Solution Preparation and Standardization Sample Essay

By and large. there are two ways in fixing a solution. one is by fade outing a weighed sum of solid in a needed dissolver and the other is by dilution of a concentrated solution into the coveted concentration.

In thining concentrated solution. the concentration of the diluted solution can be determined by standardisation. To standardise a solution. we will necessitate to execute titration. In this experiment. we will standardise acid and basal solutions.

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In this experiment. the pupils to pupils will be able to cognize the proper manner of fixing solutions from solid and liquid reagents by utilizing the proper pieces of glasswork and equipment and to cipher the exact concentration of the prepared solution from standardisation.


The reagents that were used in this experiment were concentrated hydrochloric acid. Na hyrdoxide. Na carbonate. K acerb phthalate and phenolphthalein as index.

The pieces of glasswork that were used to execute this experiment were volumetric flasks. Erlenmeyer flasks. beakers. volumetric pipette. burette. spatula and droppers. Besides. the pieces of equipment that were used were analytical balance. top-loading balance and hot home base.

Preparation of 250 milliliter 1. 0 M Na hydrated oxide solution ( from solid )

The sum of NaOH needed to fix 1. 0M solution was calculated ( 10. 0 g NaOH ) . The computed value was weighed utilizing the top-loading balance and placed in a clean and dry 250-mL beaker. Adequate sum of distilled H2O to fade out the NaOH solid was added to the beaker and stirred. After the NaOH was wholly dissolved. the solution was so transferred to a 250-mL volumetric flask quantitatively. Enough distilled H2O was added to do the volume about 200-mL. The flask was covered and cooled down to room temperature. The solution was bulked to the grade with distilled H2O and covered. The solution was assorted by perennial shaking and inversion of the flask. Last. the solution was transferred into a dry and clean plastic bottle.

Preparation of 100 milliliter 3. 0 M hydrochloric acid ( by dilution )

The volume of 12. 1 M HCl solution needed to fix a 100 milliliter 3. 0 M HCl was calculated ( 24. 8 mL 12. 1 M HCl ) . To obtain the exact sum. pipette was used to mensurate 24. 8 milliliter of concentrated HCl solution into a 100-mL volumetric flask incorporating about 25-mL distilled H2O. Enough distilled H2O was so added to do the volume about 90-mL. The solution was assorted through swirling and the flask was covered and we let the solution cooled down to room temperature. The solution was bulked to the grade with distilled H2O and covered so assorted by perennial shaking and inversion of the flask. Last. the solution was transferred into a dry and clean plastic bottle.

Standardization of 1. 0 M NaOH and 3. 0 M HCl solution
For standardisation of 1. 0M NaOH. three clean and decently labeled 250-mL Erlenmeyer flasks were used and each flask contained 0. 1g of the primary criterion KHP to the nearest 0. 1mg. The weights were recorded. After this. 50 milliliter of distilled H2O was added and mixed by twirling to fade out the solid. 2 to 3 beads of index ( phenolphthalein ) was added and titration was performed with 1. 0M NaOH. For the standardisation of the HCl solution. we used the same process except that 0. 1g of Na2CO3 was placed in each of the three Erlenmeyer flask alternatively of primary criterion KHP and 50 milliliter of boiled distilled H2O was added and mixed by twirling to fade out the Na2CO3 solid. We recorded the initial and concluding burette reading and started titrating.

Sample Working Calculations

To calculate for the concentration of NaOH and HCl. the sample working computations were used. For NaOH.

mol NaOH = 0. 1018g KHP1 mol KHP71. 08 g KHP1 mol NaOH1 mol KHP = 0. 001432189083 mol NaOH

M NaoH = 0. 001432189083 mol NaOH0. 00055=2. 60 M NaOH

Mol HCl = 0. 0992g KHP1 mol KHP105. 988 g KHP2 mol NaOH1 mol KHP=0. 001871910028 mol HCl

M NaoH = 0. 001871910028 mol HCl0. 00065=3. 12 M HCl

Principles and Concepts

In titration. we will be able to find the volume of the NaOH and HCl solution in which our primary criterions will respond wholly. This means that we must add a stoichiometrically tantamount sum of titrant ( NaOH and HCl ) to the solution with the primary criterion. By making this. we will make the equality point but normally there is no obvious indicant that the equality point has been reached. Alternatively. we normally look for the terminal point which is indicated by a alteration in the colour of a substance added to the solution incorporating the primary criterion and an index. We used strong acerb acid and base because these substances react more wholly with an analyte than their weaker opposite numbers. By finding these volumes. we will be able to calculate for the concentrations of the NaOH and HCl solutions.

In this experiment. the phenolphthalein index was used to see at what volume these solutions will make their terminal points. Besides. phenolphthalein was used as the index because it provides a crisp terminal point with a minimum titration mistake.

The chemical reactions that occured in the titrations that we performed were the undermentioned: NaOH + KHP NaHP + KOH
Na2CO3 + 2HCl NaCl + H2CO3

We can see that 1 mol NaOH is needed to respond in 1 mol of KHP while 2 mol HCl is needed to respond in Na2CO3. The complete reaction takes topographic point as the solution with KHP turned from colorless to tap and as the Na2CO3 solution changed from violet to colorless solution. These will be really of import in finding the concentration of our NaOH and HCl solutions.

The finding of the concentration of the solution can be done by ciphering the figure of moles of NaOH and HCl that will respond in each specific sum of the primary criterion. Subsequently. we can calculate for the molar concentration by utilizing the figure of moles computed and spliting it to the net volume in the titration. This measure is illustrated in the sample working computation.

Significance of the Consequences

The concentration of NaOH based on the consequence of the experiment is 2. 82 M and the concentration of the HCl is 5. 61 M. These consequences are really much higher compared to the theoretical concentration of our solution. These may be affected by some mistakes in weighing the Na hydrated oxide pellets. swirling of the Erlenmeyer flasks. or possible mistake in titration. One of the mistakes in titration is the indeterminate mistake which originates from the limited ability of the oculus to separate the intermediate colour of the index.

In add-on to this. we can state that fixing a solution utilizing liquid reagents or by dilution is more accurate than fixing a solution utilizing solid reagents. This can be supported by the mean percent divergence of the HCl solution of 87 % as compared to the 182. 33 % mean percent divergence of the NaOH solution.


There are a batch of factors that might impact the concentration of the solution. First is the readying of solutions. Based on the consequences of this experiment. if an option is available. it is better to fix a solution by dilution than by fade outing a solid reagent. Besides. the proper manner of titration must be observed. You must non overtitrate the solution to obtain a more accurate concentration of the solution.


G. D. Christian. Analytic Chemistry. 6th Edition. John Wiley & A ; Sons. New York. Chapter 8 and 2 D. Harvey. Modern Analytical Chemistry. Mc-Graw Hill. USA. p. 274 Skoog. etal. . Fundamentalsof AnalyticalChemistry. Eighth edition. 2004. p. 338-340


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