Experiment 3 Acid and Base Titration

EXPERIMENT 3: ACID AND BASE TITRATION TITRATION Objectives:

To determine the concentration of sodium sodiu m hydroxide solution through titration technique using hydrochloric acid and sulfuric acid Concepts:

1. To determined determined the concentrati concentration on of acid and base base solution solution through titrat titration ion with standard standard solution. 2. To apply apply the the correct correct technique technique in in titrat titration. ion. 3. To carry out out acid base base titration titration using using phenolphtha phenolphthalein lein as indicat indicator or.. Introduction:

Acid base titration involves a neutralization reaction in which an acid is react with an equivalent amount of base. For the neutralization of hydrochloric acid with sodium hydroxide:

HCl + NaOH→ H2O + NaCl

 Neutralization occurs when acid and bases exist in comparable stoichiometry, stoichiometry, for instance the amount of the hydrochloric acid (mole) is equivalent with the amount of sodium hydroxide (mole). The endpoint of titration can be determined using indicator. Apparatus:

Volumetric olumetr ic flask

250 mL

Filter funnel Erlenmeyer flask  Beaker

1 250 mL

Burette

1 50 mL

Pipette

1 25 mL

Chemical Reagent a) 100 mL 1.000 × 10-2 M HCl solution b) 100 mL 1.000 × 10-2 M H2SO4 solution

c) 10 mL C solution solution containing containing NaOH NaOH (with (with pipette) pipette) d) Phenol Phenolpht phthal halien ien soluti solution on Methods: 1. 10 mL of C solution are put in volumetric flask, dilutes with distilled water to the mark and

mix thoroughly. Transfer Transfer the solution to the clean beaker beake r. Labelled solution as C. Rinse a flask with water twice. 2. Burettes are clean and rinse with 5 mL HCl solution (1.000×10-2 M) twice. Place 25 mL of 

HCl solution (1.000 × 10-2) to burette using funnel. 3. Pipettes are clean and rinse twice using C solution. Pipette 25 mL of C solution in three

Erlenmeyer flask. Adds 2 of phenolphthalein indicator. 4. The initial volumes are recorded to the nearest two decimal point. Titrate C solution with HCl

from the burette to colourless solution end point. Record the final volume reading and calculate the used acid volume.  Note: phenolphthalein colour will change from magenta (base) to colourless (acid) 5. The titration titration are are repeated repeated until the the different different volume of acid is in in the range of 0.30 for three three experiments. 6. Concentration of NaOH solution (that was in flask) and the concentration of C are calculated. 7. Steps 3 and 7 are repeated by replace HCl with H2SO4.

Result: Titration between HCl and NaOH 1. Volume of HCl need for the titration:

Trial 1: 1.4 mL Trial 2: 1.0 mL Trial 3: 1.0 mL

2. Concent Concentrat ration ion of NaOH NaOH by usin using g each trial trial volum volume: e: Trial 1: 0.0056M Trial 2: 0.004 M Trial 3: 0.004 M Average verage of of conce concent ntra rati tion on NaO NaOH H solu soluti tion on

= 0.005 0.0056 6 + 0.00 0.004 4 + 0.004 0.004 3 = 0.0045 M

Titration between H 2SO4 and NaOH 1. Volume of H2SO4 need for the titration:

Trial 1: 1.1 mL Trial 2: 0.6 mL Trial 3: 0.7 mL 2. Concentration Concentration of NaOH by using each trial volume: volume: Trial 1: 0.0088M Trial 2: 0.0048 M Trial 3: 0.0056 M Average verage of concent concentrat ration ion NaOH NaOH soluti solution on

= 0.0088 0.0088 + 0.0048 0.0048 + 0.0056 0.0056 3 = 0.0064 M

Data Analysis 1. Chemical equation for the reaction between hydrochloric acid and sodium hydroxide

HCl + NaOH →NaCl + H2O Trial 1:

n =MV n HCl = 0.1 × 1.4 = 0.14 mol

From the equation 1 mol HCl need 1 mol NaOH to form 1 mol NaCl and 1 mol of water. If the mol of HCl is 0.14M, so the mole of NaOH also 0.14 M Concentration of NaOH, M

=n/V = 0.14/ 25 = 0.0056 M

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0056 56 - 0.00 0.0045 45  100 0.0045 = 24.44

Trial 2:

n =MV n HCl = 0.1 × 1.0 = 0.10 mol From the equation 1 mol HCl need 1 mol NaOH to form 1 mol NaCl and 1 mol of water. If the mole of HCl is 0.10 mol, so the mol of NaOH is also 0.10 mol. Concentration of NaOH, M

=n/V = 0.10/ 25 = 0.0040 Molar 

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0040 40 - 0.00 0.0045 45  100 0.0045 = 11.11

Trial 3:

n =MV n HCl = 0.1 × 1.0 = 0.10 mol From the equation 1 mol HCl need 1 mol NaOH to form 1 mol NaCl and 1 mol of water. If the mole of HCl is 0.10M, so the mol of NaOH also 0.10 M

Concentration of NaOH, M

=n/V = 0.10/ 25 = 0.0040 Molar 

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0040 40 - 0.00 0.0045 45  100 0.0045 = 11.11

2. Chemical equation for the reaction between Sulfuric acid and Sodium hydroxide.

H2SO4 + 2NaOH → Na2SO4 + 2H2O Trial 1:

n =MV n H2SO4 = 0.1 × 1.1 = 0.11 M From the equation, 1 mole H2SO4 need 2 mole of the NaOH to produce 1 mole of Na2SO4 and 2 mole of water. If got 0.11 mole of H2SO4, the reaction needed 0.22 mole of NaOH. Concentration of NaOH, M

=n∕V = 0.22/ 25 = 0.0088 M

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0088 88 - 0.00 0.0064 64  100 0.0064 = 37.5

Trial 2:

n =MV n H2SO4 = 0.1 × 0.6 = 0.06 mole

From the equation, 1 mole H2SO4 need 2 mole of the NaOH to produce 1 mole of Na2SO4 and 2 mole of water.

If got 0.06 mole of H2SO4, the reaction needed 0.12 mole of NaOH. Conc Concen entr trat atio ion n of NaOH NaOH,, M

= n ∕V = 0.12 / 25 = 0.0048 M

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0048 48 - 0.00 0.0064 64  100 0.0064 = -25

Trial 3:

n =MV n H2SO4 = 0.1 × 0.7 = 0.07 mol From the equation, 1 mole H2SO4 need 2 mole of the NaOH to produce 1 mole of Na2SO4 and 2 mole of water. If got 0.07 mole of H2SO4, the reaction needed 0.14 mole of NaOH. Concentartion of NaOH, M

=n∕V = 0.14/ 25 = 0.0056 M

Stan Standa dard rd devi deviat atio ion n

= 0.00 0.0056 56 - 0.00 0.0064 64  100 0.0064 = -12.5

The result is summarized below: For the reaction between HCl with NaOH

Concentration of   NaOH (M) Standard deviation

Trial 1 0.0056

Trial 2 0.0040

Trial 3 0.0040

24.44

11.11

11.11

Trial 1 0.0088

Trial 2 0.0048

Trial 3 0.0056

3 7 .5

25.00

12.50

For the reaction between H2SO4 with NaOH

Concentration of   NaOH (M) Standard deviation

Discussion:

Titration is a technique for determining either the concentration of a solution of unknown molarity or the number of moles of a substance in a given sample. A chemical reaction is used for this purpose, and the reaction must be fast, be complete, and have a determinable end point. The reactions of strong acids and bases generally meet these criteria, and acid-base titrations are among the most important examples of this technique. In this experiment, the sample is hydrochloric acid and sulfuric acid as acid substance and sodium hydroxide as base substance where the concentration of sodium hydroxide unknown. Given that, the concentration of the both acid are 0.1 M. An indicator is used as signal the point which the titration is stopped. An acid-base indicator is a weak acid or base that has a different colour from its salt. At least one of them-the indicator or its salt-must be intensely coloured so that it can be seen even in very dilute solution. The colour of the solution is thus different depending on the acidity or basicity of the solution it is in, and when the acidity of o f a solution changes chan ges sufficiently, sufficiently, a colour change will occur. occur. In acid base reaction, the general equation is,

HA + MOH Acid



base

H2O + MA water salt

The end point is the neutral point. In end point of acid-base titration, there are produced salt and water that are neutral (pH7). For most strong acid-strong base reactions, ionic equation is: H+

+ OH-



H2O

In this experiment, the indicator that use is phenolphthalein. If we use base as titrant, and acid as solution in the Erlenmeyer flask, at the end point, the solution in the Erlenmeyer flask  will turn to light pink. If we use acid as titrant, and base as solution in the Erlenmeyer flask, at the end point, the solution in the Erlenmeyer flask will turn to colourless. For this experiment, the solution turn from the purple to colourless. co lourless.

The chemical equation is,

The reaction between HCl and NaOH, HCl + NaOH →NaCl + H2O

The reaction between H2SO4 and NaOH H2SO4 + 2NaOH → Na2SO4 + 2H2O

The different between the reactions is the ionize of the acid. HCl is monoprotic acid, while H2SO4 is diprotic acid.

A diprotic acid is an acid that yields two H+ ions per acid molecule. Examples of diprotic acids are sulfuric acid, H2SO4, and carbonic acid, H2CO3. A diprotic acid dissociates in water in two stages: (1) H2X(aq) (2) HX-(aq)

H+(aq) + HX-(aq) H+(aq) + X2-(aq)

A monoprotic acid is an acid that yields only one H+ ions per acid molecule. Example of  monoprotic acid is Hydrochloric acid. While H2SO4 is a diprotic acid that’s yield two mole of H+. The mean, when the acid dissociate with NaOH, H2SO4 will need 1 mole of H2SO4 while the  NaOH needed 2 mole. That’s mean, we need less amount of acid if we use diprotic acid. That’s why the volume of H2SO4 that has been use is less than HCl.

In this experiment, there are some mistake like use more than acid volume to titrate solution C or NaOH. The volume has pass the end point, so the volume uses is much more than needed. To overcome this problem, we can titrate slowly and shake the volumetric flask for about 30 second when the solutions show changing in colour from dark purple to colourless. The other  reason reasonss is the using of volume volumetri tricc flask flask that’ that’ss had been used used with with other other soluti solution. on. So, the concentration of newest solution will affect. To overcome this problem, we must make sure the volumetric flask is totally clean and dry. That’s problems had affect our result. That’s why our 

result is not precise and accurate to the correct value. We can see that from the standard deviation. But, we had seen the distribution of result is in range of 0.0045 M to 0.0065 M. If we refer to trial 2 and 3 from titration between HCl and solution C (NaOH), the result might be said as 0.004 M base on the standard deviation. The standard deviation is the lowest among others and can be said as accepted result. But, for the titration using H2SO4, the accepted value for  concentration of solution C (NaOH) is 0.0056 M.

Conclusion:

The concentration and mole of acid or base can be determined using titration process by a given value for one of the substance. The accepted values of concentration NaOH using HCl is 0.0040 M, but the value of concentration NaOH using H2SO4 is 0.0056 M.

Reference:

1. http://dwb.u http://dwb.unl.edu/ nl.edu/calcul calculators ators/acti /activitie vities/Di s/Diproti proticAcid.h cAcid.html tml 2. http://www http://www.bioch .biochem.nor em.northwes thwestern.ed tern.edu/holm u/holmgren/G gren/Gloss lossary/D ary/Definit efinitions/ ions/DefDefM/monoprotic_acid.html 3. Fundamental Fundamental of Chemistry Chemistry,, David E. E. Goldberg, Goldberg, Mc Mc Graw Hill, 303. 303.