is a sparingly soluble salt, the system quickly reaches an equilibrium defined by $\mathrm{AgI} (s) \rightleftharpoons \mathrm{Ag}^+ (aq) + \mathrm{I}^- (aq)$, where the concentrations of the dissociated ions are governed by an extremely small solubility product constant. Consequently, the equilibrium dictates that both $\mathrm{Ag}^+$ and $\mathrm{I}^-$ exist in the solution at very low concentrations.

The pH scale is defined by the relationship $\mathrm{pH} = -\log [\mathrm{H}^+]$, where the pH value acts as a logarithmic representation of hydrogen ion concentration. Because logarithms translate multiplication into addition or subtraction, increasing the concentration by a factor of 1000 results in a change equal to $-\log(1000)$, which calculates to $-3$. This confirms that a thousandfold increase in hydrogen ion concentration causes the pH of the solution to decrease by 3 units.

When considering the salt of a weak acid and a strong base, such as calcium lactate, the anionic part undergoes hydrolysis to produce a basic solution. Because each unit of calcium lactate dissociates into one calcium ion and two lactate ions, the concentration of the hydrolyzing lactate anions is $2 \times 0.005$ M, or 0.01 M. The $pH$ of this salt solution is determined by the formula $pH = 7 + \frac{1}{2}(pKa + \log C)$, where $C$ represents the concentration of the anion. Substituting the given values of $pKa = 5$ and $C = 0.01$ M leads to $pH = 7 + \frac{1}{2}(5 + \log(0.01))$. Given that $\log(0.01) = -2$, the expression simplifies to $pH = 7 + \frac{1}{2}(3)$, resulting in a value of 8.5, which is $85 \times 10^{-1}$.

The saturation state of a salt in solution is determined by the ion product relative to the solubility product constant, $K_{\text{sp}}$. When mixing two solutions, the final concentration of each ion is calculated by dividing the total amount of substance by the total volume of the mixture. In this scenario, mixing 25.0 mL of 0.050 M barium nitrate with 25.0 mL of 0.020 M sodium fluoride results in a total solution volume of 50.0 mL.

The concentration $[\text{Ba}^{2+}]$ becomes 0.025 M, and the concentration $[\text{F}^-]$ becomes 0.01 M. The ion product is determined by the expression [\text{Ba}^{2+}]\[\text{F}^-]^2$, which equals 0.0000025. Expressing this as$25 × 10^{-7}$allows for a straightforward comparison with the given$K_{\text{sp}}$of$5 × 10^{-7}$. The ratio is calculated as:

$\frac{[\text{Ba}^{2+}]\$\$[\text{F}^-]^2}{K_{\text{sp}}} = \frac{25 \times 10^{-7}}{5 \times 10^{-7}} = 5$

The acidity of a solution is determined by the total hydrogen ion concentration, so for a $10^{-8} \, \text{M} \, \mathrm{HCl}$ solution, the contribution of hydronium ions from the auto-ionization of water cannot be neglected, resulting in a $\mathrm{pH}$ of approximately $6.98$ rather than $8$. The conjugate base of the dihydrogen phosphate ion, $\mathrm{H}_2 \mathrm{PO}_4^{-}$, is formed by removing a single proton to yield $\mathrm{HPO}_4^{2-}$. Because the dissociation of water is an endothermic process, the ionic product of water, $K_w$, increases as the temperature rises due to the shift in equilibrium. Finally, at the half-neutralization point of a weak monoprotic acid, the concentrations of the acid and its conjugate base are identical, which simplifies the Henderson-Hasselbalch equation such that the $\mathrm{pH}$ is equal to the $pK_a$ of the acid.

A buffer solution maintains a relatively constant pH upon the addition of small amounts of strong acid or base by utilizing an equilibrium between a weak base and its conjugate acid. Adding hydrochloric acid triggers a neutralization reaction where the added $\mathrm{H}^+$ ions react with the weak base $\mathrm{NH}_3$ to produce $\mathrm{NH}_4^+$ ions. With one litre of solution, the stoichiometric changes result in 0.08 mole of $\mathrm{NH}_3$ and 0.12 mole of $\mathrm{NH}_4^+$. The Henderson-Hasselbalch equation defines the relationship for basic buffers as:

$\mathrm{pOH} = \mathrm{pK}_{\mathrm{b}} + \log \frac{[\mathrm{NH}_4^+]\$}{[\mathrm{NH}_3]}$

Substituting the available data gives:

$\mathrm{pOH} = 4.745 + \log \left( \frac{0.12}{0.08} \right)$

$\mathrm{pOH} = 4.745 + \log \left( \frac{3}{2} \right)$

Using the provided values for $\log 3$ and $\log 2$, the pOH becomes 4.745 + 0.477 - 0.301, resulting in 4.921. Calculating the pH at 298 K involves subtracting the pOH from 14, which yields 9.079. This value is equivalent to $9079 \times 10^{-3}$.

The relationship between $\text{pH}$ and $\text{pOH}$ establishes that a solution with a $\text{pH}$ of 12 contains a hydroxide ion concentration of $[\text{OH}^-] = 10^{-2}$ M. Since each formula unit of $\text{Ca}(\text{OH})_2$ releases two hydroxide ions upon complete dissociation, the molarity of the solute is calculated as $[\text{Ca}(\text{OH})_2] = 0.5 \times 10^{-2}$ M, which is $5 \times 10^{-3}$ M. Multiplying this concentration by the total volume of 100 mL provides the quantity in millimoles as $5 \times 10^{-3} \times 100 = 5 \times 10^{-1}$ mmol. Comparing this result to the required form $x \times 10^{-1}$ confirms that $x = 5$.

The acidity of the mixture is determined by the total quantity of hydrogen ions present within the final combined volume. Because hydrochloric acid is monoprotic, it contributes one proton per molecule, while sulfuric acid is diprotic and provides two protons per molecule. The total millimoles of hydrogen ions are calculated by summing $(600 \times 0.01)$ from the hydrochloric acid and $(400 \times 0.01 \times 2)$ from the sulfuric acid, which equals 14 millimoles.

With a total solution volume of 1000 mL, the hydrogen ion concentration is determined to be $14 \times 10^{-3} \text{ M}$. Applying the definition $\text{pH} = -\log[\text{H}^+]$, we arrive at $\text{pH} = -\log(14 \times 10^{-3})$, which simplifies to $\text{pH} = 3 - \log 14$. Using the provided values $\log 2 = 0.30$ and $\log 7 = 0.84$, the calculation becomes $3 - (\log 2 + \log 7) = 3 - 1.14 = 1.86$. This value is equivalent to $186 \times 10^{-2}$.

Acidic buffers maintain a constant pH through the Henderson-Hasselbalch equation, which relates the pH of the mixture to the pKa of the weak acid and the ratio of the conjugate base concentration to the acid concentration. When equal volumes of the two solutions are mixed, the molar ratio of the salt to the acid simplifies to $0.1 / 0.01 = 10$. Substituting the values into the equation $pH = pKa + \log\left(\frac{[Salt]}{[Acid]}\right)$ yields the relationship $5 = pKa + \log(10)$, which simplifies to $pKa = 4$. Converting this pKa value back to the dissociation constant gives $K_a = 10^{-4}$, which is equivalent to $10 × 10^{-5}$, resulting \in $x = 10$.

When a weak acid is partially neutralized by a strong base, the resulting mixture functions as a buffer solution containing both the weak acid and its conjugate base. The Henderson-Hasselbalch equation allows for the determination of the pH based on the acid dissociation constant and the ratio of the concentrations of the salt and the acid.

The initial amount of acetic acid is $50 \text{ mL} \times 0.1 \text{ M} = 5 \text{ mmol}$. Upon the addition of $25 \text{ mL} \times 0.1 \text{ M} = 2.5 \text{ mmol}$ of sodium hydroxide, the base reacts quantitatively with the acid. This consumption leaves $2.5 \text{ mmol}$ of unreacted acetic acid and produces $2.5 \text{ mmol}$ of sodium acetate.

Because the volume of the mixture is identical for both species, the ratio of their concentrations is equal to the ratio of their millimoles. In this specific scenario, the quantities of acid and salt are equivalent, causing the logarithmic term in the equation $pH = pK_a + \log \frac{[\mathrm{salt}]}{[\mathrm{acid}]}$ to become $\log(1)$, which is zero. Consequently, the pH is equal to the $pK_a$ value of $4.76$, which expressed as the required form is $476 \times 10^{-2}$.

The solubility product constant of a sparingly soluble salt relates the equilibrium concentrations of its constituent ions in a saturated solution. For the salt $A_2 X_3$, the dissociation follows the reaction $A_2 X_3 \rightleftharpoons 2A^{n+} + 3X^{m-}$, which leads to the solubility product expression $K_{sp} = (2s)^2(3s)^3 = 108s^5$. Given the value $K_{sp} = 1.1 \times 10^{-23}$, the molar solubility $s$ is calculated to be $10^{-5}$ mol/L.

For a very dilute solution, the molar conductivity is effectively equal to the limiting molar conductivity, determined by the ratio of specific conductance $\kappa$ to molar concentration $c$. Converting the solubility to SI units results in a concentration of $c = 10^{-2}$ mol/m³. Using the given specific conductance $\kappa = 3 \times 10^{-5}$ S m⁻¹, the limiting molar conductivity $\Lambda_m^0$ is found via the equation $\Lambda_m^0 = \frac{3 \times 10^{-5}}{10^{-2}} = 3 \times 10^{-3}$ S m² mol⁻¹. Comparing this result to the form $x \times 10^{-3}$ S m² mol⁻¹, we find $x = 3$.

The ability of water to exhibit amphoteric characteristics, acting as either a proton donor or an acceptor, is defined within the framework of the Bronsted-Lowry acid-base theory instead of the Lewis concept. As a result, the assertion attributing this behavior to the Lewis theory is incorrect. Conversely, the reason is accurate because water acts as an acid when donating a proton to $\mathrm{NH}_3$ and as a base when accepting a proton from $\mathrm{H}_2\mathrm{S}$, as seen in the following reactions:

$\mathrm{H}_2\mathrm{O} + \mathrm{NH}_3 \rightleftharpoons \mathrm{OH}^- + \mathrm{NH}_4^+$
$\mathrm{H}_2\mathrm{O} + \mathrm{H}_2\mathrm{S} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{HS}^-$

In this redox titration, potassium permanganate acts as the oxidizing agent while the oxalate ion functions as the reducing agent, ultimately converting to carbon dioxide. To determine the change, calculate the oxidation state of carbon in the oxalate ion, $\mathrm{C}_2\mathrm{O}_4^{2-}$, by assigning oxygen a value of -2, which yields the equation $2x + 4(-2) = -2$ and places carbon at an oxidation state of 3. Upon forming $\mathrm{CO}_2$, the equation $x + 2(-2) = 0$ confirms that carbon reaches an oxidation state of 4. The difference between these states demonstrates that the oxidation number of carbon increases by 1.

The dissociation of the strong base $\mathrm{NaOH}$ establishes the hydroxide ion concentration $[\mathrm{OH}^-]$ at 0.001 M. Using the ionic product of water, defined as [\mathrm{H}^+]\[\mathrm{OH}^-]$ = 10^{-14}$, the hydrogen ion concentration is calculated as$[\mathrm{H}^+] = 10^{-11}$M. The final$\mathrm{pH}$is then determined by the formula$\mathrm{pH} = -\log [\mathrm{H}^+]$, which gives$-\log(10^{-11}) = 11$.

The equilibrium constant for a net chemical reaction is obtained by multiplying the equilibrium constants of the elementary steps that compose it. This principle applies to polyprotic acids where the overall dissociation process is the sum of successive ionization stages. For oxalic acid, the first dissociation step represents the loss of one proton, while the second step represents the loss of a second proton from the hydrogen oxalate ion. When these two individual chemical equations are added together, the resulting net reaction represents the total dissociation of the acid into two hydrogen ions and the oxalate ion.

Mathematically, the ionization constant for the first step is given by K_{a_{1}} = \frac{[\mathrm{H}^{+}]\$[\mathrm{HC}_2\mathrm{O}_4^{-}]$}{[\mathrm{H}_2\mathrm{C}2\mathrm{O}4]}$, and for the second step, it is$K{a{2}} = \frac{[\mathrm{H}^{+}]$$[\mathrm{C}_2\mathrm{O}_4^{2-}]$}{[\mathrm{HC}_2\mathrm{O}_4^{-}]$}$. Multiplying these two expressions leads to the cancellation of the$[\mathrm{HC}_2\mathrm{O}_4^{-}]$term, simplifying the product to$\frac{[\mathrm{H}^{+}]^{2}$[\mathrm{C}_2\mathrm{O}_4^{2-}]$}{[\mathrm{H}_2\mathrm{C}2\mathrm{O}4]}$, which is the expression for the overall ionization constant$K{a{3}}$. Consequently, the relationship between these constants is defined as $K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}$