Masterclass: 25 Solved JEE Advanced Numericals on Titrations & Volumetric Analysis
Conquer the Equivalent Concept! This exhaustive guide features highly complex multi-step problems on Double Titrations, Back Titrations, Iodometry, and Complexometry. Click "View Solution" to reveal the step-by-step breakdown.
Volumetric Analysis demands a flawless understanding of the Law of Chemical Equivalence. Before attempting these rigorous problems, ensure you can accurately calculate the n-factor across disproportionation reactions, acidic/basic mediums, and complex multi-salt mixtures.
View Solution
Step 1: Understand Indicator Chemistry
Phenolphthalein (Ph) operates at a higher pH ($\approx 8-10$). It signals the complete neutralization of strong base ($NaOH$) and exactly half of the carbonate ($Na_2CO_3 \rightarrow NaHCO_3$).
Methyl Orange (MeOH) operates at a lower pH ($\approx 3-4$). It signals the complete neutralization of the remaining bicarbonate ($NaHCO_3 \rightarrow CO_2 + H_2O$).
Step 2: Setup Equations
Let equivalents of $NaOH$ be $a$ and total equivalents of $Na_2CO_3$ (for complete neutralization) be $b$.
For Ph endpoint ($V_1 = 30 \text{ mL}$): $a + \frac{b}{2} \propto 30$
For additional MeOH endpoint ($V_2 = 10 \text{ mL}$): This volume is strictly for neutralizing the remaining $NaHCO_3$.
$\frac{b}{2} \propto 10 \implies b \propto 20 \text{ mL}$ of $0.1 \text{ M } HCl$.
Step 3: Solve for 'a'
$a + 10 = 30 \implies a \propto 20 \text{ mL}$ of $0.1 \text{ M } HCl$.
Step 4: Calculate Masses
Equivalents of $NaOH$ ($a$) = $N \times V = 0.1 \text{ N} \times 0.020 \text{ L} = 0.002 \text{ eq}$.
Mass of $NaOH$ = $0.002 \text{ eq} \times 40 \text{ g/eq} = 0.08 \text{ g}$.
Equivalents of $Na_2CO_3$ ($b$) = $N \times V = 0.1 \text{ N} \times 0.020 \text{ L} = 0.002 \text{ eq}$.
Equivalent weight of $Na_2CO_3$ = $106 / 2 = 53 \text{ g/eq}$.
Mass of $Na_2CO_3$ = $0.002 \text{ eq} \times 53 \text{ g/eq} = 0.106 \text{ g}$.
View Solution
Step 1: Phenolphthalein Endpoint
Phenolphthalein indicates neutralization of half of $Na_2CO_3$ ($Na_2CO_3 \rightarrow NaHCO_3$). It does not react with original $NaHCO_3$.
Let molarity of $Na_2CO_3$ be $M_1$. Its n-factor for this half-reaction is $1$.
$M_1 \times 1 \times 20 \text{ mL} = 0.1 \text{ N} \times 8.0 \text{ mL}$
$20 M_1 = 0.8 \implies M_1 = 0.04 \text{ M}$.
Step 2: Methyl Orange Endpoint
Methyl orange indicates complete neutralization of both salts ($Na_2CO_3 \rightarrow CO_2$ and $NaHCO_3 \rightarrow CO_2$).
Let molarity of $NaHCO_3$ be $M_2$. n-factor for $Na_2CO_3$ is $2$, and for $NaHCO_3$ is $1$.
Total equivalents of acid = $(M_1 \times 2 \times 20) + (M_2 \times 1 \times 20)$.
$0.1 \times 25.0 = (0.04 \times 2 \times 20) + 20 M_2$
$2.5 = 1.6 + 20 M_2$
$0.9 = 20 M_2 \implies M_2 = 0.045 \text{ M}$.
View Solution
Step 1: Calculate Total Acid Equivalents
Total milliequivalents (meq) of $HCl$ added = $50.0 \text{ mL} \times 0.50 \text{ N} = 25.0 \text{ meq}$.
Step 2: Calculate Excess Acid (Back Titration)
The excess acid was neutralized by $NaOH$.
meq of excess $HCl$ = meq of $NaOH$ = $20.0 \text{ mL} \times 0.10 \text{ N} = 2.0 \text{ meq}$.
Step 3: Calculate Acid Reacted with Antacid
Reacted $HCl = \text{Total} - \text{Excess} = 25.0 - 2.0 = 23.0 \text{ meq}$.
Therefore, meq of $Mg(OH)_2$ in the sample = $23.0 \text{ meq} = 0.023 \text{ eq}$.
Step 4: Calculate Mass and Percentage
n-factor of $Mg(OH)_2 = 2$. Equivalent weight = $58.3 / 2 = 29.15 \text{ g/eq}$.
Mass of $Mg(OH)_2 = 0.023 \text{ eq} \times 29.15 \text{ g/eq} = 0.67045 \text{ g}$.
$\text{Mass } \% = \left(\frac{0.67045}{1.50}\right) \times 100 = 44.7\%$.
View Solution
Step 1: Calculate Total Moles of $H_2SO_4$
Oleum is a mixture of $H_2SO_4$ and free $SO_3$. Dilution converts all $SO_3$ into $H_2SO_4$.
Equivalents of $NaOH$ used = $45.0 \text{ mL} \times 1.0 \text{ N} = 45.0 \text{ meq} = 0.045 \text{ eq}$.
Total equivalents of $H_2SO_4$ formed = $0.045 \text{ eq}$.
Since $H_2SO_4$ is dibasic ($n=2$), total moles of $H_2SO_4 = 0.045 / 2 = 0.0225 \text{ moles}$.
Step 2: Calculate Mass of Total $H_2SO_4$
Mass = $0.0225 \text{ mol} \times 98 \text{ g/mol} = 2.205 \text{ g}$.
This means $2.0 \text{ g}$ of oleum upon adding water yielded $2.205 \text{ g}$ of $H_2SO_4$.
Step 3: Define Oleum Labeling
Oleum labeling is the mass of pure $H_2SO_4$ obtained from exactly $100 \text{ g}$ of the oleum sample.
If $2.0 \text{ g}$ oleum yields $2.205 \text{ g}$ $H_2SO_4$,
Then $100 \text{ g}$ oleum yields $= \left(\frac{2.205}{2.0}\right) \times 100 = 110.25 \text{ g}$.
View Solution
Step 1: Calculate Total Acid Equivalents
$H_2SO_4$ n-factor = $2$. Normality = $0.4 \text{ N}$.
Total meq of $H_2SO_4 = 50 \text{ mL} \times 0.4 \text{ N} = 20 \text{ meq}$.
Step 2: Calculate Excess Acid (Neutralized by NaOH)
$NaOH$ n-factor = $1$. Normality = $0.5 \text{ N}$.
meq of excess $H_2SO_4$ = meq of $NaOH = 30 \text{ mL} \times 0.5 \text{ N} = 15 \text{ meq}$.
Step 3: Calculate Acid Reacted with Ammonia
Reacted acid = Total - Excess = $20 - 15 = 5 \text{ meq}$.
Therefore, meq of $NH_3$ produced = $5 \text{ meq}$.
Since $NH_3$ has an n-factor of $1$, millimoles of $N$ atoms = $5 \text{ mmol}$.
Step 4: Mass and Percentage of Nitrogen
Mass of N = $5 \times 10^{-3} \text{ mol} \times 14 \text{ g/mol} = 0.07 \text{ g}$.
$\% \text{ N} = \left(\frac{0.07}{0.25}\right) \times 100 = 28\%$.
View Solution
Step 1: Understand Iodometric Reactions
Reaction 1: $2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2$
Reaction 2: $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$
Step 2: Apply the Law of Equivalence
Equivalents of $Cu^{2+}$ = Equivalents of $I_2$ liberated = Equivalents of $Na_2S_2O_3$ used.
n-factor for $Cu^{2+} \rightarrow Cu^+$ is $1$.
n-factor for $S_2O_3^{2-} \rightarrow 0.5 S_4O_6^{2-}$ is $1$.
meq of $Na_2S_2O_3 = 20.0 \text{ mL} \times 0.10 \text{ N} = 2.0 \text{ meq}$.
Therefore, meq of $CuSO_4 = 2.0 \text{ meq}$ in $25.0 \text{ mL}$.
Step 3: Calculate Molarity and Strength
Since n-factor = 1, Molarity = Normality.
$M = \frac{2.0 \text{ meq}}{25.0 \text{ mL}} = 0.08 \text{ M}$.
Strength per Liter = $M \times \text{Molar Mass} = 0.08 \text{ mol/L} \times 249.5 \text{ g/mol} = 19.96 \text{ g/L}$.
View Solution
Step 1: Identify the Redox Change
Reaction: $C_6H_8O_6 + I_2 \rightarrow C_6H_6O_6 + 2HI$.
Ascorbic acid loses 2 Hydrogen atoms (2 electrons). Thus, n-factor of Ascorbic Acid = $2$.
Iodine ($I_2$) goes from $0$ to $-1$ (2 atoms). n-factor of $I_2 = 2$.
Step 2: Calculate Equivalents of Iodine
Normality of $I_2 = M \times n = 0.05 \times 2 = 0.10 \text{ N}$.
meq of $I_2 = 25.0 \text{ mL} \times 0.10 \text{ N} = 2.5 \text{ meq} = 0.0025 \text{ eq}$.
Step 3: Calculate Mass of Ascorbic Acid
Equivalents of Ascorbic acid = $0.0025 \text{ eq}$.
Equivalent weight = $176 / 2 = 88 \text{ g/eq}$.
Mass = $0.0025 \text{ eq} \times 88 \text{ g/eq} = 0.22 \text{ g}$.
Step 4: Percentage Purity
$\% = (0.22 / 1.00) \times 100 = 22.0\%$.
View Solution
Step 1: Determine n-factor of $FeC_2O_4$
In $FeC_2O_4$, both $Fe^{2+}$ and $C_2O_4^{2-}$ are oxidized simultaneously.
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
Total electron loss per molecule = $1 + 2 = 3$. So, n-factor = $3$.
Step 2: Determine n-factor of $KMnO_4$
In acidic medium, $MnO_4^- \rightarrow Mn^{2+}$ (change of $5e^-$). n-factor = $5$.
Step 3: Law of Equivalence
Equivalents of $FeC_2O_4$ = Equivalents of $KMnO_4$
$(\text{Moles} \times n)_1 = (\text{Molarity} \times n \times \text{Volume in L})_2$
$0.05 \times 3 = 0.1 \times 5 \times V$
$0.15 = 0.5 \times V \implies V = \frac{0.15}{0.5} = 0.3 \text{ L} = 300 \text{ mL}$.
View Solution
Step 1: Analyze Mohr's Salt
Only the $Fe^{2+}$ ion oxidizes to $Fe^{3+}$. The ammonium and sulfate ions are spectator ions in this redox reaction.
n-factor of Mohr's salt = $1$.
Moles of Mohr's salt = $3.92 / 392 = 0.01 \text{ moles}$.
Equivalents of Mohr's salt = $0.01 \times 1 = 0.01 \text{ eq}$.
Step 2: Analyze Dichromate
$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$. Oxidation state changes from $+6$ to $+3$ for two atoms.
n-factor of $K_2Cr_2O_7$ = $6$.
Step 3: Law of Equivalence
Equivalents of $K_2Cr_2O_7$ = Equivalents of Mohr's salt = $0.01 \text{ eq}$.
$N \times V(\text{L}) = 0.01$
$N \times 0.020 = 0.01 \implies N = 0.5 \text{ N}$.
Molarity = Normality / n-factor = $0.5 / 6 = 0.0833 \text{ M}$.
View Solution
Step 1: Write the chemical equation
With hot, concentrated base, halogens disproportionate into halide and halate ions.
$3I_2 + 6OH^- \rightarrow 5I^- + IO_3^- + 3H_2O$
Step 2: Calculate individual n-factors for $I_2$
Reduction: $I_2^0 \rightarrow 2I^{-1}$. Change = $|0 - (-1)| \times 2 = 2$. ($n_{red} = 2$).
Oxidation: $I_2^0 \rightarrow 2I^{+5}$. Change = $|0 - 5| \times 2 = 10$. ($n_{ox} = 10$).
Step 3: Calculate the total n-factor
Using the disproportionation formula:
$n_{total} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}} = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3}$.
Step 4: Calculate Equivalent Weight
$E = \frac{M}{n_{total}} = \frac{M}{5/3} = \frac{3M}{5}$.
View Solution
Step 1: Understand EDTA Stoichiometry
EDTA forms a 1:1 complex with $Ca^{2+}$ and $Mg^{2+}$ ions regardless of their charge or the EDTA ionization state. Thus, moles of EDTA = moles of $M^{2+}$.
Moles of EDTA used = $0.01 \text{ M} \times 0.015 \text{ L} = 1.5 \times 10^{-4} \text{ moles}$.
Total moles of $Ca^{2+}/Mg^{2+}$ in $100 \text{ mL}$ = $1.5 \times 10^{-4} \text{ moles}$.
Step 2: Convert to $CaCO_3$ Equivalents
To express hardness, we treat all the metal ions as if they were $CaCO_3$.
Molar mass of $CaCO_3 = 100 \text{ g/mol}$.
Mass of $CaCO_3$ equivalent = $1.5 \times 10^{-4} \text{ moles} \times 100 \text{ g/mol} = 1.5 \times 10^{-2} \text{ g} = 15 \text{ mg}$.
Step 3: Calculate ppm (mg/L)
We have $15 \text{ mg}$ of hardness in $100 \text{ mL}$ ($0.100 \text{ L}$) of water.
Hardness = $\frac{15 \text{ mg}}{0.100 \text{ L}} = 150 \text{ mg/L} = 150 \text{ ppm}$.
View Solution
Step 1: The Chemical Reaction
$Ag^+ + Cl^- \rightarrow AgCl \downarrow$ (white).
At the endpoint, excess $Ag^+$ reacts with the indicator: $2Ag^+ + CrO_4^{2-} \rightarrow Ag_2CrO_4 \downarrow$ (reddish-brown).
Step 2: Calculate Moles of $Cl^-$
Moles of $AgNO_3$ used = $0.1 \text{ M} \times 0.080 \text{ L} = 0.008 \text{ moles}$.
Due to 1:1 stoichiometry, moles of pure $NaCl$ = $0.008 \text{ moles}$.
Step 3: Calculate Mass of Pure $NaCl$
Molar mass of $NaCl$ = $58.5 \text{ g/mol}$.
Mass = $0.008 \text{ mol} \times 58.5 \text{ g/mol} = 0.468 \text{ g}$.
Step 4: Percentage Purity
$\% \text{ Purity} = \left(\frac{0.468}{0.585}\right) \times 100 = 80\%$.
View Solution
Step 1: Total $Ag^+$ Added
Initial millimoles of $Ag^+ = 50.0 \times 0.1 = 5.0 \text{ mmol}$.
Step 2: Excess $Ag^+$ (Back Titrated)
Reaction: $Ag^+ + SCN^- \rightarrow AgSCN \downarrow$
Millimoles of $SCN^-$ used = $20.0 \times 0.05 = 1.0 \text{ mmol}$.
Excess $Ag^+ = 1.0 \text{ mmol}$.
Step 3: $Ag^+$ Reacted with Chloride
Reacted $Ag^+ = \text{Total} - \text{Excess} = 5.0 - 1.0 = 4.0 \text{ mmol}$.
Therefore, original $Cl^- = 4.0 \text{ mmol}$.
Step 4: Calculate Mass of Chloride
Molar mass of $Cl = 35.5 \text{ g/mol} = 35.5 \text{ mg/mmol}$.
Mass = $4.0 \text{ mmol} \times 35.5 \text{ mg/mmol} = 142 \text{ mg}$.
View Solution
Step 1: Before Equivalence Point
Excess $Cl^-$ ions are in solution. The precipitated $AgCl$ particles adsorb $Cl^-$ ions onto their surface, creating a primary negatively charged layer ($AgCl \cdot Cl^-$). The negatively charged Fluorescein anion ($Ind^-$) is repelled by this surface.
Step 2: At/After Equivalence Point
The first tiny excess drop of $AgNO_3$ adds excess $Ag^+$ to the solution. The $AgCl$ particles now adsorb $Ag^+$ ions, creating a primary positively charged layer ($AgCl \cdot Ag^+$).
Step 3: The Color Change
This positively charged surface strongly attracts and adsorbs the negatively charged Fluorescein indicator anions ($Ind^-$) into the secondary layer. The physical deformation of the indicator's electron cloud upon adsorption onto the crystal lattice drastically alters its light absorption properties, turning the precipitate a brilliant pinkish-red.
View Solution
Step 1: Calculate Normality of $KMnO_4$
In acidic medium, n-factor of $KMnO_4 = 5$.
Normality = $0.05 \text{ M} \times 5 = 0.25 \text{ N}$.
Step 2: Calculate Normality of $H_2O_2$
$N_1 V_1 = N_2 V_2$
$N \times 20 = 0.25 \times 40 \implies 20 N = 10 \implies N = 0.5 \text{ N}$.
n-factor of $H_2O_2 = 2$. So Molarity = $0.5 / 2 = 0.25 \text{ M}$.
Step 3: Convert Molarity to Volume Strength
Reaction: $2H_2O_2 \rightarrow 2H_2O + O_2$.
$2 \text{ moles}$ $H_2O_2$ yield $22.4 \text{ L}$ $O_2$ at STP. So $1 \text{ mole}$ yields $11.2 \text{ L}$.
Volume Strength = Molarity $\times 11.2$
Volume Strength = $0.25 \times 11.2 = 2.8 \text{ V}$.
View Solution
Step 1: Analyze First Titration (Only $Fe^{2+}$ reacts)
$Fe^{3+}$ cannot be oxidized further. The $KMnO_4$ reacts only with original $Fe^{2+}$.
$KMnO_4$ Normality = $0.02 \times 5 = 0.1 \text{ N}$.
Equivalents of $Fe^{2+}$ = $0.1 \text{ N} \times 0.020 \text{ L} = 0.002 \text{ eq}$.
Since n-factor of $Fe^{2+} = 1$, moles of $Fe^{2+} = 0.002 \text{ moles}$ in $50 \text{ mL}$.
Molarity of $Fe^{2+} = 0.002 / 0.050 = 0.04 \text{ M}$.
Step 2: Analyze Second Titration (Total Iron)
Zinc reduces $Fe^{3+}$ to $Fe^{2+}$. Now, ALL iron is in the $Fe^{2+}$ state.
Equivalents of Total Iron = $0.1 \text{ N} \times 0.050 \text{ L} = 0.005 \text{ eq}$.
Total moles of Iron in $50 \text{ mL} = 0.005 \text{ moles}$.
Step 3: Calculate $Fe^{3+}$
Moles of original $Fe^{3+} = \text{Total Moles} - \text{Original } Fe^{2+}$
Moles of $Fe^{3+} = 0.005 - 0.002 = 0.003 \text{ moles}$ in $50 \text{ mL}$.
Molarity of $Fe^{3+} = 0.003 / 0.050 = 0.06 \text{ M}$.
View Solution
Step 1: Identify Oxidation State Changes
In $As_2S_3$, Arsenic is $+3$ and Sulfur is $-2$.
In $H_3AsO_4$, Arsenic is $+5$.
In $H_2SO_4$, Sulfur is $+6$.
Step 2: Calculate electron loss per atom
$As^{3+} \rightarrow As^{5+}$ (Loss of $2 e^-$ per As atom).
$S^{2-} \rightarrow S^{6+}$ (Loss of $8 e^-$ per S atom).
Step 3: Sum for the Entire Molecule
One molecule of $As_2S_3$ contains 2 As atoms and 3 S atoms.
Total loss from As = $2 \times 2 = 4 e^-$.
Total loss from S = $3 \times 8 = 24 e^-$.
Total electrons lost per molecule = $4 + 24 = 28$.
n-factor = $28$.
Step 4: Equivalent Weight
$E = \frac{M}{28}$.
View Solution
Step 1: The Chemistry of the Neutral Medium
In neutral or faintly alkaline media, $KMnO_4$ is reduced to solid $MnO_2$ ($+7 \rightarrow +4$). Thus, n-factor of $KMnO_4 = 3$.
Thiosulfate ($S_2O_3^{2-}$) is oxidized completely to Sulfate ($SO_4^{2-}$), not tetrathionate! (Tetrathionate forms with weak oxidants like $I_2$).
In $S_2O_3^{2-}$, average S is $+2$. In $SO_4^{2-}$, S is $+6$.
Change per S atom = $4$. Total for 2 atoms = $8$. n-factor of Thiosulfate = $8$.
Step 2: Apply Equivalence
$N_1 V_1 = N_2 V_2$
$(M_{\text{thio}} \times 8) \times 100 = (0.1 \times 3) \times 40$
$800 M_{\text{thio}} = 12$
$M_{\text{thio}} = \frac{12}{800} = 0.015 \text{ M}$.
View Solution
Step 1: Calculate Hypo Equivalents
n-factor of Hypo is 1. Normality = $0.05 \text{ N}$.
meq of Hypo = $40 \text{ mL} \times 0.05 \text{ N} = 2.0 \text{ meq}$.
Therefore, meq of $Cl_2$ liberated in $100 \text{ mL}$ = $2.0 \text{ meq}$.
Step 2: Scale up to Total Volume
Total meq of $Cl_2$ in $1.0 \text{ L}$ ($1000 \text{ mL}$) = $2.0 \times 10 = 20 \text{ meq} = 0.020 \text{ eq}$.
Step 3: Calculate Mass of Available Chlorine
Equivalent weight of $Cl_2$ gas = Molar mass / 2 = $71 / 2 = 35.5 \text{ g/eq}$.
Mass of $Cl_2 = 0.020 \text{ eq} \times 35.5 \text{ g/eq} = 0.71 \text{ g}$.
Step 4: Percentage Available Chlorine
$\% = \left(\frac{0.71}{7.1}\right) \times 100 = 10\%$.
View Solution
Step 1: Identify Equivalence Point Volume
$M_1 V_1 = M_2 V_2 \implies 0.2 \times 50 = 0.2 \times V_2 \implies V_2 = 50 \text{ mL}$.
Total volume at equivalence = $50 + 50 = 100 \text{ mL}$.
Step 2: Calculate Salt Concentration
Moles of $CH_3COONa$ formed = $0.2 \text{ M} \times 0.050 \text{ L} = 0.01 \text{ moles}$.
Concentration $C = 0.01 \text{ mol} / 0.100 \text{ L} = 0.1 \text{ M}$.
Step 3: Apply Salt Hydrolysis Formula
$CH_3COONa$ is a WA-SB salt. The solution is basic.
$pH = 7 + \frac{1}{2}(pK_a + \log_{10} C)$
$pK_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$
$\log_{10}(0.1) = -1$
$pH = 7 + \frac{1}{2}(4.74 - 1) = 7 + \frac{1}{2}(3.74) = 7 + 1.87 = 8.87$.
View Solution
Step 1: Analyze the 30% Neutralization Point
At $30\%$ neutralization, the ratio of Salt formed to unreacted Acid is $30:70$ (or $3:7$).
Apply Henderson-Hasselbalch: $pH = pK_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
$5.2 = pK_a + \log_{10}(3/7)$
$\log_{10}(3) \approx 0.477$, $\log_{10}(7) \approx 0.845$.
$\log_{10}(3/7) = 0.477 - 0.845 = -0.368$.
$5.2 = pK_a - 0.368 \implies pK_a = 5.2 + 0.368 = 5.568$.
Step 2: Analyze the 50% Neutralization Point
At exactly $50\%$ neutralization (the half-equivalence point), $[\text{Salt}] = [\text{Acid}]$.
Therefore, $\log_{10} \frac{[\text{Salt}]}{[\text{Acid}]} = \log_{10}(1) = 0$.
At this specific point, $pH = pK_a$.
View Solution
Step 1: The First Endpoint ($V_1$)
Methyl orange (pH $\approx 4$) signals the neutralization of the first acidic proton.
$H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$.
Here, 1 mole of base neutralizes 1 mole of acid. The volume used is $V_1$.
Step 2: The Second Endpoint ($V_2$)
Phenolphthalein (pH $\approx 9$) signals the neutralization of the second acidic proton.
$NaH_2PO_4 + NaOH \rightarrow Na_2HPO_4 + H_2O$.
The starting material for this step is $NaH_2PO_4$. Since all $H_3PO_4$ became $NaH_2PO_4$, the moles are identical. It requires exactly 1 mole of base to remove the second proton.
Step 3: Conclusion
Because the stoichiometry is 1:1 for both individual proton removals, the volume required for the first step must exactly equal the additional volume required for the second step.
Note: The third proton is so weakly acidic that it cannot be titrated directly in aqueous solution without highly specialized methods.
View Solution
Step 1: The Equivalent Chain
The beauty of the Winkler method is the equivalent chain: $O_2$ oxidizes $Mn^{2+}$ to $Mn^{4+}$. Acidification makes $Mn^{4+}$ oxidize $I^-$ to $I_2$. The $I_2$ is titrated with Hypo.
Equivalents of $O_2$ = Equivalents of Hypo used.
Step 2: Calculate Equivalents of Hypo
n-factor of Hypo is $1$.
meq of Hypo = $20 \text{ mL} \times 0.01 \text{ N} = 0.2 \text{ meq}$.
Step 3: Calculate Mass of Oxygen
n-factor of $O_2$ ($0 \rightarrow -2$ for two atoms) = $4$.
Equivalent weight of $O_2 = 32 / 4 = 8 \text{ g/eq} = 8 \text{ mg/meq}$.
Mass of $O_2$ = $0.2 \text{ meq} \times 8 \text{ mg/meq} = 1.6 \text{ mg}$.
Step 4: Calculate ppm
We have $1.6 \text{ mg}$ of $O_2$ in $500 \text{ mL}$ ($0.5 \text{ L}$) of water.
ppm = mg / L = $1.6 / 0.5 = 3.2 \text{ mg/L}$.
View Solution
Step 1: The Titration Reaction
$Ag^+ + SCN^- \rightarrow AgSCN \downarrow$ (White precipitate).
At the endpoint, excess $SCN^-$ reacts with $Fe^{3+}$ to form the blood-red $[Fe(SCN)]^{2+}$ complex.
Step 2: Calculate Moles of Silver
Because the stoichiometry is 1:1, moles of $Ag^+$ = moles of $KSCN$.
Moles of $KSCN$ = $0.1 \text{ M} \times 0.400 \text{ L} = 0.04 \text{ moles}$.
Therefore, moles of pure Silver ($Ag$) in the coin = $0.04 \text{ moles}$.
Step 3: Calculate Mass and Percentage
Mass of pure $Ag = 0.04 \text{ mol} \times 108 \text{ g/mol} = 4.32 \text{ g}$.
$\% \text{ Purity} = \left(\frac{4.32}{5.0}\right) \times 100 = 86.4\%$.
View Solution
Step 1: Acid-Base Titration ($NaOH$)
Only the Oxalic acid reacts with $NaOH$. Sodium oxalate is already a neutral salt regarding basic titration.
$H_2C_2O_4$ is dibasic (n-factor = 2).
Let Molarity of $H_2C_2O_4 = x$.
$(x \times 2) \times 10 = 0.1 \times 1 \times 20 \implies 20x = 2 \implies x = 0.1 \text{ M}$.
Step 2: Redox Titration ($KMnO_4$)
Both Oxalic acid and Sodium oxalate contain the oxalate ion ($C_2O_4^{2-}$), which oxidizes to $CO_2$. The n-factor for both molecules in redox is $2$.
Let Molarity of $Na_2C_2O_4 = y$.
Total equivalents of oxalate = Equivalents of $KMnO_4$.
$KMnO_4$ n-factor in acid = $5$.
$[(x \times 2) + (y \times 2)] \times 10 = 0.1 \times 5 \times 16$
$20x + 20y = 8$.
Step 3: Solve for y
We know $x = 0.1$.
$20(0.1) + 20y = 8 \implies 2 + 20y = 8 \implies 20y = 6 \implies y = 0.3 \text{ M}$.
Step 4: Calculate the Ratio
Ratio of Moles (or Molarities) of $H_2C_2O_4 : Na_2C_2O_4 = x : y = 0.1 : 0.3 = 1 : 3$.
Mastering the Erlenmeyer Flask
Congratulations on conquering these 25 highly advanced numericals on Titrations and Volumetric Analysis. This chapter represents the absolute synthesis of stoichiometry, redox chemistry, and ionic equilibrium. By mastering the distinction between an acid-base n-factor and a redox n-factor for the exact same molecule (like Oxalic acid), and learning how to chain equivalence through complex back-titrations and Iodometry, you have built the ultimate physical chemistry toolkit. Keep track of your milliequivalents, and visit Chemca.in for more elite masterclasses!
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