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What would the ppm of CaCO3 be in a 15 mL sample that was titrated with...
1. A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA solution. What is the EDTA molarity? 2. A 250.0 mL water sample requires 30.85 mL of the EDTA solution from question 1 to reach the calmagite endpoint. What was the molarity of the hard metal ions in the water sample? 3. If the metal ions in the water sample of question 2 are assumed to be Ca2+ from CaCO3, express the concentration in ppm...
A 50.00 mL aliquot of a hard water sample is titrated with 15.00 mL of 0.01280 M EDTA. A second 50.00 mL aliquot portion is made sufficiently alkaline with NaOH to precipitate the Mg ions. The precipitate was filtered and the filtrate was titrated with 10.00 mL of 0.01280 EDTA. Calculate the ppm Mg and Ca in the water sample.
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30 Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10....
Calculations of hardness as CaCO3 in ppm and the Standard deviation and % Relative deviation. I need help step by step on how to calculate these. Trial 1 50.00 33.52 Trial 2 Trial 3 50.00 31.92 Trial 4 Volume of water sample used (mL) Volume of EDTA titrant used (mL) Average Conc. of 50.00 32.47 EDTA (M)4 x10 Hardness as CaCOs (ppm) Average Hardness as CaCOs (ppm) Standard Deviation %Relative Standard Deviation 95% Cl Trial 1 50.00 33.52 Trial 2...
QUESTION 4 A 500.0 mL water sample is titrated with 15.00 mLof 0.1040 M EDTA. How many mg of Caco3 are in the sample?
23. A 25.00 mL water sample is titrated with a 0.0120 M EDTA solution. The equivalence point is reached when 17.0 mL of the EDTA is added. The hardness in the sample was ppm CаCОз. a) 817 b) 219 c) 120 d) 346 e) none of the above
four drops of indicator was added to a 50.00 mL sample of pond water. The sample was titrated with an EDTA solution that was determined to be 0.01421M. The initial burette reading was 1.16mL. When the color or the solution changed from red to blue, the burette reading was 23.96mL. a) calculate the volume of the EDTA solution added during the titration b) calculate the molarity of calcium in the water sample c) express the concentration of the water sample...
Chapter 15 Determination of the Hardness of Water 3, A 100-mL. sample of hard water is titrated with the EDTA solution in Problem 2. The same amount of Mg' is added as previously, and the volume of EDTA required is 31.84 mL a What volume of EDTA is used in titrating the Ca" in the hard water? mL h How many moles of EDTA are there in that volume? moles c. How many moles of Ca are there in the...
A mixture of CaCO3 and MgCO3 weighing 1.00 g consumes 21.86 mL of 1N HCl, with methyl orange as indicator. One gram of this same sample is dissolved in HCl and diluted to one liter. 100 ml is consumed 21.87 mL of a solution of EDTA. EDTA solution with a water hardness is determined. 100 mL of water are taken and consumed 10 mL of EDTA. Express the hardness in ppm of Ca.
Hardness in groundwater is due to the presence of metal ions, primarily Mg2 and Ca2t. Hardness is generally reported as ppm CaCO3. To measure water hardness, a sample of groundwater is titrated with EDTA, a chelating agent, in the presence of the indicator eriochrome black T, symbolized here as In. Eriochrome black T, a weaker chelating agent than EDTA, is red in the presence of Ca2 and turns blue when Ca2+ is removed. red blue Ca(ln)2+-EDTA → Ca(EDTA,-+ + In...