What is the equivalence volume when 0.0500 M EDTA is titrated with 100.0 mL of 0.0500 M M 100.55 mL 100.0 mL 100.05 mL 100.09 mL buffered to a pH of 9.00?
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- answer the following: Let us assume that there is a 0.020 M EDTA titrant solution. Calculate the mass of pure dry CaCO3 standard to use such that the volume of titrant needed to reach the endpoint will be about 35mL if one mole of calcium carbonate reacts with one mole EDTA. Given choices: 0.050 g 0.060 g 0.070 g 0.080 gThe data from my observations is given below: Concentration of Standardized NaOH - 1.00 mol/L Mass concentration of acetic acid - 2.40 g/L Trial 1 Trial 2 Measured pH of the acetic acid solution 3.07 3.05 Mass of acetic acid solution taken for titration 20.0 g 20.0 g Initial buret reading of NaOH titrant 0.00 mL 0.00 mL Final buret reading of NaOH titrant 0.82 mL 0.79 mL Net volume of NaOH 0.82 mL 0.79 mL Millimoles of NaOH to end point of titration 0.82 mmol 0.79 mmol Millimoles of acetic acid in sample 0.82 mmol 0.79 mmol Molar concentration of acetic acid solution 0.043 mol/L 0.041 mol/L Calculated molar mass of acetic acid g/mol g/mol Average molar concentration of acetic acid for the samples titrated: 0.042 mol/L Average pH: 3.06 Question: Calculate the corresponding [H3O+], compute the corresponding concentration of A- and HA, and calculate the dissociation constant Ka Thank you!A RbOH solution is titrated four (4) times against potassium hydrogen phthalate (KHP; FW=204.224) samples to the Phenolphthalein endpoint. Using the data below, determine the concentration of the RbOH solution? g of KHP Volume of Base Required 0.5373 g 42.49 mL 0.5856 g 43.88 mL 0.5790 g 48.56 mL 0.5856 g 44.60 mL (Report your answer as "mean +/- std dev") M What is the percent relative standard deviation? % What is the 99% Confidence Interval for the concentration of the solution (population mean)?
- You want to measure the concentration of carbonate (CO32-) in a mildly basic solution by using an EDTA back titration. CaCO3 has a Ksp of 5x10-9. You add 50.00 mL of 0.3484 M CaCl2 to 500.0ml of sample and filter the solution to remove the precipitate. You then take 250.0 mL of the filtered solution and titrate with 0.1786 M EDTA. You require 23.72 mL to reach the endpoint. What is the concentration of carbonate in the original sample?Answer if true or false , if false, put the correct answer 1. There is 11.34 mmol in 0.2011 of 0.5604 M HgCl2 2. The x axis of sigmoidal titration curve is p function of the analyte or titrant 3.The concentration of the secondary standard is relative to a primary standard 4. There is 110 mmol in 79.8 mL of 0.1379 M NH4VO3 (116.98 g/mol)At equivalence point in the titration of M2+ with EDTA, A. [M2+] = [Y] B. [M2+] = [Y4-] C. No answer D. [M2+] = [MY2-] E. [MY2] = [Y]
- Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution.MnO4- + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2(g) + 8H2OCalculate the molar concentration of the KMnO4 solution.Titration of 50.00 mL of 0.04715 M Na2CO3 required 39.25 mL of potassium permanganate solution. 2MnO4- + 5H2C2O4+ + 6H+ 2Mn2+ + 10CO2 (g) + 8H2O Calculate the molar concentration of KMnO4 solution (6)The end point of a titration was reached after 22.2 mL22.2 mL of 0.050 M0.050 M disodium EDTA titrant was dispensed into a solution containing the zinc ion. Calculate the moles of disodium EDTA used. moles of disodium EDTA: = mol
- 2,5 ml volume has taken from an “hypothetic” solution which includes (3+) Sb and (3+) Fe and at the titration with 0.1004 N KMnO4, the wasted amount has found as 16,4 mL. The other 2,5 mL that has taken, has reduced with Zn after that, this 2,5 mL solution has titrates with the same KMnO4 solution solution and the wasted amount is 26,5mL. With these datas find the %concentrations of the ions at the solution.Calculate the pH at 0.0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 50.0 mL of 0.100 M NH3 with 0.100 M HCl. Pls answer thank youAt 40.0 mL 0.0050 M Sr2 + pH 10, 0.0100 M EDTA is titrated. Calculate the pSr for the equivalence point. Kfor (SrY) = 4.3x108, α4 = 0.35