ChemCA Educational Portal
The Exhaustive Guide to Double Titration
Welcome to the ultimate masterclass on Double Titration (Mixture Analysis). In the JEE Advanced examination, simple acid-base titrations are rarely tested. Instead, examiners construct elaborate scenarios involving mixtures of bases with varying basic strengths. To solve these, you must understand the exact $pH$ at which different indicators operate and how polyprotic species neutralize in distinct, sequential stages.
1. The Concept of Double Titration
Double titration is an analytical technique used to determine the exact composition of a mixture containing two different bases (or two different acids) in a single solution. The most classically tested mixtures in the JEE syllabus involve combinations of strong bases like Sodium Hydroxide ($NaOH$), and salts of weak diprotic acids like Sodium Carbonate ($Na_2CO_3$) and Sodium Bicarbonate ($NaHCO_3$).
Why can't we just use a single titration? Because $NaOH$, $Na_2CO_3$, and $NaHCO_3$ do not all neutralize at the same $pH$.
- $NaOH$ is a strong base. It neutralizes completely at a relatively high $pH$.
- $Na_2CO_3$ contains the carbonate ion ($CO_3^{2-}$), which is a strong conjugate base of a very weak acid ($HCO_3^-$). It accepts its first proton easily at a high $pH$.
- $NaHCO_3$ contains the bicarbonate ion ($HCO_3^-$), which is a much weaker base. It requires a highly acidic environment (low $pH$) to accept a proton and convert into $H_2O$ and $CO_2$.
To differentiate between these components, we utilize two different indicators that change color at different $pH$ thresholds. By measuring the volume of standard acid required to reach the first indicator's endpoint, and then the additional volume required to reach the second indicator's endpoint, we can algebraically solve for the amount of each component in the mixture.
2. The Chemistry of Indicators: Phenolphthalein and Methyl Orange
An acid-base indicator is typically a weak organic acid (let's call it $HIn$) that exhibits a different color than its conjugate base ($In^-$). $$ HIn (Color A) \rightleftharpoons H^+ + In^- (Color B) $$
The human eye can detect a color change when the ratio $[In^-]/[HIn]$ is approximately $10:1$ or $1:10$. According to the Henderson-Hasselbalch equation ($pH = pK_a + \log\frac{[In^-]}{[HIn]}$), the visual transition range of an indicator is approximately $pH = pK_{a(indicator)} \pm 1$.
Phenolphthalein (Ph)
- Type: Weak organic acid
- $pH$ Working Range: $8.2 - 10.0$
- Color Change: Pink (Basic) $\rightarrow$ Colorless (Acidic)
- What it detects: Complete neutralization of strong bases ($NaOH$).
- Crucial Fact: It detects only the half-neutralization of $Na_2CO_3$. At $pH \approx 8.3$, $CO_3^{2-}$ has been converted to $HCO_3^-$. Phenolphthalein turns colorless here. It cannot detect the neutralization of $NaHCO_3$.
Methyl Orange (MeOH)
- Type: Weak organic base
- $pH$ Working Range: $3.1 - 4.4$
- Color Change: Yellow (Basic) $\rightarrow$ Red (Acidic)
- What it detects: Complete neutralization of all bases.
- Crucial Fact: It triggers at a very acidic $pH$. By the time the solution reaches $pH \approx 4$, strong bases are completely neutralized, $Na_2CO_3$ is fully neutralized to $CO_2 + H_2O$, and $NaHCO_3$ is also fully neutralized.
3. Mixture 1: $NaOH$ and $Na_2CO_3$
Let's titrate a mixture of $NaOH$ and $Na_2CO_3$ with a standard strong acid, say $HCl$. Let the molarity of $HCl$ be $M$.
Stage 1: Titration with Phenolphthalein
We add a few drops of Phenolphthalein to the mixture. The solution is highly basic, so it's deep pink. We start adding $HCl$ from the burette. Because $NaOH$ is a much stronger base than $Na_2CO_3$, the $HCl$ preferentially reacts with $NaOH$ first.
Once all $NaOH$ is consumed, the $HCl$ starts reacting with $Na_2CO_3$. Because Phenolphthalein changes color around $pH 8.3$, it will signal an endpoint the moment $CO_3^{2-}$ is converted into $HCO_3^-$.
Let the volume of $HCl$ required to make Phenolphthalein colorless be $v_1 \text{ mL}$.
Equivalents of $HCl$ ($v_1$) = Eq of $NaOH$ + $\frac{1}{2}$ Eq of $Na_2CO_3$
Stage 2: Continuing with Methyl Orange
The solution is now colorless. It contains $NaCl$ (neutral) and the newly formed $NaHCO_3$ (weakly basic). We now add a few drops of Methyl Orange to this exact same flask. The solution turns yellow.
We continue adding $HCl$ from the burette. The $HCl$ now reacts with the $NaHCO_3$ that was formed in Stage 1.
Let the *additional* volume of $HCl$ required from this point to make Methyl Orange turn red be $v_2 \text{ mL}$.
This $v_2$ volume of acid was solely used to neutralize the $NaHCO_3$ that came entirely from the half-neutralized $Na_2CO_3$.
Equivalents of $HCl$ ($v_2$) = $\frac{1}{2}$ Eq of $Na_2CO_3$
The Algebraic Resolution:
From the above two equations, we can isolate the components:
- Volume of acid required for complete $Na_2CO_3$ = $2 \times v_2$
- Volume of acid required for complete $NaOH$ = $v_1 - v_2$
4. Mixture 2: $Na_2CO_3$ and $NaHCO_3$
This is a mixture of a moderate base and a weak base. Let's apply the same continuous titration logic.
Stage 1: With Phenolphthalein
When $HCl$ is added, it will react with the stronger base, $Na_2CO_3$, first. Phenolphthalein will change color as soon as $Na_2CO_3$ is converted to $NaHCO_3$. The original $NaHCO_3$ present in the mixture remains entirely untouched during this stage because the $pH$ hasn't dropped low enough for it to react while $CO_3^{2-}$ is still present.
Let the volume of $HCl$ for the Phenolphthalein endpoint be $v_1 \text{ mL}$.
Equivalents of $HCl$ ($v_1$) = $\frac{1}{2}$ Eq of $Na_2CO_3$
Stage 2: Continuing with Methyl Orange
Now, we add Methyl Orange. The flask currently contains:
1. The $NaHCO_3$ that was newly formed from the $Na_2CO_3$.
2. The original $NaHCO_3$ that was initially present in the mixture.
We continue adding $HCl$. The acid will now neutralize ALL of the bicarbonate present.
Let the *additional* volume of $HCl$ required for the Methyl Orange endpoint be $v_2 \text{ mL}$.
Equivalents of $HCl$ ($v_2$) = $\frac{1}{2}$ Eq of $Na_2CO_3$ (newly formed) + Eq of original $NaHCO_3$
The Algebraic Resolution:
- Volume of acid required for complete $Na_2CO_3$ = $2 \times v_1$
- Volume of acid required for original $NaHCO_3$ = $v_2 - v_1$
5. The Impossible Mixture: $NaOH$ and $NaHCO_3$
A Classic JEE Trap
You will never be asked to analyze a mixture of $NaOH$ and $NaHCO_3$ in a single solution. Why? Because they undergo a spontaneous acid-base reaction with each other immediately upon mixing in water!
$NaOH + NaHCO_3 \rightarrow Na_2CO_3 + H_2O$
If you mix them, they react until the limiting reagent is exhausted. The resulting solution will either be a mixture of $NaOH$ + $Na_2CO_3$ (if $NaOH$ was in excess) or $Na_2CO_3$ + $NaHCO_3$ (if $NaHCO_3$ was in excess). It will never be a mixture of $NaOH$ and $NaHCO_3$.
6. Analytical Variations: Continuous vs. Separate Aliquots
In numerical problems, the examiner can present the data in two fundamentally different ways. Reading the question carefully is paramount.
Method A: Continuous Titration (Same Flask)
The mixture is taken in one flask. Phenolphthalein is added, and acid is run in until colorless (Volume $v_1$). Without changing the flask, Methyl Orange is added, and more acid is run in from the burette (Additional Volume $v_2$).
This is the method we discussed in Sections 3 and 4. The formulas derived there apply directly.
Method B: Separate Aliquots (Two Flasks)
The original mixture solution is divided into two equal parts (aliquots).
Flask 1: Titrated with Phenolphthalein using $V_x$ volume of acid.
Flask 2: Titrated from scratch using Methyl Orange using $V_y$ volume of acid.
Logic for Flask 2 (Methyl Orange from the start):
Because Methyl Orange's endpoint is at $pH \sim 4$, it does not care about halfway points. By the time it turns red, ALL bases in the flask ($NaOH$, $Na_2CO_3$, and $NaHCO_3$) have been completely neutralized to their lowest forms.
- For $NaOH + Na_2CO_3$ mixture: $V_y \equiv \text{Eq}(NaOH) + \text{Eq}(Na_2CO_3)$
- For $Na_2CO_3 + NaHCO_3$ mixture: $V_y \equiv \text{Eq}(Na_2CO_3) + \text{Eq}(NaHCO_3)$
7. 10 Advanced JEE Level Solved Problems
Problem 1: Continuous NaOH/Na2CO3
A solution contains a mixture of $NaOH$ and $Na_2CO_3$. $25 \text{ mL}$ of this mixture requires $20 \text{ mL}$ of $0.1 \text{ M}$ $HCl$ for titration when phenolphthalein is used as an indicator.
To the same resulting solution, a few drops of methyl orange are added. It now requires an additional $5 \text{ mL}$ of the same $0.1 \text{ M}$ $HCl$ to reach the red endpoint.
Calculate the mass of $NaOH$ and $Na_2CO_3$ in exactly $1 \text{ Liter}$ of the original mixture. (Molar mass: $NaOH = 40$, $Na_2CO_3 = 106$)
Reveal Detailed Solution
Step 1: Identify the method and variables.
This is a continuous titration on the same flask.
$v_1$ (volume for Ph endpoint) = $20 \text{ mL}$.
$v_2$ (additional volume for MeOH endpoint) = $5 \text{ mL}$.
Molarity of $HCl$ = $0.1 \text{ M}$ (Normality = $0.1 \text{ N}$ since n-factor=1).
Step 2: Analyze $Na_2CO_3$.
From theory, $v_2$ corresponds to the neutralization of $NaHCO_3$ formed from $Na_2CO_3$.
So, $\frac{1}{2}$ Eq of $Na_2CO_3 \equiv v_2$ volume of acid.
Volume of acid for complete $Na_2CO_3$ = $2 \times v_2 = 2 \times 5 = 10 \text{ mL}$.
Milli-equivalents (meq) of $Na_2CO_3$ in $25 \text{ mL}$ aliquot = $N_{\text{acid}} \times V_{\text{acid for } Na_2CO_3}$
meq of $Na_2CO_3 = 0.1 \times 10 = 1.0 \text{ meq}$.
Step 3: Analyze $NaOH$.
Volume of acid for complete $NaOH = v_1 - v_2 = 20 - 5 = 15 \text{ mL}$.
meq of $NaOH$ in $25 \text{ mL}$ aliquot = $0.1 \times 15 = 1.5 \text{ meq}$.
Step 4: Scale up to 1 Liter ($1000 \text{ mL}$).
The $25 \text{ mL}$ aliquot is $1/40$th of a Liter ($1000 / 25 = 40$).
Total meq of $Na_2CO_3$ in $1 \text{ L} = 1.0 \text{ meq} \times 40 = 40 \text{ meq} = 0.040 \text{ eq}$.
Total meq of $NaOH$ in $1 \text{ L} = 1.5 \text{ meq} \times 40 = 60 \text{ meq} = 0.060 \text{ eq}$.
Step 5: Calculate Masses.
Equivalent weight of $NaOH = 40 / 1 = 40 \text{ g/eq}$.
Mass of $NaOH$ in $1 \text{ L} = 0.060 \times 40 = \mathbf{2.4 \text{ g}}$.
Equivalent weight of $Na_2CO_3 = 106 / 2 = 53 \text{ g/eq}$.
Mass of $Na_2CO_3$ in $1 \text{ L} = 0.040 \times 53 = \mathbf{2.12 \text{ g}}$.
Answers: NaOH = 2.4 g, Na2CO3 = 2.12 g
Problem 2: Separate Aliquots Method
A $500 \text{ mL}$ solution contains a mixture of $NaOH$ and $Na_2CO_3$.
Two equal aliquots of $50 \text{ mL}$ each are taken in separate flasks (Flask A and Flask B).
Flask A is titrated with $0.2 \text{ M}$ $HCl$ using phenolphthalein, requiring $30 \text{ mL}$.
Flask B is titrated with the same $0.2 \text{ M}$ $HCl$ but using methyl orange from the very beginning, requiring $40 \text{ mL}$.
Calculate the molarity of $NaOH$ and $Na_2CO_3$ in the original mixture.
Reveal Detailed Solution
Step 1: Understand Flask A (Phenolphthalein).
Ph indicates complete $NaOH$ and half $Na_2CO_3$.
meq of acid = $30 \times 0.2 = 6.0 \text{ meq}$.
Equation 1: $\text{meq}(NaOH) + \frac{1}{2}\text{meq}(Na_2CO_3) = 6.0$
Step 2: Understand Flask B (Methyl Orange).
MeOH from the start indicates complete neutralization of EVERYTHING.
meq of acid = $40 \times 0.2 = 8.0 \text{ meq}$.
Equation 2: $\text{meq}(NaOH) + \text{meq}(Na_2CO_3) = 8.0$
Step 3: Solve the simultaneous equations.
Subtract Eq 1 from Eq 2:
$\frac{1}{2}\text{meq}(Na_2CO_3) = 8.0 - 6.0 = 2.0 \text{ meq}$.
Total meq of $Na_2CO_3$ in $50 \text{ mL} = 4.0 \text{ meq}$.
Substitute back into Eq 2:
$\text{meq}(NaOH) + 4.0 = 8.0 \implies \text{meq}(NaOH) = 4.0 \text{ meq}$.
Step 4: Calculate Molarity.
These meq are present in a $50 \text{ mL}$ aliquot.
Molarity = Moles / Volume(L) = millimoles / Volume(mL).
For $NaOH$: n-factor = 1. So meq = mmol.
mmol of $NaOH = 4.0$.
Molarity of $NaOH$ = $4.0 \text{ mmol} / 50 \text{ mL} = \mathbf{0.08 \text{ M}}$.
For $Na_2CO_3$: n-factor = 2. So mmol = meq / 2.
mmol of $Na_2CO_3 = 4.0 / 2 = 2.0 \text{ mmol}$.
Molarity of $Na_2CO_3$ = $2.0 \text{ mmol} / 50 \text{ mL} = \mathbf{0.04 \text{ M}}$.
Answers: [NaOH] = 0.08 M, [Na2CO3] = 0.04 M
Problem 3: The Na2CO3 / NaHCO3 Duo
A solution contains a mixture of $Na_2CO_3$ and $NaHCO_3$. $20 \text{ mL}$ of this solution requires $12 \text{ mL}$ of $0.1 \text{ N}$ $HCl$ for the phenolphthalein endpoint.
After the phenolphthalein endpoint is reached, methyl orange is added to the same flask, and an additional $25 \text{ mL}$ of $0.1 \text{ N}$ $HCl$ is required to reach the red endpoint.
Calculate the amount of $Na_2CO_3$ and $NaHCO_3$ in grams per liter of the solution.
Reveal Detailed Solution
Step 1: Analyze the Continuous Titration logic for this specific mixture.
Stage 1 ($v_1 = 12 \text{ mL}$): Only half of $Na_2CO_3$ reacts.
$\frac{1}{2}$ Eq of $Na_2CO_3 \equiv 12 \text{ mL}$ of acid.
Therefore, volume of acid for complete $Na_2CO_3 = 2 \times 12 = 24 \text{ mL}$.
Stage 2 ($v_2 = 25 \text{ mL}$): Both the newly formed $NaHCO_3$ (from $Na_2CO_3$) and original $NaHCO_3$ react.
Volume for newly formed $NaHCO_3$ = Volume of Stage 1 = $12 \text{ mL}$.
Volume for original $NaHCO_3$ = Total Stage 2 Volume - Volume for newly formed.
Volume for original $NaHCO_3 = 25 - 12 = 13 \text{ mL}$.
Step 2: Calculate milli-equivalents in 20 mL aliquot.
Normality of $HCl = 0.1 \text{ N}$.
meq of $Na_2CO_3 = 0.1 \times 24 = 2.4 \text{ meq}$.
meq of $NaHCO_3 = 0.1 \times 13 = 1.3 \text{ meq}$.
Step 3: Scale to 1 Liter and calculate mass.
Scaling factor = $1000 \text{ mL} / 20 \text{ mL} = 50$.
Total meq in $1 \text{ L}$:
$Na_2CO_3 = 2.4 \times 50 = 120 \text{ meq} = 0.12 \text{ eq}$.
$NaHCO_3 = 1.3 \times 50 = 65 \text{ meq} = 0.065 \text{ eq}$.
Equivalent Weights:
$Na_2CO_3$ (n-factor=2) = $106 / 2 = 53 \text{ g/eq}$.
$NaHCO_3$ (n-factor=1) = $84 / 1 = 84 \text{ g/eq}$.
Mass in 1 Liter (Strength):
$Na_2CO_3 = 0.12 \times 53 = \mathbf{6.36 \text{ g/L}}$.
$NaHCO_3 = 0.065 \times 84 = \mathbf{5.46 \text{ g/L}}$.
Answers: Na2CO3 = 6.36 g/L, NaHCO3 = 5.46 g/L
Problem 4: Percentage Composition with Inert Material
A $2.0 \text{ g}$ sample of a solid mixture containing $NaOH$, $Na_2CO_3$, and inert impurities is dissolved in water and made up to $250 \text{ mL}$.
$25 \text{ mL}$ of this solution is titrated with $0.1 \text{ M}$ $H_2SO_4$. Using phenolphthalein, the endpoint comes at $15 \text{ mL}$. Using methyl orange from the start in a separate $25 \text{ mL}$ aliquot, the endpoint comes at $20 \text{ mL}$.
Calculate the mass percentage of $NaOH$ and $Na_2CO_3$ in the original solid mixture.
Reveal Detailed Solution
Step 1: Check the Titrant!
The titrant is $0.1 \text{ M}$ $H_2SO_4$. Since it's dibasic, Normality ($N$) = $0.1 \times 2 = 0.2 \text{ N}$.
Step 2: Setup Aliquot Equations ($25 \text{ mL}$ samples).
Aliquot 1 (Ph): $\text{meq}(NaOH) + \frac{1}{2}\text{meq}(Na_2CO_3) = V_{Ph} \times N = 15 \times 0.2 = 3.0 \text{ meq}$.
Aliquot 2 (MeOH): $\text{meq}(NaOH) + \text{meq}(Na_2CO_3) = V_{MeOH} \times N = 20 \times 0.2 = 4.0 \text{ meq}$.
Step 3: Solve for meq in 25 mL.
Subtract Eq 1 from Eq 2:
$\frac{1}{2}\text{meq}(Na_2CO_3) = 4.0 - 3.0 = 1.0 \text{ meq} \implies \text{meq}(Na_2CO_3) = 2.0 \text{ meq}$.
Substitute back: $\text{meq}(NaOH) + 1.0 = 3.0 \implies \text{meq}(NaOH) = 2.0 \text{ meq}$.
Step 4: Scale to Total Volume ($250 \text{ mL}$).
Multiplier = $250 / 25 = 10$.
Total meq of $Na_2CO_3 = 2.0 \times 10 = 20 \text{ meq} = 0.02 \text{ eq}$.
Total meq of $NaOH = 2.0 \times 10 = 20 \text{ meq} = 0.02 \text{ eq}$.
Step 5: Calculate Mass and Percentage.
Mass of $Na_2CO_3 = 0.02 \text{ eq} \times 53 \text{ g/eq} = 1.06 \text{ g}$.
Mass of $NaOH = 0.02 \text{ eq} \times 40 \text{ g/eq} = 0.80 \text{ g}$.
Total sample mass = $2.0 \text{ g}$.
$\% Na_2CO_3 = (1.06 / 2.0) \times 100 = \mathbf{53\%}$.
$\% NaOH = (0.80 / 2.0) \times 100 = \mathbf{40\%}$.
(The remaining 7% is the inert impurity).
Answers: Na2CO3 = 53%, NaOH = 40%
Problem 5: Reverse Engineering Volumes
A solid mixture weighing $1.22 \text{ g}$ containing equimolar amounts of $Na_2CO_3$ and $NaHCO_3$ is dissolved in water. It requires $V_1 \text{ mL}$ of $0.1 \text{ M}$ $HCl$ for the phenolphthalein endpoint. When methyl orange is added after the first endpoint, an additional $V_2 \text{ mL}$ of the same acid is required.
Calculate the exact values of $V_1$ and $V_2$. (Molar masses: $Na_2CO_3 = 106$, $NaHCO_3 = 84$)
Reveal Detailed Solution
Step 1: Use the "equimolar" condition to find moles.
Let moles of $Na_2CO_3$ = $x$, then moles of $NaHCO_3$ = $x$.
Mass = $(x \times 106) + (x \times 84) = 1.22 \text{ g}$
$190x = 1.22 \implies x = 1.22 / 190 = 0.00642 \text{ moles} = 6.42 \text{ millimoles}$.
So, we have $6.42 \text{ mmol}$ of $Na_2CO_3$ and $6.42 \text{ mmol}$ of $NaHCO_3$.
meq of $Na_2CO_3$ (n-factor 2) = $6.42 \times 2 = 12.84 \text{ meq}$.
meq of $NaHCO_3$ (n-factor 1) = $6.42 \times 1 = 6.42 \text{ meq}$.
Step 2: Simulate the Phenolphthalein Titration ($V_1$).
Only half of $Na_2CO_3$ reacts. Original $NaHCO_3$ doesn't react.
meq of acid = $\frac{1}{2} \times \text{meq}(Na_2CO_3) = \frac{1}{2} \times 12.84 = 6.42 \text{ meq}$.
$N \times V_1 = 6.42 \implies 0.1 \times V_1 = 6.42 \implies \mathbf{V_1 = 64.2 \text{ mL}}$.
Step 3: Simulate the additional Methyl Orange Titration ($V_2$).
The remaining half of $Na_2CO_3$ (now as $NaHCO_3$) reacts + the original $NaHCO_3$ reacts.
meq of acid = $\frac{1}{2} \times \text{meq}(Na_2CO_3) + \text{meq}(\text{original } NaHCO_3)$.
meq of acid = $6.42 + 6.42 = 12.84 \text{ meq}$.
$N \times V_2 = 12.84 \implies 0.1 \times V_2 = 12.84 \implies \mathbf{V_2 = 128.4 \text{ mL}}$.
Answers: V1 = 64.2 mL, V2 = 128.4 mL
Problem 6: Water Alkalinity Analysis
Alkalinity of water is due to $OH^-$, $CO_3^{2-}$, and $HCO_3^-$. A $100 \text{ mL}$ sample of hard water requires $10.0 \text{ mL}$ of $0.02 \text{ N}$ $H_2SO_4$ to reach the phenolphthalein endpoint. Another $100 \text{ mL}$ aliquot of the same water requires $30.0 \text{ mL}$ of the same acid to reach the methyl orange endpoint.
Identify the species causing alkalinity and calculate their concentration in mg/L (ppm). Assume only two compatible species can exist together.
Reveal Detailed Solution
Step 1: Identify compatible species based on volumes.
Let $P$ = volume for Phenolphthalein endpoint = $10.0 \text{ mL}$.
Let $M$ = volume for Methyl Orange endpoint (from start) = $30.0 \text{ mL}$.
Analysis of relationships:
If $P = M$, only $OH^-$ is present.
If $P = M/2$, only $CO_3^{2-}$ is present.
If $P = 0$, only $HCO_3^-$ is present.
Here, $P = 10$, $M = 30$. So, $P < M/2$ (since $10 < 15$).
This condition implies a mixture of Carbonate and Bicarbonate. ($OH^-$ and $HCO_3^-$ are incompatible).
Step 2: Calculate equivalents for each species.
For $CO_3^{2-}$, the volume of acid required for complete neutralization = $2 \times P = 20.0 \text{ mL}$.
For $HCO_3^-$, the volume of acid required = $M - 2P = 30.0 - 20.0 = 10.0 \text{ mL}$.
Step 3: Calculate concentrations in mg/L.
For Carbonate ($CO_3^{2-}$):
meq = $N \times V = 0.02 \times 20.0 = 0.4 \text{ meq}$.
Eq. weight of $CO_3^{2-}$ (n-factor 2) = $60 / 2 = 30 \text{ mg/meq}$.
Mass in $100 \text{ mL} = 0.4 \times 30 = 12 \text{ mg}$.
Concentration = $12 \text{ mg} / 0.1 \text{ L} = \mathbf{120 \text{ mg/L (ppm)}}$.
For Bicarbonate ($HCO_3^-$):
meq = $N \times V = 0.02 \times 10.0 = 0.2 \text{ meq}$.
Eq. weight of $HCO_3^-$ (n-factor 1) = $61 / 1 = 61 \text{ mg/meq}$.
Mass in $100 \text{ mL} = 0.2 \times 61 = 12.2 \text{ mg}$.
Concentration = $12.2 \text{ mg} / 0.1 \text{ L} = \mathbf{122 \text{ mg/L (ppm)}}$.
Answers: CO3(2-) = 120 ppm, HCO3(-) = 122 ppm
Problem 7: The Thermal Decomposition Trap
A $5.0 \text{ g}$ mixture of $NaHCO_3$ and $Na_2CO_3$ is strongly heated in a crucible until no further weight loss occurs. The residue weighs $3.45 \text{ g}$.
The residue is then dissolved in water to make $100 \text{ mL}$ solution. What volume of $0.5 \text{ N}$ $HCl$ will be required to completely neutralize $20 \text{ mL}$ of this solution using methyl orange indicator?
Reveal Detailed Solution
Step 1: Analyze the chemistry of heating.
When heated, $Na_2CO_3$ is stable and does not decompose.
However, $NaHCO_3$ decomposes: $2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) \uparrow + CO_2(g) \uparrow$
The weight loss is entirely due to the escape of $H_2O$ and $CO_2$.
Step 2: Understand the Residue.
After heating, the residue is 100% pure $Na_2CO_3$. It consists of the original $Na_2CO_3$ plus the new $Na_2CO_3$ formed from the decomposition of bicarbonate.
We don't actually need to calculate the initial composition to answer the final question! This is a classic JEE trap to waste your time.
The total mass of the residue = $3.45 \text{ g}$ of pure $Na_2CO_3$.
Step 3: Titration of the Residue.
This $3.45 \text{ g}$ is dissolved in $100 \text{ mL}$.
Molar mass of $Na_2CO_3 = 106$. Equivalent weight = $53$.
Total equivalents of $Na_2CO_3 = 3.45 / 53 = 0.06509 \text{ eq} = 65.09 \text{ meq}$.
We take a $20 \text{ mL}$ aliquot (which is $1/5$th of the $100 \text{ mL}$ solution).
meq of $Na_2CO_3$ in $20 \text{ mL} = 65.09 / 5 = 13.018 \text{ meq}$.
Methyl orange indicates complete neutralization.
meq of Acid = meq of Base
$0.5 \times V = 13.018 \implies V = \mathbf{26.036 \text{ mL}}$.
Answer: 26.04 mL
Problem 8: Potassium Salts Mixture
A $4.0 \text{ g}$ mixture of $KOH$, $K_2CO_3$, and $KCl$ was dissolved in water to make $250 \text{ mL}$ solution.
A $50 \text{ mL}$ portion required $25 \text{ mL}$ of $0.1 \text{ M}$ $HCl$ for the phenolphthalein endpoint.
Another $50 \text{ mL}$ portion required $35 \text{ mL}$ of the same $HCl$ for the methyl orange endpoint from the start.
Calculate the mass percentage of each component. (Molar masses: $KOH = 56$, $K_2CO_3 = 138$, $KCl = 74.5$)
Reveal Detailed Solution
Step 1: Set up the equations for Separate Aliquots.
$KCl$ is a neutral salt; it doesn't react with $HCl$.
Aliquot 1 (Ph): $\text{meq}(KOH) + \frac{1}{2}\text{meq}(K_2CO_3) = 25 \times 0.1 = 2.5 \text{ meq}$.
Aliquot 2 (MeOH): $\text{meq}(KOH) + \text{meq}(K_2CO_3) = 35 \times 0.1 = 3.5 \text{ meq}$.
Step 2: Solve for meq in 50 mL.
Subtracting: $\frac{1}{2}\text{meq}(K_2CO_3) = 1.0 \implies \text{meq}(K_2CO_3) = 2.0 \text{ meq}$.
Substituting back: $\text{meq}(KOH) + 1.0 = 2.5 \implies \text{meq}(KOH) = 1.5 \text{ meq}$.
Step 3: Scale to 250 mL and find masses.
Multiplier = $250 / 50 = 5$.
Total meq of $K_2CO_3 = 2.0 \times 5 = 10 \text{ meq} = 0.01 \text{ eq}$.
Total meq of $KOH = 1.5 \times 5 = 7.5 \text{ meq} = 0.0075 \text{ eq}$.
Eq weight of $K_2CO_3 = 138 / 2 = 69 \text{ g/eq}$. Mass = $0.01 \times 69 = 0.69 \text{ g}$.
Eq weight of $KOH = 56 / 1 = 56 \text{ g/eq}$. Mass = $0.0075 \times 56 = 0.42 \text{ g}$.
Step 4: Find KCl and Percentages.
Mass of $KCl = \text{Total Mass} - (Mass_{K_2CO_3} + Mass_{KOH}) = 4.0 - (0.69 + 0.42) = 4.0 - 1.11 = 2.89 \text{ g}$.
$\% K_2CO_3 = (0.69 / 4.0) \times 100 = \mathbf{17.25\%}$.
$\% KOH = (0.42 / 4.0) \times 100 = \mathbf{10.5\%}$.
$\% KCl = (2.89 / 4.0) \times 100 = \mathbf{72.25\%}$.
Answers: K2CO3 = 17.25%, KOH = 10.5%, KCl = 72.25%
Problem 9: Molarity and Mole Fractions
A $1 \text{ L}$ solution contains a mixture of $Na_2CO_3$ and $NaHCO_3$. $10 \text{ mL}$ of this solution requires $10 \text{ mL}$ of $0.1 \text{ M}$ $H_2SO_4$ to decolorize phenolphthalein. Upon further addition of methyl orange and continuing the titration, $20 \text{ mL}$ more of the same acid is required to turn the solution red.
Calculate the mole fraction of $Na_2CO_3$ with respect to total basic solutes in the mixture.
Reveal Detailed Solution
Step 1: Check the Acid Normality.
$H_2SO_4$ is $0.1 \text{ M}$, so Normality = $0.2 \text{ N}$.
Step 2: Continuous Titration Logic.
Stage 1 ($v_1 = 10 \text{ mL}$): $\frac{1}{2}\text{meq}(Na_2CO_3) = 10 \times 0.2 = 2.0 \text{ meq}$.
Therefore, total meq of $Na_2CO_3$ in $10 \text{ mL} = 4.0 \text{ meq}$.
Stage 2 ($v_2 = 20 \text{ mL}$ additional): This neutralizes the newly formed $NaHCO_3$ + original $NaHCO_3$.
Total meq of acid in Stage 2 = $20 \times 0.2 = 4.0 \text{ meq}$.
We know meq of newly formed $NaHCO_3$ = $\frac{1}{2}\text{meq}(Na_2CO_3)$ = $2.0 \text{ meq}$.
Therefore, meq of original $NaHCO_3 = 4.0 - 2.0 = 2.0 \text{ meq}$.
Step 3: Convert meq to Millimoles.
$Na_2CO_3$ (n-factor 2): mmol = meq / 2 = $4.0 / 2 = 2.0 \text{ mmol}$.
$NaHCO_3$ (n-factor 1): mmol = meq / 1 = $2.0 / 1 = 2.0 \text{ mmol}$.
Step 4: Calculate Mole Fraction.
The solution contains equimolar amounts of $Na_2CO_3$ and $NaHCO_3$ ($2.0 \text{ mmol}$ each).
Mole fraction of $Na_2CO_3$ = $\frac{Moles_{Na2CO3}}{Moles_{Na2CO3} + Moles_{NaHCO3}}$
$X_{Na_2CO_3} = \frac{2.0}{2.0 + 2.0} = \frac{2}{4} = \mathbf{0.5}$.
Answer: 0.5
Problem 10: The Ultimate Mixture Trap
A chemist tries to prepare a mixture by dissolving $2.0 \text{ g}$ of $NaOH$ and $4.2 \text{ g}$ of $NaHCO_3$ in enough water to make $1 \text{ L}$ of solution.
He then takes $100 \text{ mL}$ of this solution and titrates it with $0.1 \text{ M}$ $HCl$ using phenolphthalein. What volume of $HCl$ will be required?
Reveal Detailed Solution
Step 1: Recognize the incompatibility.
As discussed in Section 5, $NaOH$ and $NaHCO_3$ react immediately:
$NaOH + NaHCO_3 \rightarrow Na_2CO_3 + H_2O$
Step 2: Calculate initial moles.
Moles of $NaOH$ initial = $2.0 / 40 = 0.05 \text{ moles}$.
Moles of $NaHCO_3$ initial = $4.2 / 84 = 0.05 \text{ moles}$.
Step 3: Analyze the Limiting Reagent reaction.
They are present in exact 1:1 stoichiometric ratio. Both are completely consumed.
Moles of $Na_2CO_3$ formed = $0.05 \text{ moles}$.
So, the $1 \text{ L}$ flask actually just contains $0.05 \text{ moles}$ of pure $Na_2CO_3$.
Step 4: Analyze the Titration.
We take $100 \text{ mL}$ aliquot ($1/10$th of the solution).
Moles of $Na_2CO_3$ in aliquot = $0.05 / 10 = 0.005 \text{ moles} = 5 \text{ mmol}$.
meq of $Na_2CO_3 = 5 \text{ mmol} \times 2 = 10 \text{ meq}$.
The titration is done with phenolphthalein.
With Ph, $Na_2CO_3$ only undergoes half neutralization.
meq of acid required = $\frac{1}{2}\text{meq}(Na_2CO_3) = 10 / 2 = 5 \text{ meq}$.
Molarity of $HCl = 0.1 \text{ M}$ ($0.1 \text{ N}$).
$0.1 \times V = 5 \implies V = \mathbf{50 \text{ mL}}$.
Answer: 50 mL
© 2026 ChemCA Educational Portal. For the most elite JEE Advanced Preparation.
No comments:
Post a Comment