100 ml boiled cooled and filtered water sample takes 9.6 ml of M/50 EDTA in titration. The Permanent hardness of the water sample in terms of ppm of CaCO3 equivalent is
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. 100 ml boiled cooled and filtered water sample takes 9.6 ml of M/50 EDTA in titration. The
Permanent hardness of the water sample in terms of ppm of CaCO3 equivalent is
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- What is the acid/base ratio? Why is it important in the titrimetric process? (Particularly in Double Indicator Titration Method in Determination of % SO3in a Soluble Sample)3 L contaminated air 50 mL 0.0116 M in an air pollution analysisCarbon dioxide (CO2) BaCO3 is passed through Ba (OH) 2 solution.is precipitated as. Excess of base, next to phenol phthalate (f.f.) indicatorIt is titrated with 23.6 mL of 0.0108 M HCl. CO2 in this air sampleCalculate its concentration in ppm. (Density of CO2Take it as 1.98 g / L. C = 12, O = 16 g / mol).A silver nitrate soluti on contains 14.77 g of primary standard AGNO3 , in 1.00 L what volume of this solution is needed toreact completely with 50.00 ml. of 0.01808 M Na2S. 1. O10.40 ml. 2. O 20.79 ml. 6.10 ml. 4. O12.31 ml. 5. O18.21 ml. 3.
- A 270.00 mL solution of 0.00150 M AB4 is added to a 380.00 mL solution of 0.00250 M C3D4. What is pQsp for A3D4?On titrating 20 ml of 0.1 M KCl with 0.1 M AgNO3. (For AgCI, Ksp = 1.82 X 10- 10). 1.The volume of AgNO3 titrant need to reach the equivalence point is ………...............ml. 2.At the equivalence point the value of pAg = ……………………. 3.pAg after addition of 25 ml AgNO3= …………………………………It is desired to determine the concentration of chlorides in a powdered milk following the method of Volhard, for such case 1.5g of milk is weighed, it is heated in a muffle until obtaining ashes, which are treated with acid water with some drops of HNO3, the solution is filtered and the obtained filtrate is collected in an Erlenmeyer and 20 mL of AgNO3 0 is added. 1M excess silver is titrated with 0.05M potassium thiocyanate, if 8 mL of KSCN has been used up at the time of the indicator turn. Calculate the % of chloride in the milk powder.
- 11. A 0.485 g sample containing chloride was titrated with 36.8 mL of 0.1060 M AgNO3 stock solution. Express the results of this analysis in:a) Percent by weight of Clb) Weight percent Cl2c) Percent by weight of CaCl2d) Weight percent of AlC135) 3 L contaminated air 50 mL 0.0116 M in an air pollution analysis Carbon dioxide (CO2) BaCO3 is passed through Ba (OH) 2 solution. as precipitated. Excess of base, next to phenol phthalate (f.f.) indicator It is titrated with 23.6 mL of 0.0108 M HCl. CO2 in this air sample Calculate its concentration in ppm. (Density of CO2 Take it as 1.98 g / L. C = 12, O = 16 g / mol).A 0.527g sample of mixture containing Na2CO3 and NaHCO3 andinert impurities is titrated with 0.109M HCI, requiring 15.7mL to reach thephenolphthalein endpoint and a total of 43.8mL to reach the modified ?methylorange endpoint, what is the percent each of Na2CO3 in mixture
- 5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.A 3.00 mL aliquot of 0.001 M NaSCN is diluted to 25.0 mL with 0.1 M HNO3. How many moles of SCN- are present?ng resourc.. (27) Could've Been - [References] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC7H502 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.40 M C2H;NH2 (K½ = 5.6 × 10¬4) titrated by 0.10 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.70 M HC1 titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =