Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The pH of a buffer solution involving a weak base is determined by the equation pH = 14 - pK_b + \log \left( \frac{[B]}{[BH^+]\} \right)$, where$[B]$is the concentration of the unreacted weak base and$[BH^+]$is the concentration of the salt formed. Since$0.02x$moles of acid react with$0.02x$moles of base to produce$0.02x$moles of salt, the number of moles of base remaining is$0.02y - 0.02x$ in the total volume of 100 mL.

Substituting the given values $pH = 9$ and $pK_b = 5.699$ yields $9 = 14 - 5.699 + \log \left( \frac{0.02y - 0.02x}{0.02x} \right)$, which simplifies to $0.699 = \log \left( \frac{y-x}{x} \right)$. Given that $\log 5 = 0.699$, the relation becomes $\frac{y-x}{x} = 5$, which leads to $y = 6x$. Combining this with the volume constraint $x + y = 100$ results in $7x = 100$, providing $x \approx 14.3$ mL and $y \approx 85.7$ mL.

For a weak diprotic acid $\mathrm{H}_2\mathrm{X}$, the concentration of the divalent anion $[X^{2-}]$ is approximately equal to the second ionization constant $K_{a2}$, provided that the first ionization constant $K_{a1}$ is significantly larger than $K_{a2}$ and the acid is weak. First, calculate the hydrogen ion concentration from the first ionization equilibrium: $[\mathrm{H}^+] = \sqrt{K_{a1} C} = \sqrt{2.5 \times 10^{-8} \times 0.1} = 5 \times 10^{-5} \mathrm{M}$. Using the equilibrium expression for the second ionization step, K_{a2} = \frac{[\mathrm{H}^+]\$[X^{2-}]$}{[\mathrm{HX}^-]$}$, and recognizing that$[\mathrm{H}^+] \approx [\mathrm{HX}^-]$because the contribution of the second dissociation to the total hydrogen ion concentration is negligible, the expression simplifies to$[X^{2-}] = K_{a2}$. Given$K_{a2} = 1.0 \times 10^{-13} \mathrm{M}$, converting this to the required format of$\times 10^{-15} \mathrm{M}$yields$1.0 \times 10^{-13} = 100 \times 10^{-15} \mathrm{M}$, resulting in an integer value of 100.

The Henderson-Hasselbalch equation describes the pH of a buffer solution composed of a weak acid and its conjugate base. Neutralization occurs when the milliequivalents of acid equal those of the base, so 20 mL of acetic acid requires 28.4 mL of 0.1 M NaOH for full reaction. Solving for the acid concentration gives $M = \frac{28.4 \times 0.1}{20}$, resulting in 0.142 M.

When mixing 20 mL of 0.142 M acetic acid with 14.2 mL of 0.1 M NaOH, the initial millimoles of acid are 2.84, while the base provides 1.42 millimoles. Because NaOH is a strong base, it reacts completely with an equivalent amount of acetic acid to form sodium acetate. This process consumes 1.42 millimoles of acetic acid, leaving 1.42 millimoles of unreacted acid and producing 1.42 millimoles of the conjugate base.

With equal quantities of the weak acid and its conjugate base, the concentration ratio in the Henderson-Hasselbalch equation is unity. Using the expression
$\text{pH} = \text{pK}_a + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right)$
the logarithmic term becomes zero, meaning the pH of the mixture is equal to the $\text{pK}_a$ value of 4.75.

The complete dissociation of the strong acid HCl produces a high concentration of $\mathrm{H}^{+}$ ions, which suppresses the ionization of $\mathrm{H}_2\mathrm{A}$ via the common ion effect. Consequently, the equilibrium concentration of $\mathrm{H}^{+}$ is effectively 0.1 M, and the concentration of the undissociated $\mathrm{H}_2\mathrm{A}$ remains approximately 0.1 M.

Applying the first dissociation equilibrium, we use the expression K_{a1} = \frac{[\mathrm{H}^{+}]\$[\mathrm{HA}^{-}]$}{[\mathrm{H}_2\mathrm{A}]}$. Inserting the given values into this equation yields$8.1 × 10^{-8} = \frac{0.1 × $[\mathrm{HA}^{-}]$}{0.1}$. Since the 0.1 M terms \in the numerator and denominator cancel out, the concentration of$\mathrm{HA}^{-}$is$8.1 × 10^{-8} , \mathrm{M}$.

Precipitation of the metal sulfide $YS$ occurs when the ionic product $[Y^{2+}][S^{2-}]$ reaches the solubility product constant, $K_{\text{sp}}(YS) = 4 \times 10^{-16}$. Given the metal ion concentration $[Y^{2+}] = 0.01\,\text{M}$, the required sulfide ion concentration to initiate precipitation is $[S^{2-}] = \frac{4 \times 10^{-16}}{0.01} = 4 \times 10^{-14}\,\text{M}$.

The sulfide ion concentration is controlled by the dissociation of saturated $\mathrm{H}_2\mathrm{S}$ in an acidic solution, where the equilibrium constant is K_a = \frac{[H^+]^2\[S^{2-}]$}{[H_2S]} = 10^{-21}$. Substituting$[H_2S] = 0.1,\text{M}$and$[S^{2-}] = 4 \times 10^{-14},\text{M}$into the expression yields$[H^+]^2 = \frac{10^{-21} \times 0.1}{4 \times 10^{-14}} = 2.5 \times 10^{-9} = 25 \times 10^{-10}$. Taking the square root gives$[H^+] = 5 \times 10^{-5},\text{M}$, leading to a$\text{pH} = -\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.7 = 4.3$. The nearest integer value is 4.

To determine the relative pH values, we must calculate the net concentration of $\mathrm{H}^+$ or $\mathrm{OH}^-$ ions for each mixture, keeping in mind that acidic solutions have a pH below 7, neutral solutions equal 7, and basic solutions exceed 7.

For mixture A, 10 mL of 0.2 M $\mathrm{Ca(OH)_2}$ provides 4 mmol of $\mathrm{OH}^-$, reacting with 2.5 mmol of $\mathrm{H}^+$ from 25 mL of 0.1 M $\mathrm{HCl}$ to leave excess base. Mixture B involves 0.2 mmol each of $\mathrm{H}^+$ from $\mathrm{H_2SO_4}$ and $\mathrm{OH}^-$ from $\mathrm{Ca(OH)_2}$, resulting in complete neutralization and a pH of 7. Mixture C provides 2 mmol of $\mathrm{H}^+$ from $\mathrm{H_2SO_4}$ and 1 mmol of $\mathrm{OH}^-$ from $\mathrm{KOH}$, leaving 1 mmol of excess acid.

Comparing these results, the acidic solution in C has the lowest pH, followed by the neutral solution in B, and finally the basic solution in A, establishing the order as $C < B < A$.

Before any base is added to the weak monoprotic acid, the concentration of hydrogen ions depends on the acid dissociation constant. For a weak acid, the pH is calculated using the relation $pH = \frac{1}{2}(pK_a - \log c)$. Substituting the given values, with an acid concentration of 0.2 M and a $pK_a = 3.3$, the expression becomes $pH = \frac{1}{2}(3.3 - \log 0.2)$. Since the logarithm of 0.2 is -0.7, the calculation simplifies to $pH = \frac{1}{2}(3.3 + 0.7)$, resulting in an initial $pH = 2.0$.

Adding 10 mL of 0.2 M NaOH initiates a neutralization reaction, forming an acidic buffer solution. Initially, the system contains 4 mmol of HX, and the addition of 2 mmol of NaOH consumes an equal amount of the acid, leaving 2 mmol of HX and forming 2 mmol of its conjugate base salt. In this buffer system, the pH is determined by the Henderson-Hasselbalch equation $pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$. Plugging in the values, we obtain $pH = 3.3 + \log \frac{2}{2}$, which simplifies to $pH = 3.3 + 0$, giving a final $pH = 3.3$.

Sodium dichromate, $\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7$, is inherently hygroscopic and absorbs atmospheric moisture, which prevents it from maintaining a constant mass and disqualifies it from being classified as a primary standard in titrimetric analysis. Furthermore, phenolphthalein functions as a weak organic acid, typically represented as $\mathrm{HIn}$, which dissociates in basic media to produce its characteristic colored form, contradicting the claim that it acts as a weak base that dissociates in acidic environments.

Solubility product constants, or $K_{sp}$, represent the equilibrium between a solid salt and its dissolved ions in a saturated solution. For the dissociation of $\mathrm{Ag}_2\mathrm{CrO}_4$, the equilibrium expression in terms of its molar solubility $s_1$ is given by
$4s_1^3 = 32x$
This expression simplifies to $s_1 = 2\sqrt[3]{x}$. For silver bromide, the dissociation constant $K_{sp}$ relates to its molar solubility $s_2$ via the equation
$s_2^2 = 4y$
Solving this yields $s_2 = 2\sqrt{y}$. Consequently, the ratio of the molarity of $\mathrm{Ag}_2\mathrm{CrO}_4$ to the molarity of $\mathrm{AgBr}$ is expressed as
$\frac{s_1}{s_2} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$

The pH of a buffer solution is governed by the Henderson-Hasselbalch equation, where for a basic buffer $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Given $pH = 9.25$, the $pOH$ is calculated as $14 - 9.25 = 4.75$, which equals the given $pK_b$ of $4.75$, implying that the ratio of salt to base concentrations must be equal to 1 for the condition to hold.

For option B, the initial number of moles of $NH_4OH$ is $0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1$ mol and the moles of $HCl$ are $0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05$ mol. The neutralization reaction $NH_4OH + HCl \rightarrow NH_4Cl + H_2O$ consumes the limiting reagent $HCl$ completely, resulting in $0.05$ mol of $NH_4Cl$ salt and $0.1 - 0.05 = 0.05$ mol of unreacted $NH_4OH$. Since the number of moles of salt and base are equal, their concentrations are also equal within the same volume, satisfying the condition $\log(1) = 0$ and thus yielding the required buffer solution.

The dissolution of the sparingly soluble salt $M_3A_2$ reaches an equilibrium represented by the equation $M_3A_2(s) \rightleftharpoons 3M^{2+}(aq) + 2A^{3-}(aq)$. Given the solubility is $x \, \text{g L}^{-1}$ and the molar mass is $y \, \text{g mol}^{-1}$, the molar solubility $s$ is calculated by dividing the mass concentration by the molar mass, resulting in $s = \frac{x}{y} \, \text{mol L}^{-1}$.

From the dissociation stoichiometry, the equilibrium concentration of the cation is $3s$ and the anion is $2s$. The solubility product constant $K_{sp}$ is defined as the product of the ionic concentrations raised to their stoichiometric coefficients:
$K_{sp} = \$[M^{2+}]^3 \$[A^{3-}]^2 = (3s)^3 \cdot (2s)^2 = 27s^3 \cdot 4s^2 = 108s^5$

The ratio of the molar concentration of the anion to the solubility product is determined by dividing $2s$ by $108s^5$:
$\frac{[A^{3-}]\$}{K_{sp}} = \frac{2s}{108s^5} = \frac{1}{54s^4}$
Substituting the molar solubility $s = \frac{x}{y}$ back into this expression provides the final result:
$\frac{1}{54} \cdot \frac{1}{(x/y)^4} = \frac{1}{54} \cdot \frac{y^4}{x^4}$

Phenolphthalein behaves as a pH-sensitive indicator that changes its molecular structure depending on the acidity or basicity of the environment. In the presence of excess $NaOH$, the compound undergoes a specific reaction that causes it to lose its typical pink pigment and become entirely colourless. This phenomenon highlights how high concentrations of strong bases push the indicator beyond its standard transition range into a state where it no longer reflects light in the visible spectrum.

For a weak electrolyte dissociating according to the equation $A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$, we determine the equilibrium constant $K$ by evaluating the molar concentrations at equilibrium. Starting with an initial concentration $c$ and degree of dissociation $\alpha$, the concentration of undissociated $A_x B_y$ is $c(1-\alpha)$, while the concentrations of the dissociated ions $A^{y+}$ and $B^{x-}$ are $c x \alpha$ and $c y \alpha$, respectively. Substituting these values into the law of mass action expression gives $K = \frac{ (c x \alpha)^x (c y \alpha)^y }{ c(1-\alpha) }$.

Rearranging and simplifying the numerator terms results in $K = c^{x+y-1} x^x y^y \frac{ \alpha^{x+y} }{ 1-\alpha }$. Because the electrolyte is weak, the degree of dissociation is typically small enough that the term $1-\alpha$ can be approximated as 1, simplifying the expression to $K \approx c^{x+y-1} x^x y^y \alpha^{x+y}$. Solving for $\alpha$ by isolating the variable leads to $\alpha = \left( \frac{ K }{ c^{x+y-1} x^x y^y } \right)^{ \frac{1}{x+y} }$.

The dissociation equilibrium for a weak acid $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$ is characterized by the acid dissociation constant K_a = \frac{[\mathrm{H}^{+}]\$[\mathrm{A}^{-}]$}{[\mathrm{HA}]}$. Given$\mathrm{p}K_a = 4$, the value of$K_a$is$10^{-4}$. Since the degree of dissociation is negligible, the concentration of undissociated$\mathrm{HA}$at equilibrium is approximately equal to the initial concentration$c = 10 \text{ mM} = 10^{-2} \text{ M}$, leading to the simplified relation$[\mathrm{H}^{+}] = \sqrt{K_a \cdot c}$. Substituting the given values,$[\mathrm{H}^{+}] = \sqrt{10^{-4} \cdot 10^{-2}} = \sqrt{10^{-6}} = 10^{-3} \text{ M}$. The$\mathrm{pH}$is calculated using the definition$\mathrm{pH} = -\log$[\mathrm{H}^{+}]$ = -\log(10^{-3})$, which results in 3.

The solubility product, denoted as $K_{sp}$, measures the extent to which an ionic compound dissociates in a saturated solution, where lower values indicate significantly lower solubility. Mercuric sulfide ($\mathrm{HgS}$) and lead sulfide ($\mathrm{PbS}$) possess extremely low solubility due to their exceptionally stable crystal lattices, making them the least soluble compounds in this set. Silver bromide ($\mathrm{AgBr}$) is sparingly soluble, while calcium hydroxide ($\mathrm{Ca}(\mathrm{OH})_2$) exhibits the highest solubility product among these options, resulting in the increasing order $\mathrm{HgS} < \mathrm{PbS} < \mathrm{AgBr} < \mathrm{Ca}(\mathrm{OH})_2$.

The van't Hoff factor ($i$) for a solute undergoing dissociation is defined by the equation $i = 1 + \alpha(n - 1)$, where $\alpha$ is the degree of dissociation and $n$ is the number of ions produced per formula unit. For the salt $\mathrm{MX}_3$, the dissociation follows the reaction $\mathrm{MX}_3 \rightarrow \mathrm{M}^{3+} + 3\mathrm{X}^-$, resulting in $n = 4$ ions. Substituting the given values $i = 2$ and $n = 4$ into the equation gives $2 = 1 + \alpha(4 - 1)$, which simplifies to $1 = 3\alpha$. Solving for $\alpha$ yields $0.3333$, which corresponds to $33.33\%$, and the nearest integer is 33.

The pH of the solution is 10.0, implying a pOH of $14.0 - 10.0 = 4.0$, which corresponds to a hydroxide ion concentration of $[\mathrm{OH}^-] = 10^{-4}$ M. Because $\mathrm{Mg(OH)_2}$ dissociates as $\mathrm{Mg(OH)_2} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{OH}^-$, the required molarity of the solute is [\mathrm{Mg(OH)_2}] = \frac{[\mathrm{OH}^-]\}{2} = 5 × 10^{-5}$M. Given the molar mass of 58 g/mol and a volume of 1.0 L, the mass calculation is$\text{mass} = (5 × 10^{-5} × 1.0 × 58) = 2.9 × 10^{-3}$g. Converting this to milligrams yields$2.9 × 10^{-3} × 1000 = 2.9$ mg, which rounds to the nearest integer of 3.

The dissociation of a weak acid $\mathrm{HX}$ is governed by the equilibrium expression K_a = \frac{[\mathrm{H}^+]\$[\mathrm{X}^-]$}{[\mathrm{HX}]{\text{eq}}}$, where$[\mathrm{HX}]{\text{eq}} = C - [\mathrm{H}^+]$represents the concentration of undissociated acid at equilibrium and$C$is the total analytical concentration. For the diluted solution with a pH of 6, the hydrogen ion concentration is$[\mathrm{H}^+] = 10^{-6} \ \mathrm{M}$. Substituting the known values$K_a = 4 × 10^{-10}$and$[\mathrm{H}^+] = 10^{-6} \ \mathrm{M}$ into the equilibrium expression gives:

$4 \times 10^{-10} = \frac{(10^{-6})^2}{C - 10^{-6}}$

Rearranging the equation to solve for the analytical concentration $C$ leads to $C - 10^{-6} = \frac{10^{-12}}{4 \times 10^{-10}}$, which simplifies to $C - 10^{-6} = 0.25 \times 10^{-2} = 25 \times 10^{-4} \ \mathrm{M}$. The total concentration $C$ is therefore $25 \times 10^{-4} + 10^{-6} = 25.01 \times 10^{-4} \ \mathrm{M}$. Matching this result to the form $x \times 10^{-4} \ \mathrm{M}$ and taking the nearest integer results in $x = 25$.

The ionization of a weak base in water is characterized by its dissociation constant, $K_b$. In aqueous solutions, the $\text{pH}$ and $\text{pOH}$ are linked by the relationship $\text{pH} + \text{pOH} = 14$. With a given $\text{pH}$ of 9, the $\text{pOH}$ is 5, implying a hydroxide ion concentration of $[\text{OH}^-] = 10^{-5}$ M.

Since the degree of ionization of the base can be neglected with respect to unity, the hydroxide ion concentration relates to the initial concentration $C$ through the equation $[\text{OH}^-] = \sqrt{K_b \cdot C}$. Converting the given 1 mM concentration into molarity yields $C = 10^{-3}$ M. Substituting these values into the equilibrium formula provides:
$10^{-5} = \sqrt{K_b \cdot 10^{-3}}$
Squaring both sides leads to $10^{-10} = K_b \cdot 10^{-3}$. Dividing by $10^{-3}$ results in $K_b = 10^{-7}$. Comparing this to the format $10^{-x}$ confirms that $x$ is 7.

Solubility equilibrium relies on the specific stoichiometry of a salt's dissociation in water. When $\mathrm{Cr}(\mathrm{OH})_3$ dissolves, it releases one $\mathrm{Cr}^{3+}$ ion for every three $\mathrm{OH}^{-}$ ions; defining the molar solubility as $s$ results in a chromium ion concentration of $s$ and a hydroxide ion concentration of $3s$. Substituting these values into the equilibrium expression K_{\text{sp}} = \[\mathrm{Cr}^{3+}]$$[\mathrm{OH}^{-}]^3$produces$K_{\text{sp}} = (s)(3s)^3$, which simplifies to$27s^4$. Algebraically isolating$s$requires dividing the$K_{\text{sp}}$value by 27 and taking the fourth root, leading to the expression$\sqrt[4]{\frac{1.6 × 10^{-30}}{27}}$.