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How do you prepare a standard solution

In addition to this, we can say that preparing a solution using liquid reagents or by dilution is more accurate than preparing a solution using solid reagents. This can be supported by the average percent deviation of the HCl solution of 87% as compared to the 182.33% average percent deviation of the NaOH solution.

Solution Preparation And Standardization

Generally, there are two ways in preparing a solution, one is by dissolving a weighed amount of solid in a required solvent and the other is by dilution of a concentrated solution into the desired concentration.

In diluting concentrated solution, the concentration of the diluted solution can be determined by standardization. To standardize a solution, we will need to perform titration. In this experiment, we will standardize acid and base solutions.

In this experiment, the students to students will be able to know the proper way of preparing solutions from solid and liquid reagents by using the proper pieces of glassware and equipment and to calculate the exact concentration of the prepared solution from standardization.

PROCEDURE

The reagents that were used in this experiment were concentrated hydrochloric acid, sodium hyrdoxide, sodium carbonate, potassium acid phthalate and phenolphthalein as indicator.

The pieces of glassware that were used to perform this experiment were volumetric flasks, Erlenmeyer flasks, beakers, volumetric pipette, burette, spatula and droppers. Also, the pieces of equipment that were used were analytical balance, top-loading balance and hot plate.

Preparation of 250 mL 1.0 M sodium hydroxide solution (from solid)

The amount of NaOH needed to prepare 1.0M solution was calculated (10.0 g NaOH).

The computed value was weighed using the top-loading balance and placed in a clean and dry 250-mL beaker. Enough amount of distilled water to dissolve the NaOH solid was added to the beaker and stirred. After the NaOH was completely dissolved, the solution was then transferred to a 250-mL volumetric flask quantitatively. Enough distilled water was added to make the volume about 200-mL. The flask was covered and cooled down to room temperature. The solution was bulked to the mark with distilled water and covered. The solution was mixed by repeated shaking and inversion of the flask. Lastly, the solution was transferred into a dry and clean plastic bottle.

3 pages, 1441 words

The Essay on Dconcentrations Of Solutions Determine The Mass Of A Potato

Introduction: The way to get the full results of this lab was through the process of osmosis. Osmosis is the movement of water across a membrane into a more concentrated solution to reach an equilibrium. When regarding cells osmosis has three different terms that are used to describe their concentration. The first of these words is isotonic. Cells in an isotonic solution show that the water has no .

Preparation of 100 mL 3.0 M hydrochloric acid (by dilution)

The volume of 12.1 M HCl solution needed to prepare a 100 mL 3.0 M HCl was calculated (24.8 mL 12.1 M HCl).

To obtain the exact amount, pipette was used to measure 24.8 mL of concentrated HCl solution into a 100-mL volumetric flask containing about 25-mL distilled water. Enough distilled water was then added to make the volume about 90-mL. The solution was mixed through swirling and the flask was covered and we let the solution cooled down to room temperature. The solution was bulked to the mark with distilled water and covered then mixed by repeated shaking and inversion of the flask. Lastly, the solution was transferred into a dry and clean plastic bottle.

Standardization of 1.0 M NaOH and 3.0 M HCl solution

For standardization of 1.0M NaOH, three clean and properly labeled 250-mL Erlenmeyer flasks were used and each flask contained 0.1g of the primary standard KHP to the nearest 0.1mg. The weights were recorded. After this, 50 mL of distilled water was added and mixed by swirling to dissolve the solid. 2 to 3 drops of indicator (phenolphthalein) was added and titration was performed with 1.0M NaOH. For the standardization of the HCl solution, we used the same procedure except that 0.1g of Na2CO3 was placed in each of the three Erlenmeyer flask instead of primary standard KHP and 50 mL of boiled distilled water was added and mixed by swirling to dissolve the Na2CO3 solid. We recorded the initial and final burette reading and started titrating.

Sample Working Calculations

To compute for the concentration of NaOH and HCl, the sample working calculations were used. For NaOH,

1 page, 353 words

The Essay on Standardizing A Solution Of Potassium Hydroxide

Abstract: In this lab, a prepared solution of Potassium Hydroxide will be standardized. The solution we will be standardized by performing multiple calculations to ensure the upmost accuracy. The acid used for this titration will be KHP (C4H5KO4). Phenolphthalein we be added to the beaker of the dissolved acid before the titration will be added. The titration will stop when the solution turns pink .

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 determine the volume of the NaOH and HCl solution in which our primary standards will react completely. This means that we must add a stoichiometrically equivalent amount of titrant (NaOH and HCl) to the solution with the primary standard. By doing this, we will reach the equivalence point but usually there is no obvious indication that the equivalence point has been reached. Instead, we usually look for the end point which is indicated by a change in the color of a substance added to the solution containing the primary standard and an indicator. We used strong acid acid and base because these substances react more completely with an analyte than their weaker counterparts. By determining these volumes, we will be able to compute for the concentrations of the NaOH and HCl solutions.

In this experiment, the phenolphthalein indicator was used to see at what volume these solutions will reach their end points. Also, phenolphthalein was used as the indicator because it provides a sharp end point with a minimal titration error.

The chemical reactions that occured in the titrations that we performed were the following: NaOH + KHP NaHP + KOH

Na2CO3 + 2HCl NaCl + H2CO3

We can see that 1 mol NaOH is needed to react in 1 mol of KHP while 2 mol HCl is needed to react in Na2CO3. The complete reaction takes place as the solution with KHP turned from colorless to pink and as the Na2CO3 solution changed from violet to colorless solution. These will be very important in determining the concentration of our NaOH and HCl solutions.

The determination of the concentration of the solution can be done by calculating the number of moles of NaOH and HCl that will react in each specific amount of the primary standard. Subsequently, we can compute for the molarity by using the number of moles computed and dividing it to the net volume in the titration. This step is illustrated in the sample working calculation.

2 pages, 969 words

The Essay on Determination Of Zinc And Nickel Concentration

I. Introduction In this experiment, the zinc and nickel contents of unknowns were tested using two methods. In the first method, nickel and zinc were separated through ion-exchange chromatography and analyzed through chelometric titration. In the second method, the unknown was analyzed through the atomic absorption spectroscopy (AAS) of the mixture. In and ion-exchange column, the ions are .

Significance of the Results

The concentration of NaOH based on the result of the experiment is 2.82 M and the concentration of the HCl is 5.61 M. These results are very much higher compared to the theoretical concentration of our solution. These may be affected by some errors in weighing the sodium hydroxide pellets, swirling of the Erlenmeyer flasks, or possible error in titration. One of the errors in titration is the indeterminate error which originates from the limited ability of the eye to distinguish the intermediate color of the indicator.

In addition to this, we can say that preparing a solution using liquid reagents or by dilution is more accurate than preparing a solution using solid reagents. This can be supported by the average percent deviation of the HCl solution of 87% as compared to the 182.33% average percent deviation of the NaOH solution.

Conclusion

There are a lot of factors that might affect the concentration of the solution. First is the preparation of solutions. Based on the results of this experiment, if an option is available, it is better to prepare a solution by dilution than by dissolving a solid reagent. Also, the proper way of titration must be observed. You must not overtitrate the solution to obtain a more accurate concentration of the solution.

References

G. D. Christian, Analytical Chemistry, 6th Edition, John Wiley & 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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How do you prepare a standard solution?

How do you prepare a standard solution?

Preparation of a standard solution by weighing method

  • A solution whose concentration is accurately known is called a standard solution.
  • A standard solution can be prepared by weighing method in the following way.
    (a) The mass of solute needed is calculated and weighed.
    (b) The solute is dissolved in some distilled water in a beaker.
    (c) The solution is transferred into a volumetric flask.
    (d) More distilled water is added to obtain the required volume. The flask is stoppered and shaken.
  • For example, to prepare 1.0 dm 3 of 0.50 mol dm -3 aqueous sodium hydroxide.
  • A very accurate standard alkaline solution cannot be prepared using sodium hydroxide as the primary standard. This is because
    (a) sodium hydroxide is deliquescent. It absorbs moisture and dissolves to form a solution.
    (b) sodium hydroxide may not be pure. It reacts with carbon dioxide from the air:
    2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
    Hence, the concentration of this alkaline solution is not accurate.
  • Pure anhydrous sodium carbonate, Na2CO3, is used to prepare a primary standard alkaline solution.
  • Solid organic acids such as oxalic acid, H2C2O4.2H2O, is used to prepare primary standard acidic solution.
  • A standard solution can also be prepared by dilution method.

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Preparation of a standard solution by dilution method

  • A standard solution can also be made by dilution. Bench acids such as hydrochloric acid, sulphuric acid and nitric acid are all prepared by diluting the commercial concentrated acids (stock solutions) with varying amounts of distilled water.
  • Adding water to a concentrated solution:
    (a) changes the concentration of the solution
    (b) does not change the number of moles of solute present
  • This means that the total number of moles of solute particles before dilution is equal to the total number of moles of solute particles after the dilution is carried out.
    Moles of solute before dilution = moles of solute after dilution
  • A dilution equation can be derived as shown below.

Since adding water does not change the number of moles of solute particles,
M1V1 = M2V2

Figure shows how Example is carried out in the laboratory.

  • The volume of stock solution required must be measured out accurately. Use a pipette to measure out exactly 25.0 cm 3 of stock solution. Use a burette to measure out any other volumes that less than 50.0 cm 3 .
  • For example, to prepare 500 cm 3 of 0.15 mol dm -3 sulphuric acid from a stock solution of 2.0 mol dm -3 sulphuric acid.
    (a) Calculate the volume of stock solution required.

(b) Fill a burette with the stock solution.
Run out 37.5 cm 3 of stock solution into a 500 cm 3 volumetric flask containing some distilled water.

1. Calculate the volume of concentrated nitric acid, 18 mol dm -3 which is required to prepare 5.0 dm 3 of 2.0 mol dm -3 nitric acid.
Solution:

2. What volume of 2.0 mol dm -3 sulphuric acid is needed to prepare 100 cm 3 of 0.5 mol dm -3 sulphuric acid?
Solution:

Solutions

Now dissolve 58.5 grams of NaCl in distilled water and make the solution to one liter.

Solutions: Molar Solutions preparation

Normal solution

  • The normal solution is defined as the gram equivalent weight per liter of the solution (solvent).
    • Normal solution = gram equivalent weight of solute/liter of the solution (solvent) = Eq.wt/L.
    • Gram equivalent weight = Gram molecular weight/valency.
    • NaCl gram molecular weight = 58.5 g
    • Valency =1
    • 58.5/1 = 58.5 gram equivalent weight.

    Example

    To make a 1 N sodium chloride solution:

    • The molecular weight of NaCl is 58.5.
    • Gram equivalent weight of NaCl = molecular weight/1 (valency).
      • So dissolve 58.5 grams of NaCl in distilled water and make up to one liter.
      • Dissolve 58.5 grams of NaCl in distilled water to make one liter.

      Percent solution

      1. This is per hundred part of the total solution.
      2. There are three possibilities for a percent solution.
      3. Weight/weight:
        1. It is a percentage of solute in 100 grams of final solution equal to solute + solvent.
        2. e.g., For the 5% solution, take 5 grams of NaCl dissolved in 95 grams of water, around 95 mL.
        1. 5 grams of NaCl dissolved in water, and the volume made 100 ml is called a 5% solution of NaCl.
        1. It is composed of two solutions. e.g., if we take 5 mL of acid and dilute it to 100 mL of water will be a 5% solution of that acid.

        Dilution

        1. It is very common to prepare the dilution of the serum with a high concentration of chemicals like urea in the blood if it is above 300 mg/dL.
        2. If we make a dilution of serum like this:
          1. Serum = 1 ml
          2. Diluting fluid 4 mL
          3. This will be a dilution of 1:5 (1+4 =5).

          Procedure for dilution

          1. This dilution can be made from the stronger solution by this formula:

          Solution diluting formula

          1. Example dilution of sodium hydroxide 1:
            1. Make 250 mL of sodium hydroxide solution of 0.25 mol/L from a solution of 0.4 mol/L solutions.
              1. C = o.25 mol/L
              2. V = 250 mL
              3. S = 0.4 mol/L
              4. Calculation = 0.25 x 250 / 0.4 = 156.25 mL
              1. Make 500 ml of HCL acid, 0.01 mol/L from a 1 mol/L acid.
                1. C = 0.01
                2. V = 500
                3. S = 1
                4. Calculation = 0.01 x 500 / 1 = 5 mL of HCL acid.
                1. Make 5 ml of 1 in 10 dilutions of the serum.
                  1. C = 1:10
                  2. Volume = 5 mL
                  3. S = 1
                  4. Calculations = 1/10 x 5 / 1 = 0.5 mL of the serum and 4.5 mL of the saline = 0.5 : 4.5 = 1:10 dilution

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