Masterclass: 25 Solved JEE Advanced Numericals on Ionic Equilibrium
Conquer the complexities of acids, bases, and salts! This exhaustive guide features highly rigorous multi-step problems on Buffer Solutions, Salt Hydrolysis, Solubility Products, and the exact pH of dilute solutions. Click "View Solution" to reveal the step-by-step breakdown.
Ionic Equilibrium is the ultimate test of approximation logic in physical chemistry. Before attempting these rigorous problems, ensure you clearly understand when to ignore $x$ with respect to $C$, how the Common Ion Effect drastically suppresses dissociation, and when the auto-ionization of water ($10^{-14}$) becomes mathematically significant.
View Solution
Step 1: Understand the Trap
A common mistake is taking $pH = -\log_{10}(10^{-8}) = 8$. An acid can NEVER have a basic pH (> 7) at $298 \text{ K}$. At such low concentrations, the $H^+$ contributed by the auto-ionization of water cannot be ignored.
Step 2: Setup the Equations
Let the concentration of $H^+$ from water be $x \text{ M}$.
Therefore, $[OH^-] = x \text{ M}$.
Total $[H^+] = 10^{-8} \text{ (from HCl)} + x \text{ (from water)}$.
Step 3: Apply the Ionic Product of Water ($K_w$)
$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$
$(10^{-8} + x)(x) = 10^{-14}$
$x^2 + 10^{-8}x - 10^{-14} = 0$
Step 4: Solve the Quadratic Equation
$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2}$
$x = \frac{-10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} = \frac{-10^{-8} + \sqrt{401 \times 10^{-16}}}{2}$
$x = \frac{-10^{-8} + 20.02 \times 10^{-8}}{2} = \frac{19.02 \times 10^{-8}}{2} = 9.51 \times 10^{-8} \text{ M}$.
Step 5: Calculate Final pH
Total $[H^+] = 10^{-8} + 9.51 \times 10^{-8} = 10.51 \times 10^{-8} = 1.051 \times 10^{-7} \text{ M}$.
$pH = -\log_{10}(1.051 \times 10^{-7}) = 7 - \log_{10}(1.051) = 7 - 0.021 = 6.98$.
View Solution
Step 1: Check Approximation Validity
Ostwald's dilution law: $K_a = \frac{C\alpha^2}{1-\alpha}$.
If $K_a \le 10^{-4}$ and $C$ is relatively large ($0.1 \text{ M}$), $\alpha$ is very small, so $1 - \alpha \approx 1$.
$K_a \approx C\alpha^2 \implies \alpha = \sqrt{\frac{K_a}{C}}$.
Step 2: Calculate $\alpha$
$\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = \sqrt{1.8 \times 10^{-4}} = \sqrt{1.8} \times 10^{-2}$.
Since $\sqrt{1.8} \approx 1.34$, $\alpha = 1.34 \times 10^{-2}$ or $1.34\%$. (Approximation is highly valid since $\alpha < 5\%$).
Step 3: Calculate pH
$[H^+] = C\alpha = 0.1 \times 1.34 \times 10^{-2} = 1.34 \times 10^{-3} \text{ M}$.
Alternatively: $[H^+] = \sqrt{K_a C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{18 \times 10^{-7}} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$.
$pH = -\log_{10}(1.34 \times 10^{-3}) = 3 - \log_{10}(1.34) = 3 - 0.127 = 2.87$.
View Solution
Step 1: Understand Isohydric Solutions
When two weak acids are mixed, they mutually suppress each other's dissociation due to the common ion effect of $H^+$. The total $[H^+]$ is the sum of contributions from both acids.
Step 2: Apply the Mixture Formula
For a mixture of two weak acids with concentrations $C_1$ and $C_2$, the total $H^+$ concentration is given by:
$[H^+] = \sqrt{K_{a1}C_1 + K_{a2}C_2}$
Step 3: Substitute and Solve
$[H^+] = \sqrt{(1.8 \times 10^{-4} \times 0.1) + (1.8 \times 10^{-5} \times 0.2)}$
$[H^+] = \sqrt{(1.8 \times 10^{-5}) + (0.36 \times 10^{-5})}$
$[H^+] = \sqrt{2.16 \times 10^{-5}} = \sqrt{21.6 \times 10^{-6}}$
$[H^+] = 4.65 \times 10^{-3} \text{ M}$.
View Solution
Step 1: First Dissociation
$H_2S \rightleftharpoons H^+ + HS^-$
Since $K_{a1} \gg K_{a2}$, virtually all $H^+$ comes from the first step.
$[H^+] \approx [HS^-] = \sqrt{K_{a1}C} = \sqrt{10^{-7} \times 0.1} = \sqrt{10^{-8}} = 10^{-4} \text{ M}$.
Step 2: Second Dissociation
$HS^- \rightleftharpoons H^+ + S^{2-}$
$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$
Step 3: Analyze Concentrations
From Step 1, we know $[H^+] \approx 10^{-4} \text{ M}$ and $[HS^-] \approx 10^{-4} \text{ M}$.
Substitute these into the $K_{a2}$ expression:
$1.2 \times 10^{-13} = \frac{(10^{-4})[S^{2-}]}{10^{-4}}$
The $10^{-4}$ terms cancel out completely!
Golden Rule: For any weak diprotic acid where $K_{a1} \gg K_{a2}$, the concentration of the secondary divalent anion is exactly equal to $K_{a2}$, independent of the initial acid concentration.
View Solution
Step 1: Identify the Salt
$NH_4Cl$ is a salt of a Weak Base ($NH_4OH$) and a Strong Acid ($HCl$). Its aqueous solution will undergo cationic hydrolysis ($NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$), making the solution acidic.
Step 2: Use the Hydrolysis Formula for pH
For a SA-WB salt, $pH = 7 - \frac{1}{2}(pK_b + \log_{10} C)$
Step 3: Calculate $pK_b$
$pK_b = -\log_{10}(1.8 \times 10^{-5}) = 5 - \log_{10}(1.8) = 5 - 0.255 = 4.745$.
Step 4: Substitute into pH formula
$C = 0.1 \implies \log_{10} C = -1$.
$pH = 7 - \frac{1}{2}(4.745 + (-1)) = 7 - \frac{1}{2}(3.745)$
$pH = 7 - 1.8725 = 5.127$.
View Solution
Step 1: Identify the Salt and Hydrolysis Type
$CH_3COONa$ is a WA-SB salt. It undergoes anionic hydrolysis ($CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$), making the solution basic.
Step 2: Calculate Hydrolysis Constant ($K_h$)
$K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.55 \times 10^{-10}$.
Step 3: Calculate Degree of Hydrolysis ($h$)
Since $K_h$ is very small, $h = \sqrt{\frac{K_h}{C}}$.
$h = \sqrt{\frac{5.55 \times 10^{-10}}{0.05}} = \sqrt{1.11 \times 10^{-8}} = 1.05 \times 10^{-4}$. ($0.01\%$, approximation valid).
Step 4: Calculate pH
For a WA-SB salt, $pH = 7 + \frac{1}{2}(pK_a + \log_{10} C)$.
$pK_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$.
$\log_{10} C = \log_{10}(0.05) = \log_{10}(5 \times 10^{-2}) = 0.699 - 2 = -1.301$.
$pH = 7 + \frac{1}{2}(4.74 - 1.301) = 7 + \frac{1}{2}(3.439) = 7 + 1.72 = 8.72$.
View Solution
Step 1: Identify the Salt
This is a WA-WB salt. Both the cation and anion hydrolyze.
Step 2: Formula for pH of WA-WB salt
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Step 3: Analyze the unique property
Notice that the concentration ($C$) does not appear anywhere in the formula! The pH of a weak acid-weak base salt is completely independent of its concentration (assuming it is not so dilute that water's auto-ionization dominates).
Step 4: Calculate
Since $K_a = K_b = 1.8 \times 10^{-5}$, then $pK_a = pK_b = 4.74$.
$pH = 7 + \frac{1}{2}(4.74 - 4.74) = 7 + 0 = 7.0$.
Step 5: Effect of Dilution
Because the formula lacks a $C$ term, diluting the solution to $0.01 \text{ M}$ will not change the pH. It remains exactly 7.0.
View Solution
Step 1: Identify the Buffer
A mixture of a weak acid ($CH_3COOH$) and its conjugate base salt ($CH_3COONa$) forms an acidic buffer.
Step 2: Apply Henderson-Hasselbalch Equation
$pH = pK_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Step 3: Substitute and Calculate
$pH = 4.74 + \log_{10} \frac{0.5}{0.2}$
$pH = 4.74 + \log_{10} (2.5)$
$\log_{10}(2.5) \approx 0.398$.
$pH = 4.74 + 0.398 = 5.138$.
View Solution
Step 1: Understand Buffer Action
The added strong acid ($H^+$) will react completely with the basic component of the buffer (the Acetate ion, $CH_3COO^-$) to form more weak acid ($CH_3COOH$).
Reaction: $CH_3COO^- + H^+ \rightarrow CH_3COOH$.
Step 2: Stoichiometry of the Reaction
Initial moles (in $1 \text{ L}$): $[Salt] = 0.5 \text{ mol}$, $[Acid] = 0.2 \text{ mol}$.
Added $H^+ = 0.1 \text{ mol}$.
The $0.1 \text{ mol}$ of $H^+$ consumes $0.1 \text{ mol}$ of Salt and creates $0.1 \text{ mol}$ of Acid.
New $[Salt] = 0.5 - 0.1 = 0.4 \text{ M}$.
New $[Acid] = 0.2 + 0.1 = 0.3 \text{ M}$.
Step 3: Apply Henderson-Hasselbalch Equation
$pH = pK_a + \log_{10} \frac{[\text{New Salt}]}{[\text{New Acid}]}$
$pH = 4.74 + \log_{10} \frac{0.4}{0.3} = 4.74 + \log_{10}(1.33)$
$\log_{10}(1.33) \approx 0.124$.
$pH = 4.74 + 0.124 = 4.864$.
Check: The pH dropped from 5.14 to 4.86 upon adding a strong acid, demonstrating excellent buffer capacity.
View Solution
Step 1: Convert pH to pOH
For a basic buffer, the equation uses pOH.
$pOH = 14.0 - pH = 14.0 - 10.0 = 4.0$.
Step 2: Apply Henderson-Hasselbalch for Basic Buffer
$pOH = pK_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]}$
$4.0 = 4.74 + \log_{10} \frac{[\text{Salt}]}{0.1}$
Step 3: Solve for [Salt]
$\log_{10} \frac{[\text{Salt}]}{0.1} = 4.0 - 4.74 = -0.74$
$\frac{[\text{Salt}]}{0.1} = 10^{-0.74} = 0.182$
$[\text{Salt}] = 0.182 \times 0.1 = 0.0182 \text{ M}$.
Step 4: Calculate Mass of $NH_4Cl$
Since volume is $1.0 \text{ L}$, we need $0.0182 \text{ moles}$ of $NH_4Cl$.
Mass = Moles $\times$ Molar Mass = $0.0182 \text{ mol} \times 53.5 \text{ g/mol} = 0.9737 \text{ g}$.
View Solution
Step 1: Write the Dissociation Equation
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$
Step 2: Relate Solubility ($S$) to Ion Concentration
If solubility is $S$, then at equilibrium:
$[Ba^{2+}] = S$
$[SO_4^{2-}] = S$
Step 3: Write $K_{sp}$ Expression and Calculate
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = (S)(S) = S^2$
$K_{sp} = (1.05 \times 10^{-5})^2 = 1.1025 \times 10^{-10}$.
View Solution
Part (a): Solubility in Pure Water
$AgCl(s) \rightleftharpoons Ag^+ + Cl^-$
$K_{sp} = [Ag^+][Cl^-] = (S)(S) = S^2$
$S = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ M}$.
Part (b): Solubility in $0.1 \text{ M } NaCl$
$NaCl$ is a strong electrolyte, completely yielding $[Cl^-] = 0.1 \text{ M}$.
Let the new solubility of $AgCl$ be $S'$.
Total $[Cl^-] = 0.1 \text{ (from NaCl)} + S' \text{ (from AgCl)}$.
Since $S'$ is incredibly small, total $[Cl^-] \approx 0.1 \text{ M}$.
$K_{sp} = [Ag^+][Cl^-]$
$1.8 \times 10^{-10} = (S')(0.1)$
$S' = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ M}$.
Insight: The common $Cl^-$ ion forced the equilibrium backward, dropping the solubility of $AgCl$ by a factor of 10,000!
View Solution
Step 1: Set up simultaneous equilibria
Let solubility of $AgCl = x$ and solubility of $AgBr = y$.
$AgCl \rightleftharpoons Ag^+ + Cl^-$
$AgBr \rightleftharpoons Ag^+ + Br^-$
Total $[Ag^+]$ in the beaker $= x + y$.
$[Cl^-] = x$, $[Br^-] = y$.
Step 2: Write $K_{sp}$ equations
$K_{sp}(AgCl) = (x+y)x = 10^{-10}$ --- (Eq 1)
$K_{sp}(AgBr) = (x+y)y = 5 \times 10^{-13}$ --- (Eq 2)
Step 3: Analyze via approximation
Because $K_{sp}(AgCl) \gg K_{sp}(AgBr)$, $AgCl$ is vastly more soluble. Therefore, almost all the $Ag^+$ in solution comes from $AgCl$. Thus, $x \gg y$, meaning $(x+y) \approx x$.
Eq 1 simplifies to: $(x)x \approx 10^{-10} \implies x \approx 10^{-5} \text{ M}$.
Total $[Ag^+] \approx 10^{-5} \text{ M}$.
Step 4: Solve for $y$ ($Br^-$ concentration)
Substitute total $[Ag^+]$ into Eq 2:
$[Ag^+][Br^-] = 5 \times 10^{-13}$
$(10^{-5})y = 5 \times 10^{-13}$
$y = 5 \times 10^{-8} \text{ M}$.
View Solution
Step 1: Calculate new concentrations after mixing
When equal volumes ($50 \text{ mL}$ each) are mixed, the total volume doubles ($100 \text{ mL}$), so concentrations are exactly halved.
New $[Ba^{2+}] = \frac{0.02}{2} = 0.01 \text{ M} = 10^{-2} \text{ M}$.
New $[SO_4^{2-}] = \frac{0.02}{2} = 0.01 \text{ M} = 10^{-2} \text{ M}$.
Step 2: Calculate Ionic Product ($Q_{sp}$)
$Q_{sp} = [Ba^{2+}][SO_4^{2-}] = (10^{-2})(10^{-2}) = 10^{-4}$.
Step 3: Compare $Q_{sp}$ with $K_{sp}$
$K_{sp} = 1.1 \times 10^{-10}$.
Since $Q_{sp} > K_{sp}$ ($10^{-4} \gg 10^{-10}$), the solution is massively supersaturated. To restore equilibrium, the excess ions will crash out of solution.
View Solution
Step 1: Calculate $[Ag^+]$ required to start precipitation for each
For $AgCl$: $[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ M}$.
For $Ag_2CrO_4$: $[Ag^+] = \sqrt{\frac{K_{sp}}{[CrO_4^{2-}]}} = \sqrt{\frac{1.1 \times 10^{-12}}{0.1}} = \sqrt{1.1 \times 10^{-11}} = 3.3 \times 10^{-6} \text{ M}$.
Step 2: Identify first precipitate
Since $1.8 \times 10^{-9} < 3.3 \times 10^{-6}$, $AgCl$ requires less $Ag^+$ to reach its solubility limit. Thus, $AgCl$ precipitates first.
Step 3: Concentration of $Cl^-$ when $Ag_2CrO_4$ begins to precipitate
$Ag_2CrO_4$ begins precipitating exactly when $[Ag^+]$ reaches $3.3 \times 10^{-6} \text{ M}$.
At this specific $[Ag^+]$, what is the $[Cl^-]$ forced to be by the $AgCl$ equilibrium?
$[Cl^-] = \frac{K_{sp}(AgCl)}{[Ag^+]} = \frac{1.8 \times 10^{-10}}{3.3 \times 10^{-6}} = 5.45 \times 10^{-5} \text{ M}$.
Insight: The $[Cl^-]$ dropped from $0.1 \text{ M}$ to $0.0000545 \text{ M}$ before the chromate even started precipitating. This proves $>99.9\%$ of the chloride can be separated cleanly!
View Solution
Step 1: Combine the two simultaneous equilibria
Dissolution: $AgCl(s) \rightleftharpoons Ag^+ + Cl^-$ $\quad (K_{sp})$
Complexation: $Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$ $\quad (K_f)$
Overall Net Reaction: $AgCl(s) + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+ + Cl^-$
Step 2: Calculate overall equilibrium constant ($K_{net}$)
When equations are added, constants are multiplied.
$K_{net} = K_{sp} \times K_f = (1.8 \times 10^{-10}) \times (1.6 \times 10^7) = 2.88 \times 10^{-3}$.
Step 3: Setup ICE Table for the overall reaction
Let solubility be $S$.
Initial $[NH_3] = 1.0 \text{ M}$. Products = $0$.
Equilibrium $[NH_3] = 1.0 - 2S$.
Equilibrium $[[Ag(NH_3)_2]^+] = S$.
Equilibrium $[Cl^-] = S$.
Step 4: Solve for S
$K_{net} = \frac{S \cdot S}{(1.0 - 2S)^2} = \frac{S^2}{(1.0 - 2S)^2}$
$2.88 \times 10^{-3} = \left( \frac{S}{1.0 - 2S} \right)^2$
Take the square root of both sides ($\sqrt{0.00288} \approx 0.0537$):
$0.0537 = \frac{S}{1.0 - 2S}$
$0.0537 - 0.1074S = S$
$1.1074S = 0.0537 \implies S = 0.0485 \text{ M}$.
View Solution
Step 1: Understand Amphoteric nature of $HCO_3^-$
The bicarbonate ion can act as an acid (donating its proton to become $CO_3^{2-}$) or as a base (accepting a proton to become $H_2CO_3$).
Step 2: Formula for ampholyte pH
For an intermediate amphoteric ion derived from a polyprotic acid, the pH is practically independent of concentration and is given by the average of the two relevant $pK_a$ values:
$pH = \frac{pK_{a1} + pK_{a2}}{2}$
Step 3: Calculate the $pK_a$ values
$pK_{a1} = -\log_{10}(4.3 \times 10^{-7}) = 7 - \log_{10}(4.3) = 7 - 0.633 = 6.367$
$pK_{a2} = -\log_{10}(5.6 \times 10^{-11}) = 11 - \log_{10}(5.6) = 11 - 0.748 = 10.252$
Step 4: Calculate pH
$pH = \frac{6.367 + 10.252}{2} = \frac{16.619}{2} = 8.31$.
View Solution
Step 1: Analyze Total $[H^+]$
$HCl$ is a strong acid and dissociates $100\%$, providing $[H^+] = 0.1 \text{ M}$.
$HCN$ is a very weak acid. Its dissociation is heavily suppressed by the common ion effect from the strong $HCl$.
Total $[H^+] = 0.1 \text{ (from HCl)} + x \text{ (from HCN)} \approx 0.1 \text{ M}$.
Step 2: Calculate $[CN^-]$
Write the equilibrium for $HCN$: $HCN \rightleftharpoons H^+ + CN^-$
$K_a = \frac{[H^+][CN^-]}{[HCN]}$
Substitute the total $[H^+]$ from the strong acid:
$4.0 \times 10^{-10} = \frac{(0.1)[CN^-]}{0.1}$
Step 3: Solve
The $0.1$ cancels out perfectly on the right side.
$[CN^-] = 4.0 \times 10^{-10} \text{ M}$.
View Solution
Step 1: Identify Equivalence Point Chemistry
At equivalence, all $CH_3COOH$ is neutralized to $CH_3COONa$.
Volume of $0.1 \text{ M } NaOH$ required to neutralize $50 \text{ mL}$ of $0.1 \text{ M }$ acid is exactly $50 \text{ mL}$.
Total volume = $50 + 50 = 100 \text{ mL}$.
Step 2: Calculate Concentration of the Salt
Moles of salt formed = Moles of acid initially present = $0.1 \text{ M} \times 0.05 \text{ L} = 0.005 \text{ mol}$.
New concentration $C = \frac{0.005 \text{ mol}}{0.100 \text{ L}} = 0.05 \text{ M}$.
Step 3: Apply Salt Hydrolysis Formula
$CH_3COONa$ is a WA-SB salt. The solution is basic.
$pH = 7 + \frac{1}{2}(pK_a + \log_{10} C)$
$pK_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$
$\log_{10}(0.05) = \log_{10}(5 \times 10^{-2}) = 0.699 - 2 = -1.301$
$pH = 7 + \frac{1}{2}(4.74 - 1.301) = 7 + \frac{1}{2}(3.439) = 7 + 1.72 = 8.72$.
View Solution
Step 1: Understand Indicator Theory
An indicator is just a weak acid: $HIn \rightleftharpoons H^+ + In^-$.
By Henderson-Hasselbalch: $pH = pK_{In} + \log_{10}\frac{[In^-]}{[HIn]}$.
The human eye perceives a complete color change when one form is at least 10 times more concentrated than the other.
Step 2: Calculate Limits
Red limit: $[HIn]$ is 10x greater than $[In^-]$. Ratio $= 1/10$.
$pH_{red} = pK_{In} + \log_{10}(1/10) = pK_{In} - 1$.
Blue limit: $[In^-]$ is 10x greater than $[HIn]$. Ratio $= 10/1$.
$pH_{blue} = pK_{In} + \log_{10}(10) = pK_{In} + 1$.
Step 3: Calculate Range
$pK_{In} = -\log_{10}(10^{-5}) = 5.0$.
Range $= (5.0 - 1)$ to $(5.0 + 1)$.
View Solution
Step 1: Calculate $[H^+]$ in pure water
In pure water, auto-ionization produces equal amounts of $H^+$ and $OH^-$.
$[H^+] = [OH^-] = \sqrt{K_w}$
$[H^+] = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \text{ M}$.
Step 2: Calculate pH
$pH = -\log_{10}(1.0 \times 10^{-6}) = 6.0$.
Step 3: Analyze Acidic/Neutral Concept
Even though the pH is 6.0, the water is strictly Neutral. The definition of neutrality is not "pH = 7", but rather $[H^+] = [OH^-]$. Since both are equal at $10^{-6} \text{ M}$, it remains neutral. The entire pH scale shifts downwards at higher temperatures because the auto-ionization of water is an endothermic process.
View Solution
Step 1: Reaction Stoichiometry
Let $V$ be the volume of $NaOH$ added in mL.
Initial millimoles of $HCOOH = 100 \times 0.5 = 50 \text{ mmol}$.
Added millimoles of $NaOH = V \times 0.2 = 0.2V \text{ mmol}$.
The $NaOH$ converts $HCOOH$ into $HCOONa$ (Salt).
Moles of Salt formed $= 0.2V$.
Moles of Acid remaining $= 50 - 0.2V$.
Step 2: Apply Henderson-Hasselbalch
$pH = pK_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Since they share the same total volume, we can use the ratio of millimoles directly.
$4.04 = 3.74 + \log_{10} \frac{0.2V}{50 - 0.2V}$
Step 3: Solve for V
$0.30 = \log_{10} \frac{0.2V}{50 - 0.2V}$
Taking antilog ($10^{0.30} \approx 2.0$):
$2.0 = \frac{0.2V}{50 - 0.2V}$
$100 - 0.4V = 0.2V$
$0.6V = 100 \implies V = \frac{100}{0.6} = 166.67 \text{ mL}$.
View Solution
Step 1: Understand the Double Equilibrium
$CaF_2$ dissolves: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$.
Because $F^-$ is the conjugate base of a weak acid, it reacts with the abundant $H^+$ in the buffer: $F^- + H^+ \rightleftharpoons HF$.
The $H^+$ pulls the $F^-$ out of the solubility equilibrium, forcing more $CaF_2$ to dissolve!
Step 2: Establish Mass Balance
Let solubility be $S$. Then total Calcium is $S$, and total Fluorine species ($F^-$ + $HF$) is $2S$.
$[Ca^{2+}] = S$
$[F^-] + [HF] = 2S$
Step 3: Relate $[F^-]$ to $[HF]$ using $K_a$
$K_a = \frac{[H^+][F^-]}{[HF]} \implies [HF] = \frac{[H^+][F^-]}{K_a}$.
Substitute into mass balance:
$[F^-] + \frac{[H^+][F^-]}{K_a} = 2S \implies [F^-]\left(1 + \frac{[H^+]}{K_a}\right) = 2S$
$[F^-] = \frac{2S}{1 + \frac{[H^+]}{K_a}}$
Step 4: Substitute into $K_{sp}$ and Solve
$pH = 3.0 \implies [H^+] = 10^{-3} \text{ M}$.
Denominator factor: $1 + \frac{10^{-3}}{7.2 \times 10^{-4}} = 1 + \frac{10}{7.2} = 1 + 1.389 = 2.389$.
So, $[F^-] = \frac{2S}{2.389} = 0.837 S$.
$K_{sp} = [Ca^{2+}][F^-]^2 \implies 4.0 \times 10^{-11} = (S)(0.837 S)^2$
$4.0 \times 10^{-11} = 0.70 S^3$
$S^3 = 5.71 \times 10^{-11} = 57.1 \times 10^{-12}$
$S = (57.1)^{1/3} \times 10^{-4} \approx 3.85 \times 10^{-4} \text{ M}$.
View Solution
Step 1: Find maximum allowed $[S^{2-}]$
To just prevent precipitation, $Q_{sp} \le K_{sp}$.
$[Zn^{2+}][S^{2-}] = 1.0 \times 10^{-21}$
$(0.01)[S^{2-}] = 1.0 \times 10^{-21} \implies [S^{2-}] = 1.0 \times 10^{-19} \text{ M}$.
Step 2: Relate $[S^{2-}]$ to $[H^+]$ using overall $K_a$
For the overall dissociation $H_2S \rightleftharpoons 2H^+ + S^{2-}$:
$K_{overall} = K_{a1} \times K_{a2} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$
Step 3: Solve for $[H^+]$
$1.0 \times 10^{-21} = \frac{[H^+]^2 (1.0 \times 10^{-19})}{0.1}$
$[H^+]^2 = \frac{1.0 \times 10^{-22}}{1.0 \times 10^{-19}} = 1.0 \times 10^{-3}$
$[H^+] = \sqrt{10 \times 10^{-4}} = 3.16 \times 10^{-2} \text{ M}$.
Step 4: Calculate pH
$pH = -\log_{10}(3.16 \times 10^{-2}) = 2 - \log_{10}(3.16) = 2 - 0.5 = 1.5$.
If the pH is kept below 1.5 (highly acidic), the $S^{2-}$ concentration stays suppressed below the precipitation threshold.
View Solution
Part 1: Pure KCN Solution
$KCN$ is a WA-SB salt. It undergoes anionic hydrolysis ($CN^- + H_2O \rightleftharpoons HCN + OH^-$).
$K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{4.0 \times 10^{-10}} = 2.5 \times 10^{-5}$.
$h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{2.5 \times 10^{-5}}{0.1}} = \sqrt{2.5 \times 10^{-4}} = 1.58 \times 10^{-2}$ (or $1.58\%$).
Part 2: Mixing with $NH_4Cl$
When mixed, you have a mixture of a weak acid anion ($CN^-$) and a weak base cation ($NH_4^+$). They will mutually hydrolyze each other in a process called mutual hydrolysis.
Reaction: $NH_4^+ + CN^- \rightleftharpoons NH_3 + HCN$.
The equilibrium constant for this mutual reaction ($K_{mut}$) is much larger than individual $K_h$.
$K_{mut} = \frac{K_w}{K_a \times K_b} = \frac{10^{-14}}{(4.0 \times 10^{-10})(1.8 \times 10^{-5})} = \frac{10^{-14}}{7.2 \times 10^{-15}} = \frac{10}{7.2} = 1.38$.
Because $K_{mut}$ is huge compared to standard hydrolysis constants ($10^{-5}$), the reaction goes forward aggressively. The degree of hydrolysis ($h$) jumps from a mere $1.58\%$ to a massive percentage!
Calculate new $h$:
$K_{mut} = \frac{h^2}{(1-h)^2} \implies \frac{h}{1-h} = \sqrt{1.38} = 1.175$.
$h = 1.175 - 1.175h \implies 2.175h = 1.175 \implies h = 0.54$ (or $54\%$).
Mastering the Math of Acids and Bases
Congratulations on completing these 25 highly advanced numericals on Ionic Equilibrium. By conquering simultaneous solubilities, exact polyprotic derivations, the mathematical nuance of buffer capacity, and the extreme limits of the common ion effect, you have built the ultimate foundation for all of physical chemistry. Remember that in equilibrium, mastering the approximations—knowing exactly when to ignore a tiny '$x$' and when to solve the full quadratic—is the true mark of a JEE Advanced scholar!
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