volumetric.ppt

Author: J R Reid

Volumetric Analysis

IntroductionThe EquipmentThe ProcessCalculations

Introduction

Analisis volumetri merupakan teknik penetapan jumlah sampel melalui perhitungan volume Bahan bahan yang diperlukan:

Zat yang diketahui konsentrasinya yang akan bereaksi dengan senyawa yang akan dicari konsentrasinya (unknown) Indikator yang akan menunjukkan kapan reaksi selesai Alat alat yang dapat mengukur volume secara akurat.

The Equipment

Alat yang dibutuhkan:

Pipette/pipet – mengukur larutan secara tepat (pipet volume/ukur, pipet gondok)

Burette/buret – meneteskan sejumlah reagen cair dalam eksperimen yang memerlukan presisi, seperti pada eksperimen titrasi

Conical flask/ erlenmeyer – mencampurkan larutan

Wash bottles/botol semprot – berisi air suling untuk membersihkan peralatan

Funnel/corong – membantu memasukkan cairan kedalam wadah yang memiliki lubang sempit agar tidak tumpah

Volumetric flasks/labu takar – mengencerkan atau membuat larutan dengan volume tertentu secara akurat. 

The Process - Preparation

Two solutions are used:The solution of unknown concentration;The solution of known concentration – this is also known as the standard solution

Write a balanced equation for the reaction between your two chemicalsClean all glassware to be used with distilled water. The pipettes and burettes will be rinsed with the solutions you are adding to them

Buret dijepit pada clamp stand di atas erlenmeyer

Buret diisi larutan (misalnya larutan standar)

Pipet untuk memindahkan larutan ke erlenmeyer

Tambahkan indikator ke erlenmeyer

Process – The Titration

Baca volume awal larutan dalam buret Buka kran untuk memulai reaksi ke dalam erlenmeyer, goyang erlenmeyer secara kontinyu. Pada saat indikator mulai berubah warna, kecepatan aliran buret mulai diperlambat. Ketika perubahan warna permanen, hentikan aliran dan baca volume akhirnya. Volume total yang dioerlukan disebut titer. Ulangi titrasi, karena anda telah mengetahui volume sebelumnya, maka anda dapat mengulanginya dengan lebih tepat.

Calculations – Mean Titre

We will have a number of titres for each solution we analysed. The first thing we do is to calculate the mean (average) titre:

Titres = 12.6 ml 13.0 ml 13.1 ml 12.9ml

Mean = Sum of the titres / number of titres= (13.0 + 13.1 + 12.9) / 3= 13.0ml

Why did we discard the 12.6 ml reading?

Calculations – The Unknown Concentration – Preparation

1. Write down the balanced equation e.g.H2SO4 + 2NaOH → Na2SO4 + 2H2O

2. Write down everything else we know. This will be:

a. Volume of liquid in the pipetteb. Mean titre (from burette)c. The concentration of the standard solutiond. Was the standard solution in the pipette or in

the burette?

Calculations – The ‘Unknown’ Amount

3. Now calculate the amount in the standard solution you used. Use the n = cv formula. Remember: the millilitres must always be converted into litres for these formulae

4. Now that you know how many moles of the standard you used, look at the balanced equation. Would you need more or less of the ‘unknown’ substance in a balanced reaction?

If more, then how much more – two times, three times?If less, then how much less – half as much, one third?

We can calculate the amount of the unknown:We multiply if we need more i.e. 2x, 3x, …etcWe divide if we need less i.e. ½ = divide by 2, …etc

Calculations – The ‘Unknown’ Concentration

5. Now we have the volume and amount of the ‘unknown’ substance. We can now rearrange our n = cv formula to say c = n/v

Remember: All the calculations must be in litres (not millilitres)The final value must have units (molL-1) written after it

Example:

1. H2SO4 + 2NaOH → Na2SO4 + 2H2O

2. - Standard solution = NaOH (in burette) = 0.1molL-1

- Unknown concentration = H2SO4 (from 20ml pipette)- Titres = 12.6ml, 13.1ml, 13.0, 12.9ml- Average titre = (13.1+13.0+12.9) / 3 = 13.0ml

3. Amount of NaOH = cv = 0.1 x (13/1000) = 0.0013mol

4. Amount of H2SO4 = half of NaOH = 0.0013/2 = 0.00065mol

5. Concentration H2SO4 = n/v = 0.00065/(20/1000) = 0.325molL-1

Titration examples A

Titres: 12.1mL, 12.3mL, 12.1mL, 12.0mL

Known solution details: HCl in the burette, Concentration = 0.522 molL-1

Unknown solution details: NaOH15mL aliquots

Calculations:

HCl + NaOH → NaCl + H2O

Titration examples B

CH3COOH + NaOH → NaCH3COO + H2O

Titres: 17.6mL, 18.5mL, 17.4mL, 17.5mL

Known solution details: NaOH in the burette, Concentration = 0.103 molL-1

Unknown solution details: CH3COOH15mL aliquots

Calculations:

Titration examples C

2HCl + Na2CO3 → 2NaCl + H2O + CO2

Titres: 12.8mL, 12.8mL, 12.8mL, 12.9mL

Known solution details: HCl in the burette, Concentration = 0.555 molL-1

Unknown solution details: Na2CO3 25mL aliquots

Calculations:

Titration examples D

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Titres: 12.1mL, 12.3mL, 12.1mL, 12.0mL

Known solution details: NaOH in the burette, Concentration = 1.04 molL-1

Unknown solution details: H2SO4 10mL aliquots

Calculations:

Level 3 – Spot the difference…

Here is an extract from a level 3 titration assessment. It shows an example of a chemical reaction that could be used in a titration:

Hypochlorite ions react with iodide ions according to the equation; 

OCl + 2I- + 2H+ Cl + I2 + H2O The iodine produced is then titrated with standardised sodium thiosulfate

solution. It reacts according to the equation below. 

I2 + 2S2O32 2I + S4O6

2–

 Since starch turns blue in the presence of iodine, it is used as an indicator

for this final reaction. The overall equation for both reactions is: 

OCl + 2H+ + 2S2O32 Cl + S4O6

2 + H2O