analytical chemistry laboratory report experiment 2 acid base titration

Objectives:+ Understand how to utilize the acid-base titration technique to determinethe concentration of an unknown acid or base solution.+ Know how to utilize an indicator to determine

SEMESTER II, 2022-2023 INTERNATIONAL UNIVERSITY

VIETNAM NATIONAL UNIVERSITY HCMC

ANALYTICAL CHEMISTRY LABORATORY REPORT

Instructor: MSc Bui Xuan Anh Dao

EXPERIMENT 2:

ACID - BASE TITRATION

Team members:

1 Trần Thị Quỳnh Mai - BTBTIU20266

2 Nguyễn Võ Quỳnh Nhi - BTBTIU20213

3 Dương Hà Trúc Tâm - BTBTUN20013

Grade: _

Comment: _ _ _ _ _

Table of Contents

II. RESULTS

2 Titration of unknown CH 3 COOH with standardized NaOH 8

I INTRODUCTION

1 Objectives:

+ Understand how to utilize the acid-base titration technique to determine the concentration of an unknown acid (or base) solution

+ Know how to utilize an indicator to determine the end point of a titration

2 Abstract:

The acid-base titration experiment is divided into two stages:

- In the first section, you prepare and standardize a robust basic solution

- In the second portion, you will titrate an unknown weak acid with the standardized strong base to ascertain its concentration

→ To achieve great precision in titration, you must be meticulous in your usage and reading of volumetric glassware, as explained in this manual

3 Material:

10 mL pipette + pump [01] Phenolphthalein: 12 - 18 drops

250 mL volumetric flask [01] CH3COOH: 10mL ; unknown

concentration Burette and Ring stand (Burette)

Funnel

4 Procedures:

● Estimate the amount contained of Acetic Acid:

+ Standardize of NaOH by 0.1M KHP (10ml)

+ Titration of unknown Acetic acid with standardized NaOH

Step 1: Prepare standard solution:

+ 0.1M NaOH & keep in plastic bottle

+ KHP 0.1M

+ Indicator phenolphthalein

Step 2: Standardize NaOH by 0.1M KHP

+ Re-standardize of 0.1M NaOH by 10 mL 0.1M KHP

+ Do titration to standardization

+ Triplicate

+ Calculation of NaOH molar concentration

● Obtain the value of an unknown concentration sample:

+ Determine concentration of unknown Acetic acid with standardized NaOH

+ Calculation of NaOH molar concentration

II RESULTS

1 Titration of 0.10 M KHP by NaOH

a Preparation of 100 mL of KHP 0.10 M

MKHP= 204.22

VKHP= 10.0 mL = 0.01 L

C = 0.10 M

nKHP= CKHPx VKHP= 0.10 (M) x 0.01 (L) = 0.001 (mol)

mKHP= nKHPx MKHP= 0.001 (mol) x 204.22 = 0.20422 (g) = 204.22 (mg)

→ Weight of KHP solid (mg) to prepare 100 mL of KHP 0.10 M is 204.22 mg

Report: Determination of the exact concentration of NaOH solution

Initial buret

reading (mL)

Final buret

reading (mL)

Volume of NaOH

used (mL)

Picture

To determine the exact concentration of NaOH solution:

● Calculate the average volume of NaOH used in the titrations for each trial: Trial #1: (9 mL)

Trial #2: (10 mL)

Trial #3: (9 mL)

Average Volume𝑉 = 9+10+93 = 9.33 mL

● Calculate the number of moles of KHP used based on the known molarity and volume used:

Molarity of KHP = CKHP= 0.10 M

Volume of KHP used = VKHP= 10.0 mL = 0.010 L

nKHP= CKHP×VKHP= 0.10 M×0.010 L = 0.001 mol

● Calculate the number of moles of NaOH used based on the volume and molarity of NaOH:

Volume of NaOH used = VNaOH= 9.33 mL = 0.00933 L

Molarity of NaOH = CNaOH(to be determined) = x

nNaOH= CNaOH×VNaOH= x (M)×0.00933 (L) = 0.001 mol

● Solve for the Molarity of NaOH:

Molarity of NaOH = CNaOH=0.001 𝑚𝑜𝑙0.00933𝐿 = 0.107 M

Therefore, the exact concentration of the NaOH solution is 0.107 M

b Standardization of NaOH by 0.1 M KHP

Trial #1:

𝐶𝑁𝑎𝑂𝐻(1)= 𝑛𝑁𝑎𝑂𝐻

𝑉𝑁𝑎𝑂𝐻(1) = 9 𝑚𝐿 ×100.001−3 = 0 111 𝑀

Trial #2:

𝐶𝑁𝑎𝑂𝐻(2)= 𝑛𝑁𝑎𝑂𝐻

10 𝑚𝐿 ×10 −3 = 0 1 𝑀

Trial #3:

𝐶𝑁𝑎𝑂𝐻(3)= 𝑛𝑁𝑎𝑂𝐻

𝑉𝑁𝑎𝑂𝐻(3) = 0.001

9 𝑚𝐿 ×10 −3 = 0 111 𝑀

- The average concentration of NaOH obtained after 3 trials is:

3

∑ 𝐶𝑖

3 0.111+0.1+0.1113

- The standard deviation of NaOH concentration:

𝑁

∑ 𝐶𝑖−𝐶 ( )2

2 +(0.1−0.107) 2 +(0.111−0.107) 2

3−1

The desire molar concentration of NaOH is 0.1M→ µ0= 0 1 𝑀

The null hypothesis𝐻0: µ0= µ→ There is no difference between the standardized and the original NaOH molarity

→ There is significant difference between the standardized and the 𝐻: µ0≠ µ

original NaOH molarity

Using the t-test statistic:

𝑡 = 𝑥−µ0

𝑆/ 𝑁 = 0.107−0.10.006/ 3 = 2 021

At 95% confidence level, tcrit= 4.303 for 2 degree of freedom of 2 tail test

Since 2.021 < 4.303, we can not reject the null hypothesis

Therefore, there is no difference between the standardized and the original NaOH molarity

Hence, we could use the standardized value of NaOH molarity for calculation 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝑚𝑒𝑎𝑛𝑠 ×100% = 0.0060.107×100% = 0 056%

The objective of this lab is to determine the molarity of a NaOH solution by titrating it against a known, standard solution of KHP (potassium hydrogen phthalate)

The standard deviation and coefficient of variance are used to assess the precision and accuracy of the results obtained from the three trials A low standard deviation and coefficient of variance indicate that the results are reproducible and reliable

The color pink of the resulting solution at the endpoint of each trial is due to the presence of phenolphthalein, which is commonly used as an indicator in acid-base titrations Phenolphthalein is colorless in acidic solutions, but turns pink or red in basic solutions In this titration, the initial solution of KHP is acidic, but as NaOH is added, it reacts with the acid to form water and a salt Once all the KHP has reacted with the NaOH, the solution becomes basic, and phenolphthalein turns pink, indicating that the endpoint of the titration has been reached The pink color is an indication that alkalinity has been reached, and that the NaOH solution has reacted with all of the acid in the KHP solution

2 Titration of unknown CH 3 COOH with standardized NaOH

Report : Determination of the concentration of the unknown CH 3 COOH solution

Initial buret

reading (mL)

Final buret

reading (mL)

Volume of NaOH

used (mL)

Picture

From the previous experiment, we have CNaOH= 0.107 M

Trial #1:

nNaOH(1)= nCH3COOH= CNaOH.VNaOH= 0.107×3.5×10-3 = 3.745 10× -4(mol)

𝐶𝐶𝐻3𝐶𝑂𝑂𝐻(1)= 𝑛𝐶𝐻3𝐶𝑂𝑂𝐻

𝑉𝐶𝐻3𝐶𝑂𝑂𝐻 = 3.0 𝑚𝑙×103.745×10−4−3 = 0 125 𝑀

Trial #2:

nNaOH(2)= nCH3COOH= CNaOH.VNaOH= 0.107×3.0×10-3= 3.21 10× -4(mol)

𝐶𝐶𝐻3𝐶𝑂𝑂𝐻(2)= 𝑛𝐶𝐻3𝐶𝑂𝑂𝐻(2)

𝑉𝐶𝐻3𝐶𝑂𝑂𝐻(2) = 3.21×10−4

3.0 𝑚𝑙 ×10 −3 = 0 107 𝑀

Trial #3:

nNaOH(3)= nCH3COOH= CNaOH.VNaOH= 0.107×3.0×10-3= 3.21 10× -4(mol)

𝐶𝐶𝐻3𝐶𝑂𝑂𝐻(3)= 𝑛𝐶𝐻3𝐶𝑂𝑂𝐻(3)

𝑉 = 3.0 𝑚𝑙×103.21 ×10−4−3 = 0 107 𝑀

- The average concentration of CH3COOH obtained after 3 trials is:

CH3COOH= = = 0.113 M

3

∑ 𝐶𝑖

3 0.125 + 0.107 + 0.107 3

- The standard deviation of NaOH concentration:

𝑁

∑ 𝐶𝑖−𝐶 ( )2

2 +(0.107−0.113)2+(0.107−0.113)2 3−1

𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝑚𝑒𝑎𝑛𝑠 ×100% = 0.010

Discussion:

This experiment is aimed to determine the molar concentration of CH3COOH through the titration of 0.107M NaOH obtained from the standardization step

It can be seen from the table of data that the results obtained from 3 times of titration were slightly different from each other with a small standard deviation

of 0.010 Therefore, the error that occurred in this titration part was subtle

At the conclusion of each trial, the resultant solution becomes pink because phenolphthalein, a substance frequently employed as an indicator in acid-base titrations, is present In acidic solutions, phenolphthalein is colorless, but in basic solutions, it takes on a pink or red hue The CH3COOH solution used in this titration starts off acidic, but as NaOH is added, it combines with the acid to produce water and a salt Phenolphthalein becomes pink, signaling that the titration's endpoint has been achieved, when all of the CH3COOH has interacted with the NaOH and the solution has become basic Once alkalinity has been attained and all of the acid in the CH3COOH solution has been interacted with

by the NaOH solution, the solution becomes pink

III CONCLUSION

1 Titration of 0.10 M KHP by NaOH:

A low standard deviation (0.006) and coefficient of variance (0.056%) indicate that the results are reproducible and reliable

Once all the KHP has reacted with the NaOH, the solution becomes basic, and phenolphthalein turns pink, indicating that the endpoint of the titration has been reached

2 Titration of unknown CH 3 COOH with standardized NaOH:

A low standard deviation (0.010) and coefficient of variance (0.088%) indicate that the reproducibility is very high

The CH3COOH mixture utilized in this titration begins to be acidic, however when NaOH is added, it reacts with the acid to form water and a salt When all

of the CH3COOH has reacted with the NaOH and the solution has become basic, phenolphthalein becomes pink, indicating that the titration's endpoint has been reached