Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

Diluting an acid reduces its concentration while the number of moles of solute remains constant, which is described by the dilution equation $M_1 V_1 = M_2 V_2$. An HCl solution with a pH of 1 has a hydrogen ion concentration of 0.1 M. To change the pH to 2, the final concentration of the solution must be 0.01 M.

Substituting these values into the dilution equation yields the following expression:
$0.1 \times 1 = 0.01 \times V_2$
Solving this equation determines that the final volume $V_2$ is 10 L. Subtracting the original 1 L of solution from this final volume shows that 9 L of water must be added.

The ionization constant of a weak acid describes its tendency to dissociate, calculated by relating the equilibrium concentrations of the acid species to its initial concentration. Since the pH of the solution is 3, the hydrogen ion concentration is $[H^+] = 10^{-3}$ M.

With an initial concentration of 0.1 M, the degree of ionization $\alpha$ is found by dividing the hydrogen ion concentration by the initial amount, giving $\alpha = 10^{-3} / 0.1 = 10^{-2}$. Using the equilibrium relationship $K_a = C\alpha^2$, the value is determined by calculating $K_a = 0.1 \times (10^{-2})^2 = 10^{-5}$.

Precipitation of a sparingly soluble salt begins the moment the ionic product of its constituent ions equals the solubility product constant, $K_{sp}$. For magnesium hydroxide, this equilibrium is defined by the expression K_{sp} = \[\mathrm{Mg}^{2+}]$$[\mathrm{OH}^{-}]^2$. By substituting the provided values of$1.0 × 10^{-11}$for$K_{sp}$and$0.001 \mathrm{M}$ for the magnesium ion concentration into this equation, we can determine the specific hydroxide ion concentration required to trigger the formation of a solid precipitate.

Rearranging the expression gives $[\mathrm{OH}^{-}]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}}$, which simplifies to $[\mathrm{OH}^{-}]^2 = 10^{-8}$. Taking the square root results in a hydroxide ion concentration of $10^{-4} \mathrm{M}$, corresponding to a pOH of $4$. Since the sum of pH and pOH is $14$ at $25^{\circ}\mathrm{C}$, subtracting the pOH value from $14$ indicates that precipitation initiates at a pH of $10$.

Precipitation of a sparingly soluble salt occurs when the ionic product of the constituent ions exceeds the solubility product constant, defined as K_{sp} = \[Ag^+]$[Br^-]$. Since silver nitrate acts as a strong electrolyte in solution, the concentration of silver ions is fixed at 0.05 M. To determine the threshold concentration needed to initiate precipitation, we rearrange the equilibrium expression to$[Br^-] = \frac{K_{sp}}{[Ag^+]$}$. By substituting the given values, we find that the required bromide ion concentration is$\frac{5.0 × 10^{-13}}{0.05} = 1.0 × 10^{-11} \text{ M}$. Given that the solution volume is 1 litre, this corresponds to$1.0 × 10^{-11} \text{ moles}$of bromide ions. Using the molar mass of 120 g/mol, the required mass of potassium bromide is$1.0 × 10^{-11} \text{ mol} × 120 \text{ g/mol} = 1.2 × 10^{-9} \text{ g}$.

According to the Brønsted-Lowry theory, an acid is defined as a proton donor, while a base functions as a proton acceptor. In reaction (i), $\mathrm{H_3PO_4}$ serves as the acid by donating a proton, which results in the formation of $\mathrm{H_2PO_4^{-}}$, meaning the latter is not acting as an acid in this process. Conversely, in reaction (ii), $\mathrm{H_2PO_4^{-}}$ donates a proton to $\mathrm{H_2O}$ to yield $\mathrm{HPO_4^{2-}}$, clearly demonstrating its role as an acid. Finally, in reaction (iii), $\mathrm{H_2PO_4^{-}}$ accepts a proton to form $\mathrm{H_3PO_4}$, identifying it as a base in this specific chemical interaction.

Precipitation begins the instant the ionic product of the salt equals its solubility product constant. For the sparingly soluble salt $\mathrm{BaCO}_3$, the solubility product expression is K_{\mathrm{sp}} = \[\mathrm{Ba}^{2+}]$[\mathrm{CO}_3^{2-}]$. Given that sodium carbonate dissociates completely, the solution provides carbonate ions at a concentration of 1.0 x 10^-4 M. Substituting the known values into the solubility expression gives:

$5.1 \times 10^{-9} = \$[\mathrm{Ba}^{2+}]\$ \times 1.0 \times 10^{-4}$

Solving this for the barium concentration yields a threshold value of 5.1 x 10^-5 M.

Acidity is fundamentally determined by the ability of a species to donate a proton, which is governed by the stability of its resulting conjugate base and the electronic environment of the acidic hydrogen. Comparing these four species, $\mathrm{HSO}_3\mathrm{F}$ acts as the strongest acid, a superacid where the highly electronegative fluorine atom and the multiple oxygens bonded to sulfur effectively stabilize the conjugate base. The $\mathrm{H}_3\mathrm{O}^{+}$ ion is stronger than $\mathrm{HSO}_4^{-}$ because the neutral hydronium ion dissociates more readily than the negatively charged hydrogen sulfate, while $\mathrm{HCO}_3^{-}$ is the weakest among the group due to the significant electrostatic resistance to proton loss from its negatively charged structure, resulting in the sequence $\mathrm{HCO}_3^{-} < \mathrm{HSO}_4^{-} < \mathrm{H}_3\mathrm{O}^{+} < \mathrm{HSO}_3\mathrm{F}$, or $i < iii < ii < iv$.

Salts formed from the neutralization of a weak acid and a weak base undergo hydrolysis in aqueous solution, where both the cation and the anion react with water to influence the overall hydrogen ion concentration. Because both components are weak, the resulting pH is independent of the salt concentration and is defined by the relationship:

$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_{\mathrm{w}} + \mathrm{p}K_{\mathrm{a}} - \mathrm{p}K_{\mathrm{b}})$

Taking the standard value for $\mathrm{p}K_{\mathrm{w}}$ as 14, and substituting the given values of 4.80 for $\mathrm{pK}_\mathrm{a}$ and 4.78 for $\mathrm{pK}_\mathrm{b}$, the calculation involves finding the average of the sum of 14 and 4.80 minus 4.78. This simplifies to 14.02 divided by 2, resulting in a solution pH of 7.01.

The solubility of a sparingly soluble salt is governed by the solubility product constant, $\mathrm{K}_{\mathrm{sp}}$, which balances the ion concentrations in a saturated solution. Since the dissociation reaction for $\mathrm{AgIO}_3$ releases one equivalent of each ion, the relationship simplifies to $\mathrm{K}_{\mathrm{sp}} = s^2$. Extracting the square root of $\mathrm{K}_{\mathrm{sp}} = 1.0 \times 10^{-8}$ provides a molar solubility of $1.0 \times 10^{-4} \text{ mol/L}$. Multiplying this molarity by the molecular mass of $283 \text{ g/mol}$ converts the solubility to $2.83 \times 10^{-2} \text{ g/L}$. Because the sample volume is $100 \text{ ml}$, or one-tenth of a liter, the actual mass of the solute present is $2.83 \times 10^{-3} \text{ g}$.

The behavior of a weak acid buffer is determined by the Henderson-Hasselbalch equation, pH = pK_a + \log \frac{[A^-]\}{[HA]}$. When 50$[A^-]$is equal to the concentration of the undissociated acid$[HA]$, which effectively cancels the logarithmic term to zero and makes the solution$pH$equal to the$pK_a$value of 4.5. Since the \sum of$pH$and$pOH$\in an aqueous solution is always 14, subtracting the$pH$of 4.5 from 14 yields a final$pOH$ of 9.5.

When a polyprotic acid dissociates in successive stages, the overall equilibrium constant for the complete breakdown of the molecule is the product of the individual dissociation constants for each step. This relationship holds because the net chemical reaction is the sum of the stepwise reactions, and adding chemical equations mathematically requires multiplying their respective equilibrium constants.

The overall dissociation constant is determined by multiplying the first dissociation constant, $1.0 \times 10^{-5}$, by the second, $5.0 \times 10^{-10}$. Performing the calculation $(1.0 \times 10^{-5}) \times (5.0 \times 10^{-10})$ results in an overall dissociation constant of $5.0 \times 10^{-15}$.

The pH of a solution represents the negative base-10 logarithm of the hydrogen ion concentration, defined by the relationship $\mathrm{pH} = -\log [\mathrm{H}^+]$. Rearranging this to isolate the concentration provides the expression $[\mathrm{H}^+] = 10^{-\mathrm{pH}}$. Substituting 5.4 for the pH value results in $[\mathrm{H}^+] = 10^{-5.4}$, which is equivalent to $10^{0.6} \times 10^{-6}$. Since 10 raised to the power of 0.6 is approximately 3.98, the final hydrogen ion concentration is $3.98 \times 10^{-6} \, \mathrm{mol/L}$.

When a salt with the formula $\mathrm{MX}_2$ dissolves in water, it dissociates into one $\mathrm{M}^{2+}$ cation and two $\mathrm{X}^-$ anions. Because the stoichiometry involves one $\mathrm{M}^{2+}$ ion for every molecule of $\mathrm{MX}_2$, the molar concentration of $\mathrm{M}^{2+}$ ions is equivalent to the solubility $s$ of the salt, while the concentration of $\mathrm{X}^-$ ions is $2s$. The solubility product constant is defined as:
$K_{sp} = \$[\mathrm{M}^{2+}]\$\$[\mathrm{X}^-]^2$
Substituting the molar concentrations into this expression results in:
$K_{sp} = (s)(2s)^2 = 4s^3$
Given the value $K_{sp} = 4 \times 10^{-12}$, we equate this to the derived expression:
$4s^3 = 4 \times 10^{-12}$
Solving this equation yields $s^3 = 1 \times 10^{-12}$, which simplifies to $s = 1 \times 10^{-4}$. Since the concentration of $\mathrm{M}^{2+}$ ions is defined as $s$, the concentration is $1.0 \times 10^{-4}$ M.

The fundamental principle behind conjugate acid-base pairs is that they differ by exactly one proton, so finding the conjugate base simply requires removing an $\mathrm{H}^{+}$ ion from the starting substance. When you apply this to the hydroxide ion, $\mathrm{OH}^{-}$, subtracting a hydrogen ion leaves the oxygen atom with the remaining negative charge, resulting in the oxide ion, as shown by the reaction $\mathrm{OH}^{-} \rightarrow \mathrm{H}^{+} + \mathrm{O}^{2-}$.

The Bronsted-Lowry theory defines a conjugate acid-base pair as two species that differ by exactly one proton. To find the conjugate base of any molecule or ion, you simply remove one $\mathrm{H}^{+}$ from its formula, which simultaneously decreases its net charge by one unit. Applying this process to $\mathrm{H}_2\mathrm{PO}_4^{-}$ involves taking away one hydrogen atom and subtracting one positive charge from the total, resulting in $\mathrm{HPO}_4^{2-}$.

The solubility product constant, $K_{sp}$, represents the equilibrium condition for a sparingly soluble salt, where the product of the ion concentrations, raised to their stoichiometric coefficients, remains constant. When $\text{MX}_4$ dissociates in water, it yields one $\text{M}^{4+}$ ion and four $\text{X}^{-}$ ions, implying that if the molar solubility is $s$, the ion concentrations are $[M^{4+}] = s$ and $[X^{-}] = 4s$. Applying these values to the equilibrium expression produces the following relation:

$K_{sp} = [s][4s]^4 = 256s^5$

Isolating $s$ by dividing by 256 and taking the fifth root gives the final expression $s = \left( \frac{K_{sp}}{256} \right)^{1/5}$.

The solubility product constant, $K_{sp}$, describes the equilibrium for a sparingly soluble salt by the product of its constituent ion concentrations raised to their stoichiometric powers. Dissolving $AB_2$ generates ions according to the equation $AB_2 \rightleftharpoons A^{2+} + 2B^{-}$, meaning that for a molar solubility $s$, the equilibrium concentrations are $[A^{2+}] = s$ and $[B^{-}] = 2s$. Substituting these into the solubility product expression gives K_{sp} = \[A^{2+}]$$[B^{-}]^2 = s(2s)^2 = 4s^3$. With the given solubility$s = 1.0 × 10^{-5} , \mathrm{mol} , \mathrm{L}^{-1}$, the calculation becomes$4(1.0 × 10^{-5})^3$, resulting \in$4 × 10^{-15}$.

Acids, by definition, must yield a pH value below 7, making the claim that a dilute hydrochloric acid solution results in a pH of 8 physically impossible. When dealing with extremely low concentrations of strong acids, such as $1 \times 10^{-8}$ M HCl, the contribution of hydrogen ions from the auto-ionization of water cannot be ignored because it is comparable to the concentration provided by the acid itself. The water molecules provide $[\mathrm{H}^{+}] = 10^{-7}$ M, while the acid contributes $[\mathrm{H}^{+}] = 10^{-8}$ M. Summing these values gives the total concentration as follows:

$[\mathrm{H}^{+}] = 10^{-8} + 10^{-7} = 11 \times 10^{-8} \text{ M}$

Calculating the pH using the definition $-\log [\mathrm{H}^{+}]$ yields:

$\text{pH} = -\log(11 \times 10^{-8}) = 6.9586$

Since 6.9586 is less than 7, the premise that such a solution would have a pH of 8 is clearly incorrect.

The pH scale indicates the acidity of a substance, where a decrease in value signifies a higher concentration of acidic components. During a thunderstorm, the powerful electrical discharges from lightning trigger reactions such as $N_2 + O_2 \rightarrow 2NO$, which creates various nitrogen oxides. These oxides subsequently dissolve in falling rainwater to produce acidic compounds, leading to a reduction in the pH level of the collected water compared to rain that falls without such atmospheric activity.

When $\mathrm{Mg}(\mathrm{OH})_2$ dissociates in an aqueous solution, it establishes an equilibrium defined by $\mathrm{Mg}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Mg}^{2+} + 2\mathrm{OH}^-$. If the solubility of this compound is $x$, the stoichiometry of the reaction dictates that the equilibrium concentration of magnesium ions will be $x$ and the concentration of hydroxide ions will be $2x$. The solubility product constant, $K_{\text{sp}}$, is calculated by multiplying these concentrations raised to the power of their stoichiometric coefficients, expressed as K_{\text{sp}} = \[\mathrm{Mg}^{2+}]$$[\mathrm{OH}^-]^2$. Substituting the concentrations into this formula yields$K_{\text{sp}} = (x)(2x)^2$, which simplifies to$4x^3$.