direct titration method

direct titration method

Determination of equivalent weight and molecular weight of an organic acid by direct titration method practical with theory, procedure, observation table, calculations, graph, result and precautions for B.Sc chemistry students and laboratory experiments.

Determination of Equivalent Weight and Molecular Weight of an Organic Acid by Direct Titration Method

1. Aim

किसी अज्ञात organic acid sample का equivalent weight तथा molecular weight को standard NaOH solution से direct titration method द्वारा निर्धारित करना।

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2. Theory

Organic acids सामान्यतः monobasic या polybasic होते हैं और ये alkali (NaOH) के साथ neutralization reaction देते हैं।

Reaction:$$RCOOH + NaOH \rightarrow RCOONa + H_2O$$

Neutralization के समय$$N_1 V_1 = N_2 V_2$$

जहाँ

• $N_1$N1​ = acid की normality
• $V_1$V1​ = acid का volume
• $N_2$N2​ = base की normality
• $V_2$V2​ = base का volume

Equivalent weight:$$Equivalent\ Weight = \frac{Weight\ of\ acid}{Equivalent\ of\ acid}$$

और$$Equivalent = N \times V$$

Molecular weight:$$Molecular\ Weight = Equivalent\ Weight \times Basicity$$

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3. Apparatus Required

• Burette
• Pipette
• Conical flask
• Volumetric flask
• Funnel
• Beaker
• Wash bottle

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4. Chemicals Required

• Unknown organic acid sample
• Standard NaOH solution (0.1 N)
• Phenolphthalein indicator
• Distilled water

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5. Procedure

1. लगभग 1 g organic acid sample को weigh करें।
2. इसे 100 ml volumetric flask में घोलकर distilled water से mark तक भर दें।
3. Burette में 0.1 N NaOH भरें।
4. Pipette से 20 ml acid solution लेकर conical flask में डालें।
5. इसमें 2–3 drops phenolphthalein indicator मिलाएँ।
6. Burette से NaOH से titration करें।
7. End point पर colorless से light pink color आ जाता है।
8. तीन बार titration दोहराएँ और average value लें।

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6. Observation Table

Burette Reading

Trial	Initial Reading (ml)	Final Reading (ml)	NaOH Used (ml)
1	0.0	18.5	18.5
2	0.0	18.3	18.3
3	0.0	18.4	18.4

Average burette reading$$= \frac{18.5 + 18.3 + 18.4}{3}$$$$= 18.4\ ml$$

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7. Calculation

Step 1 – Normality of Acid

$$N_1 V_1 = N_2 V_2$$$$N_1 \times 20 = 0.1 \times 18.4$$$$N_1 = \frac{0.1 \times 18.4}{20}$$$$N_1 = 0.092 N$$

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Step 2 – Equivalent Weight

Weight of acid taken = 1 g

Volume prepared = 100 ml

Acid in 20 ml$$= \frac{1}{100} \times 20$$$$= 0.2 g$$

Equivalent$$Equivalent = N \times V$$$$= 0.092 \times 0.02$$$$= 0.00184$$

Equivalent weight$$Equivalent\ Weight = \frac{0.2}{0.00184}$$$$= 108.7$$

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8. Graph

Graph plot किया जाता है

X-axis: Volume of NaOH (ml)
Y-axis: pH of solution

Graph में sharp rise at equivalence point दिखाई देता है।

pH
|
|           /
|         /
|       /
|_____/____________
       Volume of NaOH

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9. Result

Equivalent weight of given organic acid$$= 108.7$$

अगर acid monobasic है$$Molecular\ Weight = 108.7$$

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10. Precautions

1. Burette में air bubble नहीं होना चाहिए।
2. End point पर हल्का pink colour ही लेना चाहिए।
3. Indicator की अधिक मात्रा नहीं डालनी चाहिए।
4. Reading eye level पर लेनी चाहिए।
5. Accurate weighing करना चाहिए।