# Volumetric analysis – Stoichiometry

**Objective:**

To determine the exact concentration of a monobasic acid, HX.

**Materials:**

i) Monobasic acid, HX.

ii) Solid sodium hydroxide, NaOH.

**Apparatus:**

i) One volumetric flask 250 ml and stopper.

ii) Electronic balance ±0.01 g

iii) Three titration flask.

iv) One 25 cm³ pipette and pipette filler

v) One 50 cm³ burette.

vi) One retort stand and clamp.

vii) One white tile.

viii) One wash bottle filled with distilled water.

**Theory:**

Neutralization process occurred between acid and base. Then H+ ion from acid react with OH־ ion from the base to form H_{2}O.

Phenolpthalein?

**Procedures:**

### A) Preparation Of NaOH.

a. A piece of filter paper is weighted using the electronic weight.

b. A spoon of solid sodium hydroxide, NaOH is taken from the container using spatula and weighted.

c. The reading of the measurement is recorded.

d. The sodium hydroxide, NaOH is then put into the beaker.

e. A little amount of distilled water is filled into the beaker and stirred using a glass rod to dissolve the solid sodium hydroxide, NaOH.

f. The solution is then transferred into 250 cm³ volumetric flask until reach the line marked using the meniscus of the solution to determine your endpoint.

### B) Titration

a. Using the 25 cm³ pipette to pipette 25 cm³ of NaOH solution into the titration/ conical flask.

b. Three drops of phenolphthalein indicator is added into the titration flask containing 25 cm^{3} of NaOH solution.

c. Monobasic acid, HX solution is poured into the 50 ml burette. Note that milliliter, ml and centimeter cubic, cm^{3} is the same.

d. Titrate the solution as well.

e. Record your readings in the table below.

f. Repeat the titration as many as you think necessary to achieve accurate result.

## Results:

Record your titration reading in the table below:

### a. Rough

Final reading cm^{3} = _____________

Initial reading cm^{3} = _____________

Volume of HX cm^{3} = _____________

### b. Accurate First

Final reading cm^{3} = _____________

Initial reading cm^{3} = _____________

Volume of HX cm^{3} = A1

### c. Accurate Second

Final reading cm^{3} = _____________

Initial reading cm^{3} = _____________

Volume of HX cm^{3} = A2

### d. Accurate Third

Final reading cm^{3} = _____________

Initial reading cm^{3} = _____________

Volume of HX cm^{3} = A3

**Data Calculations:**

(i) Relative molecular mass, RMM of NaOH =

23.0 + 16.0 + 1.0 = 40.0 g mol^{–}^{1}

So, 40.0 g of NaOH in one mole.

(ii) To produce 1.0 Molar of NaOH, what is the mass of solid NaOH should be use?

We must know what is the volume of volumetric flask we use to dissolve the solid NaOH before we calculate.

If we use 250 ml of volumetric flask,

1 mol = 1000 ml,

x = 250 ml.

x = (250 ml x 1 mo) / 1000 ml = 0.25 mol.

So,

40 g = 1 mol,

x g = 0.25 mol.

x = (0.25 mol x 40 g) / 1 mol.

= 10 g should be added to 250 ml of volumetric flask.

(iii) Calculate your average titre value showing the suitable titration values.

Average volume : ( A1 + A2 + A3 ) cm^{3} / 3 = y cm^{3}

Average volume of HX cm^{3} = y

25.0 cm^{3} of NaOH require y cm^{3} of HX for a complete reaction.

(iv) HX + NaOH → NaX + H_{2}O

From the equation,

1 mole of HX react with 1 mole of NaOH.

Concentration of monobasic acid, HX.

M_{1}V_{1} = M_{2}V_{2},

M_{2} = M_{1}V_{1}/V_{2}

M_{2} = ( ___ × ____ ) / ____ = ______ mol dm^{–}^{3}

= ( 1.0 mol dm^{–}^{3} × 25 cm^{3} ) / y = ______ mol dm^{–}^{3 }the concentration of HX.

**Conclusion:**

y cm³ of HX react with 25.0 cm^{3} of 1.0 mol dm^{–}^{3} of NaOH to produce water and salt. The concentration of HX obtained through calculation is ______ mol dm^{–}^{3}.

**Improvements for next:**

1. The value obtained is slightly different from the theory because there is some parallax error when the experiment is carried on.

2. The colour changes of phenolphthalein….

3. Calibrate mark line, meniscus of the solution.

4. Eye level must parallel.

5. Avoid spill in eye, shake.