Moles = Molarity × volume(L)
= 0.003232×0.01086
= 3.51×10-5 mol
So moles of CaCO3 is 3.51×10-5 mol
Question 15 (1 point) Saved If a titration trial requires 10.86 mL of EDTA solution to...
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...
Question 4 (1 point) An indirect EDTA titration was performed on a water sample to determine the sulfate content. First, the analyst measured the hardness of the water by directly titrating a 50 mL aliquot with 0.00515 M EDTA. If it took 14.40 mL of the EDTA solution to reach the endpoint, what is the hardness, or concentration (in M) of metals in the water sample?
A titration was performed to standardize an EDTA solution. a) A 25.00 mL aliquot of a standard solution containing 0.01500 M Ca2+ required 42.87 mL of EDTA to reach the endpoint. The molarity of the EDTA solution is: ________ M b) That same EDTA solution was then used to titrate an unknown hard water sample. A 40.00 mL aliquot of unknown hard water required 34.21 mL of EDTA solution to reach a distinct endpoint. The concentration of Ca2+ ions (assuming...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
Please show steps for 32 and 33.
Data Table 1: EDTA Titration Volume Trial Initial EDTA Volume (mL) Final EDTA Volume (mL) Total Volume EDTA Used (mL) 1 9 mL 8.6 mL .3mL 2 8.6 6 2.6 mL 3 6 4.2 1.8mL Average Volume of EDTA Used (mL) 1.6mL Data Table 2: Water Hardness Average Volume of EDTA Used (mL) Concentration Ca2+ lons Per Liter of Water (mol/L) Water Hardness (ppm CaCO3) 1.6mL 32 Using the following equation, determine the...
The titration of 23.30 mL of HCl solution of unknown concentration requires 13.00 mL of a 0.200 M NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in M? Express your answer in moles per liter to three significant figures. M(HCl)M(HCl) = _____________
Part A The titration of 23.40 mL of HCl solution of unknown concentration requires 12.18 mL of a 0.140 M NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in M? Express your answer in moles per liter to three significant figures. We ΑΣΦ ? M(HCI) M Submit Request Answer
The end point of the Zn2+−EDTA titration was observed after 21.50 mL of 0.0500 M EDTA solution was dispensed. Determine the number of moles of zinc ion present in the sample. number of moles of zinc ion =
Question 15 10 points Save Answer In a titration, a sample of HCl required 18.4 mL of a 0.7971 M NaOH solution to reach the endpoint. Calculate moles of NaOH dispensed. (M=mol/L; mol=MxV) HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Answer to 4 sig figs. Do not include unit, mol.
The titration of 10.00 mL of an HCl solution of unknown concentration requires 50.75 ml of 0.0621 M KOH solution to reach equivalence. What is the concentration of the HCI solution in M (mol/L). give your answer to two decimal places