Double Titration
Analysis of Mixtures with Two Indicators
1. What is Double Titration?
Double titration involves the titration of a mixture of bases (typically $NaOH, Na_2CO_3, NaHCO_3$) using two different indicators, Phenolphthalein (HPh) and Methyl Orange (MeOH), in sequence or separately, to determine the composition of the mixture.
Why two indicators? The reaction happens in stages with different pH endpoints, which cannot be detected by a single indicator.
2. Working Range of Indicators
A. Phenolphthalein (HPh)
- pH Range: 8.3 - 10.0
- Color Change: Pink (Basic) $\rightarrow$ Colorless (Acidic/Neutral).
- Indicates: Completion of strong base ($NaOH$) neutralization and half neutralization of Carbonate ($CO_3^{2-} \to HCO_3^-$).
B. Methyl Orange (MeOH)
- pH Range: 3.1 - 4.4
- Color Change: Yellow (Basic) $\rightarrow$ Red (Acidic).
- Indicates: Neutralization of strong bases, carbonates, and bicarbonates completely ($HCO_3^- \to H_2CO_3$).
3. Chemical Reactions Involved
When titrating with acid ($HCl$):
1. Complete Neutralization of NaOH:
$$ NaOH + HCl \rightarrow NaCl + H_2O \quad (HPh \text{ & } MeOH \text{ work}) $$2. First Step of Carbonate (Half Neutralization):
$$ Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl \quad (HPh \text{ works}) $$At this point, pH drops to ~8.3, Phenolphthalein turns colorless.
3. Second Step of Carbonate (Complete Neutralization):
$$ NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2 \quad (MeOH \text{ works}) $$At this point, pH drops to ~4, Methyl Orange turns red.
4. Calculation Principles
Let equivalents of components be $Eq$.
With Phenolphthalein:
$Eq_{Acid} = Eq_{NaOH} + \frac{1}{2} Eq_{Na_2CO_3}$
$Eq_{Acid} = Eq_{NaOH} + \frac{1}{2} Eq_{Na_2CO_3}$
With Methyl Orange (Total):
$Eq_{Acid} = Eq_{NaOH} + Eq_{Na_2CO_3} + Eq_{NaHCO_3}$
$Eq_{Acid} = Eq_{NaOH} + Eq_{Na_2CO_3} + Eq_{NaHCO_3}$
Common Scenario: Mixture of $NaOH$ and $Na_2CO_3$
- V1 (HPh endpoint): Reacts all NaOH + 1/2 $Na_2CO_3$.
- V2 (MeOH endpoint from start): Reacts all NaOH + all $Na_2CO_3$.
- Result:
Eq of $Na_2CO_3$ corresponds to $2(V_2 - V_1)$ acid.
Eq of $NaOH$ corresponds to $V_2 - 2(V_2 - V_1)$ acid.
Practice Quiz
Test your logic on Indicator Titrations.
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