Determination of MnO2 in pyrolusite

Author name: Xiaoxiao Zeng

Learner Code: B876496

Abstract: In this paper, employing a back titration of KMnO4 method to determine the content of MnO2 in pyrolusite. The %w/w MnO2 in the sample is 29.41%.

Keywords: pyrolusite; manganese dioxide; titration；sample；determination；accuracy；precision；confidence interval.

1 Introduction

The important component of pyrolusite is manganese dioxide[1]. A sample of pyrolusite was taken from a mine at Heshang area in Guangxi province, which is a very low-grade ore. In order to utilize the low-grade pyrolusite, we must find a good method to solve the problem.

The main component of pyrolusite is manganese dioxide. Is a kind of common manganese mineral. Pyrolusite is very soft, and it softer than people fingernails. The pyrolusite color is gray-black with a metallic luster. Some pyrolusite like arborization on rock surface, it's very amusing. So it is called pseudofossils.

A new hydrometallurgical method has to be developed to produce a chemical material of manganese sulfate. Thus the approximate content of MnO2 in the pyrolusite must be determined. The content of MnO2 in the pyrolusite can be measured using one of the methods of iodimetry[1]，oxalatometry[2]，hydrogen peroxidometry[3]and ferrometry[4]，that all meet the accuracy and precision for the test of MnO2 in the pyrolusite, in theory. Oxalatometry is selected to determine manganese dioxide in the pyrolusite, basing on a standard solution of KMnO4 is stable and convenient to prepare and stock, despite of ferric ion interfering the analyses. The accuracy (relative error) and precision (standard deviation) should be ±5% and ±0.2% in this experiment, respectively.

2 Principle of experiment

Manganese dioxide is a main component in pyrolusite. As a result of manganese dioxide is a strong oxidizer. Obviously, KMnO4 cannot used directly titration method.In the absence of oxidizer conditions, manganese dioxide is hardly dissolved in acid or base. Therefore, we cannot use directly with reducing agents for titration, we adopt the back titrimetry.Under acidic conditions, an excessive sodium oxalate and a 50-mL 3.7 mol/l H2SO4 solution were added in the pyrolusite, the manganese (IV) was reduced to Mn2+ ion dissolved in water. The remaining Na2C2O4 was titrated with a standard solution KMnO4 involved in the chemical equations as follows.

MnO2+C2O42-+4H+=Mn2++2CO2↑+2H2O

2MnO4-+5C2O42-+16H+=2Mn2++10CO2↑+8H2O

According to the stoichiometric relation of Na2C2O4 to KMnO4 and the consumed volume of the titrant, we can calculate the content of manganese dioxide in the sample of pyrolusite.

3 Reagents and apparatus

Reagents:The standard solution of 0.0208-mol·L-1 KMnO4, which is diluted with a 0.2080-mol/L stock solution of KMnO4; The analytical reagent of Na2C2O4 (AR); The solution of 3-mol·L-1H2SO4; H2SO4 (AR).

Apparatus: Some Beakers; Medicine spoon; Glass rod; An electronic analytical balance; Watch glass; Water-bath boiler; Electric cooker;Thermometer; A 25.00-mL volumetric pipette; A 250-mL Volumetric flask.

In order to get an accurate value, we must calibrate a 25-mL volumetric pipette and an electronic analytical balance.

Calibrate the equipments.

a. volumetric pipette

Weighed a dry volumetric flask (25 mL) with a stopper. Filled pipette with distilled water and delivered the water into the flask, stoppered container to avoid evaporation loss. Recorded the room temperature is 29oC and reweighed the container to obtain the weight in air of the water delivered by the pipette. Used the density of water in air (d = 0.9959 g/mL at 29oC) to calculate the volume of the pipette. According to the mass of water delivered by the pipette is 25.0196 g, the volume of the pipette is 25.1226 mL. The tolerance is 0.12 mL, that is larger than 0.06 mL, thus the pipette is production of class C.[5]

b. electronic analytical balance

Leveled the balance using the air-bubble float and leaved the balance constantly under power. When a calibration button was pushed, a 200.0000-g test weight was desired to load on the pan, the balance calibrated itself to full accuracy and returned to weighing mode within seconds.

4 Experimental

4.1 Sampling of pyrolusite

Pulverizing：Took a 1.5-kg pyrolusite from a barrel using the sampling methodology of compact solid [6] and pulverized the pyrolusite by ceramic mortar. After Blending equably, an approximately 10-g pyrolusite powder was coned and then to be flatten. A sample pyrolusite was sampled by quartering [8].

4.2 Scaling of sample of pyrolusite and standard reagents preparing standard solution.

Calibrated electronic balance and scaled samples and standard reagents; prepared standard solution.

In order to guarantee result accuracy, we must calibrate electronic before we use it.

Scaled 2.2788 grams pyrolusite sample and put it into a 250-mL beaker.

According the probable content of MnO2 scaled a 2.2059-g Na2C2O4 reagent and put it into the beaker. Then added a 50-mL 3mol·L H2SO4 solution and a 2-ml thick H2SO4 solution into the beaker.

Cover a watch glass on the beaker. Dissolving in 70~80℃ the water bath, until no longer emitted CO2 bubble. Also there had no black pellets in the residue. Generally, not surpassed 30 minutes when dissolved, in order to avoid or reduce loss of the oxalic acid. Wash the beaker's wall and the watch glass several times.

4.3 Pretreatment of pyrolusite

There were a lot of solid impurities, so we must remove it out. We used long neck funnel and filter paper to remove the solid materials from solution by filtration. We must perform trial separations of filtration. The experimental results show that MnO2 in the sample of pyrolusite had been transferred into the analyte solution (250 mL), completely.

4.4 Titration

Took the titration of a titrant VS a measured volume solution of sodium oxalate; recorded the data of volume of the consumed titrant. (Repeat 4~5 times); processed the experimental data and uated the results.

Heated conical flask in 70~80℃ the water bath, and used the KMnO4 standard solution to titrate until the solution became micro-red. No depigment for a half minute indicated the end-point had reached.

The reaction vate of MnO2 with Na2C2O4 is slow under room temperature. So we must heat up the solution. But the temperature cann't be too high. If the temperature exceeded 90℃, the H2C2O4 should be vaporized.

Transferred a 25.00-mL aliquot of the analyte solution from the 250-mL volumetric flask. Titrated the analyte solution with a 0.0208-mol/L KMnO4 titrant and required 16.73 mL to reach the end point. Similarly, repeated the titration four times and recorded the desired the volumes of the titrant, processed the data using statistical method (t Test) [7] and listed the result in Table 1. The result shows that there are not data of the volumes falling outside the confidence interval [16.73, 16.97].

Table 1 data of process for the titrant volumes

Times Iterms

1

2

3

4

5

Vs(mL)

16.73

16.97

16.81

16.80

16.88

s(mL)

16.83

s

0.09138

t0.95,4

2.776

Confidence limit

(x±t s/√n)

（16.83±0.11）

Confidence interval

[16.73,16.97]

There are two stoichiometric reactions involved in the oxalatometric titration

MnO2+C2O42-+4H+=Mn2++2CO2+2H2O 2.1

2MnO4-+5C2O42-+16H+=2Mn2++10CO2+8H2O 2.2

Due to the ratio of the moles MnO4- to the moles C2O42- is that

moles MnO4-∶moles C2O42-=2∶5.

Thus

moles C2O42-=5 moles MnO4-/2

Substituting for moles of MnO4- leaves us with an equation that is solved for the consumed moles of Na2C2O4.

Consumed moles of Na2C2O4=5×cs×Vs/2

Consumed moles of Na2C2O4=5×0.0208×16.73×10-3/2=8.7×10-4 (moles)

According to the ratio of the moles MnO2 to the moles C2O42- is that

moles MnO2∶moles C2O42-=1∶1.

So

Moles of MnO2=total moles of Na2C2O4-consumed moles of Na2C2O4×10

=total moles of Na2C2O4-consumed moles of Na2C2O4×10

=2.2059/134-8.7×10-4×10

=1.646×10-2 - 8.7×10-3

=7.76×10-3(moles)

Finally, the %w/w MnO2 in the sample is

-×100= -×100

=7.76×10-3×86.94×100/2.2788

=29.61

Similarly, calculate the %w/w MnO2 in the sample for replicated titrations are 29.15, 29.46, 29.46 and 29.31, respectively.

5 Discussion and Result

5.1 Process the data of analyses and result presentation.

Calculate the mean and the standard deviation of the set of data 29.61%, 29.15%, 29.46%, 29.46% and 29.31% are 29.41%and 0.18%. Check Table 13 in reference book [7] to obtain the t0.95 (4 degrees of freedom) of the statistic is 2.776. And now the confidence limit of the %w/w MnO2 in the sample becomes [29.14±0.15], that means that the mean of the population falls into the confidence interval 95% probability level.

5.2 Error analysis

The main errors in this determination root in the volatilization of oxalic acid and the interference of ferric ion, and that the statistical weight of the latter error is bigger than the former's. When the %w/w iron in the sample ore is 4.4, the error will be 4.7 [4]. The accuracy (relative error) is probably ±5% and the precision (standard deviation) should be ±0.2% in this experiment, but the results could meet the desire for classifying the grade of a pyrolusite. More accuracy and precision could meet ±0.5% and ±0.2% in next assignment where the analyte, which is reserved from this determination, will be measured by employing Atomic Absorption Spectrometry (AAS).

There is a near-infinite list of what might effect reaction rates. Major, common factors are Temperature, Pressure, Solvent Concentration, Solute Concentration, Sterics, Size (Surface area) Catalysts, Concentration and light .

The rate of reaction is determined by the Activation Energy (Ea, Delta G 'double dagger'). It is the pathway of the reaction that can be shorter in the presence of enzymes or catalysts.

Enzymes are biological catalysts, they speed up chemical reactions inside livig things. They are made up of proteins. Enzymes have a special shape, there is a starting chemical called a substrate that fits into the enzymes special shape.

- you may wanna consider the substrates that are in enzyme's along with the product

Enzyme Catalysis is the catalysis of chemical reactions by specialized proteins known as enzymes. Catalysis of biochemical reactions in the cell is vital due to the very low reaction rates of the uncatalysed reactions.

The mechanism of enzyme catalysis is similar in principle to other types of chemical catalysis. By providing an alternative reaction route and by stabilizing intermediates the enzyme reduces the energy required to reach the highest energy transition state of the reaction. The reduction of activation energy (Ea) increases the number of reactant molecules with enough energy to reach the activation energy and form the product.