Titration blank

A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.3172-g sample is dissolved in 50 mL of water and titrated to the Ag2Cr04 end point, requiring 36.85 mL of 0.1120 M AgNOa. A blank titration requires 0.71 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample. 

In the blank titration, the acetic anhydride is hydrolyzed to give 2 mol of acetic acid per mole of AC2O. In a sample titration, each unreacted anhydride molecule likewise yields two of acetic acid, but each reacted AC2O molecule yields only one... 

Before commencing the elution titrate 10.0 mL of the 0.3 M sodium nitrate with the standard silver nitrate solution, and retain the product of this blank titration for comparing with the colour in the titrations of the eluates. When the titre of the eluate falls almost to zero (i.e. nearly equal to the blank titration) — ca 150 mL of effluent — elute the column with 0.6M sodium nitrate. Titrate as before until no more bromide is detected (titre almost zero). A new blank titration must be made with 10.0 mL of the 0.6M sodium nitrate. 

Carry out a blank determination on the acetic anhydride/ethyl acetate solution following the above procedure without adding the carbohydrate. Use the difference between the blank titration, Vb, and the sample titration, Vs, to calculate the number of hydroxyl groups in the sugar (Note 2). 

Determination of calcium. Pipette two 25.0 mL portions of the mixed calcium and magnesium ion solution (not more than 0.01M with respect to either ion) into two separate 250 mL conical flasks and dilute each with about 25 mL of de-ionised water. To the first flask add 4 mL 8 M potassium hydroxide solution (a precipitate of magnesium hydroxide may be noted here), and allow to stand for 3-5 minutes with occasional swirling. Add about 30 mg each of potassium cyanide (Caution poison) and hydroxylammonium chloride and swirl the contents of the flask until the solids dissolve. Add about 50 mg of the HHSNNA indicator mixture and titrate with 0.01 M EDTA until the colour changes from red to blue. Run into the second flask from a burette a volume of EDTA solution equal to that required to reach the end point less 1 mL. Now add 4 mL of the potassium hydroxide solution, mix well and complete the titration as with the first sample record the exact volume of EDTA solution used. Perform a blank titration, replacing the sample with de-ionised water. 

The Karl Fischer procedure was applied to the determination of water present in hydrated salts or adsorbed on the surface of solids. The procedure, where applicable, was more rapid and direct than the commonly used drying process. A sample of the finely powdered solid, containing 5-10 millimoles (90-180 mg) of water, was dissolved or suspended in 25 mL of dry methanol in a 250-mL glass-stoppered graduated flask. The mixture was titrated with standard Karl Fischer reagent to the usual electrometric end point. A blank titration was also carried out on a 25 mL sample of the methanol used to determine what correction (if any) needed to be applied to the titre obtained with the salt. 

The amount of unreacted iodine was determined by titration with sodium thiosulfate [Thiosulfuric acid (H2S203), disodium salt], and the amount of iodine initially present was determined by a separate blank titration.3... 

A sample of polyester (ca. 1 g, exactly weighed) is dissolved in 20 mL toluene-ethanol mixture (1/1 vol.) and titrated by a solution of KOH in ethanol (0.05 mol/L) using a potentiometric titrator. A blank titration must be performed under the same conditions. Hardly soluble polyesters (e.g., PET) must be dissolved in an o-cresol-chloroform mixture or in hot benzyl alcohol.417 The result (acid content) is normally expressed in mmol COOH/g polyester but may also be given as the acid number, defined as the number of milligrams of KOH required to neutralize 1 g of polyester. [Acid number = (number of mmol COOH/g polyester) x 56.106.]... 

Furthermore, pH electrode calibration can be performed in situ by the new method [48], concurrently with the pKj determination. This is a substantial improvement in comparison to the traditional procedure of first doing a blank titration to determine the four Avdeef-Bucher parameters [24]. The traditional cosolvent methods used with sparingly soluble molecules can be considerably limited in the pH<4 region when DMSO-water solutions are used. This is no longer a serious problem, and routine blank titrations are now rarely needed in the new in situ procedure. 

Operationally, such a plot can be obtained by subtracting a titration curve containing no sample ( blank titration left curve in Fig. 3.1a) from a titration curve with sample (right curve in Fig. 3.1a) at fixed values of pH. The resultant difference plot is shown in Fig. 3.1b. The plot is then rotated (Fig. 3. Id), to emphasize that % is the dependent variable and pH is the independent variable [163], The volume differences can be converted to proton counts as described in the preceding paragraph, to obtain the final form, shown in Fig. 3.Id. 

Dissolve 0.35 g of miconazole nitrate in 75 mL of anhydrous acetic acid R, with slight heating, if necessary. Titrate with 0.1 M perchloric acid determining the end point potentiometrically, according the general method (2.2.20). Carry out a blank titration. One milliliter of 0.1 M perchloric acid is equivalent to 47.91 mg of Ci8H15Cl4N304. 

Procedure for content uniformity. Transfer 1 tablet, previously crushed or finely powdered, to a beaker, add 5 mL of hydrochloric acid and about 25 g of crushed ice, then add water to bring the total volume to about 50 mL. Proceed as directed under Nitrite Titration under the general procedure <451 >, beginning with slowly titrate , and using as the titrant 0.01 M sodium nitrite VS, freshly prepared from 0.1 M sodium nitrite. Concomitantly perform a blank titration, and make any necessary correction. Each milliliter of 0.01 M sodium nitrite is equivalent to 4.553 mg of Ci5H21N30-2H3P04. 

Blank Titration It is usually carried out to account for the possible reaction of atmospheric moisture with the titrant perchloric acid and also to check the titrant being employed to bring about the blue-green end-point. 

Procedure Weigh accurately about 0.2 g of ethacrynic acid, dissolve in 40 ml of glacial acetic acid in a 250 ml iodine flask. Add to it 20 ml of 0.1 N bromine and 30.0 ml of hydrochloric acid, immediately place in position the moistened stopper to the ffask, mix the contents vigorously and allow it to stand in a dark place for 60 minutes (to complete the reaction with bromine). Add to it 100 ml of water and 20 ml of KI Solution and titrate immediately with 0.1 sodium thiosulphate, employing freshly prepared starch solution as an indicator towards the end of the titration. Repeat an operation without the pharmaceutical substance (blank titration) thus the difference between the titrations represents the amount of bromine required by the ethacrynic acid. Each ml of 0.1 N bromine is equivalent to 0.01516 g of C13H12C1204. 

Procedure Weigh accurately 0.5 g of phenol and dissolve in sufficient water to produce 500 ml in a volumetric flask. Mix 25.0 ml of this solution with 25.0 ml of 0.1 N potassium bromate in a 250 ml iodine flask and add to it 1 g of powdered KI and 10.0 ml of dilute hydrochloric acid. Moisten the glass stopper with a few drops of KI solution and place it in position. Set it aside in a dark place for 20 minutes while shaking the contents frequently in between. Add to it 10 ml of KI solution, shake the contents thoroughly and allow it to stand in the dark for a further duration of 5 minutes. Wash the stopper and neck of the flask carefully with DW, add 10 ml chloroform and titrate with the liberated iodine with 0.1 N sodium thiosulphate using freshly prepared starch as an indicator. Carry out a blank titration simultaneously and incorporate any necessary correction, if required. Each ml of 0.1 N potassium bromate is equivalent to 0.001569 g of C6H60. 

Procedure Weigh accurately benzalkonium chloride 4.0 g and dissolve it in sufficient DW to make 100 ml. Pipette 25.0 ml into a separating funnel, add 25 ml of chloroform, 10 ml of 0.1 N NaOH and 10 ml of potassium iodide solution. Shake the contents thoroughly, allow to separate and collect the chloroform layer in another separating funnel. Treat the aqueous layer with 3 further quantities each of 10 ml of chloroform and discard the chloroform layer. To the aqueous layer add 40 ml of hydrochloric acid, cool and titrate with 0.05 M potassium iodate till the solution becomes pale brown in colour. Add 2 ml of chloroform and continue the titration until the chlorofonn layer becomes colourless. Titrate a mixture of 29 ml of water, 10 ml of KI solution and 40 ml of hydrochloric acid with 0.05 M potassium iodate under identical conditions (Blank Titration). The differences between the titrations represent the amount of 0.05 M potassium iodate required. Each ml of 0.05 M potassium iodate is equivalent to 0.0354 g of C H CIN. 

Amount of ethanolic KOH used for hydrolysis and in blank titration = 25 ml... 

The number of mg of KOH required to neutralise a blank titration of the reagents -the number of mg KOH required to neutralise excess AA + acetic acid after reaction with 1 g of the test substance. 

A sample of phenol glycerol injection was dilutee with water and an aliquot was taken and reacted with excess bromine generated from potassium bromide and potassium bromate solutions. The excess bromine remaining after reaction was reacted with potassium iodide and the liberated iodine was titrated with sodium thiosulphate. A blank titration was carried out where the same quantity of bromine was generated as was used in the titration of the diluted injection, potassium iodide was then added and the liberated iodine was titrated with sodium thiosulphate. From the following data calculate the percentage of w/v of the phenol in the injection. 

Blazek determined chlorpromazine in H2O and H2SO4 media by titration with 0.05N Ce(S04)2 [62]. The endpoint consisted of the complete color discharge that appeared during the nitration step. When determining the drug in tablets or drops, the use of ferroin indicator was recommended. A blank titration must be performed to achieve proper results. 

Active oxygen content is determined iodometrically 3 In an iodine flask, an accurately weighed sample (0.1-0.3 g.) is dissolved in 20 ml. of an acetic acid-chloroform solution (3 2 by volume), and 2 ml. of saturated aqueous potassium iodide solution is added. The flask is immediately flushed with nitrogen, stoppered, and allowed to stand at room temperature for 15 minutes. Fifty milliliters of water is then added with good mixing, and the liberated iodine is titrated with 0.1 A sodium thiosulfate, employing starch as indicator. A blank titration, which usually does not exceed 0.2 ml., is also run. One milliliter of 0.1 N sodium thiosulfate is equivalent to 0.00821 g. of tetralin hydroperoxide. 

A il of hydrochloric acid for blank titration B= ll of hydrochloric acid for sample titration N= Jormality of hydrochloric acid W eight of sample in grams, corrected for total volatility... 

Cool to 25° and filter the supernatant liquid, retaining the DPhA in the flask. Repeat the extraction with 50ml of boiling distd w exactly as above. To the combined filtrates add a few drops of phenolphtbalein indicator and observe the color. If the soln remains colorless, titrate it with 0.1N alkali and if, it. turns red, titrate with 0.1N acid. Make a blank titration on the distd w used and correct the titration readings for any acidity or alkalinity found in w. Calculate the acidity to HC1 and alkalinity to NaOH- Save this soln for the next test... 

N potassium bromide-bromate so In and 5m of coned HO. Close the flask immediately with the stopper previously moistened with a drop of 10% KI soln, mix the contents by swirling and, after allowing to stand for 10+0.5 mins, add 25ml of 10% KI soln. Titrate the liberated iodine with Q.1N Na thiosulfate soln until near discoloration add 5ml of starch soln and continue titration until the blue color is completely discharged. Run concurrently a blank titration using the same reagents as above... 

Peroxide Determinations. Bawn and Williamson report two iodometric procedures for determining peracetic acid (Methods I and III below) and one method for determining total peroxide (4) (Method II). Bawn and Jolly report another method for total peroxide (5) (Method IV below). The difference between total peroxide and peracetic acid is assumed to be acetaldehyde monoperacetate (AMP). Each method was tested in our preliminary studies. Method III is preferred for peracetic acid because the results are more reproducible. In Method I a large blank titration was always observed, while in Method III the blank titration was very small. Method IV is preferred for total peroxide because it seems to be more sensitive to total peroxide and less sensitive to water content of the acetic acid solvent. 

A blank titration against a suspension of zinc oxide is necessary.4... 

The difference between the end point and the equivalence point is an inescapable titration error. By choosing a physical property whose change is easily observed (such as pH or the color of an indicator), we find that the end point can be very close to the equivalence point. We estimate the titration error with a blank titration, in which we carry out the same procedure without analyte. For example, we can titrate a solution containing no oxalic acid to see how much Mn04 is needed to produce observable purple color. We then subtract this volume of Mn04 from the volume observed in the analytical titration. 

The volume of reagent (titrant) required for stoichiometric reaction of analyte is measured in volumetric analysis. The stoichiometric point of the reaction is called the equivalence point. What we measure by an abrupt change in a physical property (such as the color of an indicator or the potential of an electrode) is the end point. The difference between the end point and the equivalence point is a titration error. This error can be reduced by subtracting results of a blank titration, in which the same procedure is carried out in the absence of analyte, or by standardizing the titrant. using the same reaction and a similar volume as that used for analyte. 

Distinguish between the terms end point and equivalence point. 7-3. How does a blank titration reduce titration error ... 

D. Lowinsohn and M. Bertotti, Coulometric Titrations in Wine Samples Determination of S(TV) and the Formation of Adducts, J. Chem. Ed 2002, 79, 103. Some species in wine in addition to sulfite react with Ij. A blank titration to correct for such reactions is described in this article. 

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