Exam Master Review Sheet - Titration & Volumetric Analysis (JEE & NEET)

CHEMCA

EXAM MASTER FORMULA SHEET

Physical Chemistry: Volumetric Analysis

Titrations, Redox Concepts & Quantitative Analysis for JEE/NEET

1 Fundamental Volumetric Laws & n-factor

The Law of Chemical Equivalence:

In any chemical reaction, substances react in the ratio of their equivalent weights. At the equivalence point:

\[ N_1 V_1 = N_2 V_2 \]

\[ \text{Equivalents}_1 = \text{Equivalents}_2 \]

Concentration Relations

Normality (N) = Molarity (M) × n-factor

Equivalent Wt (E) = Molar Mass (M) / n-factor

No. of Equivalents = Weight / E = N × V(L)

Calculation of n-factor

Acids: Basicity (number of replaceable $H^+$).
e.g., $H_3PO_4 (n=3), H_3PO_3 (n=2)$
Bases: Acidity (number of replaceable $OH^-$).
Salts: Total positive or negative charge.
e.g., $Al_2(SO_4)_3 (n=6)$
Redox: Total change in oxidation number per molecule.

2 Acid-Base Titration & Indicators

Indicators are weak organic acids or bases whose ionized and unionized forms have different colors (Ostwald's Theory). Color change occurs around $pH = pK_{In} \pm 1$.

Type of Titration	pH at Eq. Point	Suitable Indicator	Indicator pH Range
Strong Acid vs Strong Base	7	Any (Phenolphthalein or Methyl Orange)	4.0 to 10.0 (Steep curve)
Weak Acid vs Strong Base	> 7 (Basic)	Phenolphthalein (HPh)	8.2 - 10.0
Strong Acid vs Weak Base	< 7 (Acidic)	Methyl Orange (MeOH)	3.1 - 4.4
Weak Acid vs Weak Base	Depends on $K_a/K_b$	No suitable indicator	-

3 Double Titration (JEE Advanced Special)

Analysis of alkali mixtures (e.g., $NaOH + Na_2CO_3$ or $Na_2CO_3 + NaHCO_3$) titrated against standard $HCl$ using two indicators successively.

Phenolphthalein End Point ($V_1$)

HPh changes color around pH 8.2. At this point:

$NaOH + HCl \rightarrow NaCl + H_2O$ (Complete)
$Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl$ (Half neutralization)

Eq of HCl used = Eq of NaOH + $\frac{1}{2}$ Eq of $Na_2CO_3$

Methyl Orange End Point ($V_2$)

MeOH changes color around pH 3.1. At this point, total alkalinity is neutralized:

$NaHCO_3 \text{ (formed)} + HCl \rightarrow NaCl + H_2O + CO_2$
$NaHCO_3 \text{ (original)} + HCl \rightarrow NaCl + H_2O + CO_2$

Eq of HCl used (Total) = Eq of NaOH + Eq of $Na_2CO_3$ + Eq of $NaHCO_3$

4 Redox Titrations: Permanganate & Dichromate

Potassium Permanganate ($KMnO_4$)

Self-Indicator

Acts as a powerful oxidizing agent. Its equivalent weight changes depending on the medium.

Acidic Medium ($H_2SO_4$)

$MnO_4^- \xrightarrow{+5e^-} Mn^{2+}$

n-factor = 5 $E = M/5$

Neutral / Faintly Basic

$MnO_4^- \xrightarrow{+3e^-} MnO_2 \downarrow$

n-factor = 3 $E = M/3$

Strongly Alkaline

$MnO_4^- \xrightarrow{+1e^-} MnO_4^{2-}$

n-factor = 1 $E = M/1$

Note: $HCl$ is not used as a medium because $KMnO_4$ oxidizes $Cl^-$ to $Cl_2$, consuming the titrant and causing error.

Potassium Dichromate ($K_2Cr_2O_7$)

Used as a primary standard (unlike $KMnO_4$). It acts as an oxidizing agent only in acidic medium.

$Cr_2O_7^{2-} + 14H^+ + \mathbf{6e^-} \longrightarrow 2Cr^{3+} + 7H_2O$

n-factor = 6 ($E = M/6$)

5 Iodometry vs Iodimetry (JEE Adv)

Iodimetry (Direct)

Direct titration of standard $I_2$ solution against reducing agents (like Hypo, $H_2S$, $Sn^{2+}$).

$I_2 + 2Na_2S_2O_3 \longrightarrow 2NaI + Na_2S_4O_6$

Indicator: Starch (added near the end point). Deep blue color disappears at equivalence point.

Iodometry (Indirect)

An oxidizing agent (like $Cu^{2+}, H_2O_2, Cr_2O_7^{2-}$) is reacted with excess $KI$ to liberate $I_2$. The liberated $I_2$ is then titrated against standard Hypo ($Na_2S_2O_3$).

Step 1: $2Cu^{2+} + 4I^- \longrightarrow Cu_2I_2 \downarrow + \mathbf{I_2}$

Step 2: $\mathbf{I_2} + 2S_2O_3^{2-} \longrightarrow 2I^- + S_4O_6^{2-}$

Eq of Oxidizing Agent = Eq of $I_2$ liberated = Eq of Hypo used

6 Special Quantitative Calculations

Volume Strength of $H_2O_2$

"10V $H_2O_2$" means 1L of this solution yields 10L of $O_2$ at STP.

Molarity ($M$) = $\frac{\text{Volume Strength}}{11.2}$
Normality ($N$) = $\frac{\text{Volume Strength}}{5.6}$
% Strength ($w/v$) = $N \times 1.7 = \frac{17 \times \text{Vol. Str.}}{56}$

Hardness of Water

Expressed in terms of ppm of $CaCO_3$ equivalent.

$\text{ppm} = \frac{\text{Mass of } CaCO_3 \text{ eq.}}{\text{Total mass of water}} \times 10^6$

$Eq. Wt \text{ of } CaCO_3 = 100/2 = 50$

Kjeldahl's Method (% Nitrogen Estimation)

Organic compound is heated with $H_2SO_4$ to form $(NH_4)_2SO_4$, then boiled with $NaOH$ to liberate $NH_3$, which is absorbed in standard acid.

$\% N = \frac{1.4 \times N \times V}{W}$

Where: $N$ = Normality of standard acid, $V$ = Volume of acid neutralized by $NH_3$ (in mL), $W$ = Mass of organic compound (in grams).

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Chemca Formula Sheet - Titration & Volumetric Analysis