Determination of Water Hardness

A 25.00 mL unknown water sample is titrated with a standardized 0.0140 M EDTA solution.  It is determined that 8.54 mL of EDTA is required to reach the end point.  The blank titre was determined to be 1.50 mL.  Determine the concentration of Calcium in the water sample in ppm.  (mwt. of CaCO3 = 100.0892 g/mol).

Given the following information complete the last (blank) data table

Using the two images attached please write one paragraph EXPLAINING THE ANALYSIS Please please please answer as fast as possible Please please try to answer

A student titrated a 25.00 mL sample of water with 0.0200 M EDTA. The titration required 20.40 mL of EDTA. Calculate the ppm hardness in the water sample

Determine the concentration of acetic acid and vinegar

A 15.00 mL sample of hard water was titrated with a 0.0150 M EDTA solution. The titration required 26.72 mL of EDTA to reach the end point. What is the total hardness concentration of the water sample in ppm?

The amount of calcium in physiologic fluids can be determined by a complexometric titration with EDTA. In one such analysis, a 0.100-mL sample of blood serum was made basic by adding 2 drops of NaOH and titrated with 0.00376 M EDTA, requiring 0.135 mL to reach the end point. Report the concentration of calcium in the sample as miligrams of Ca per 100 mL. (Ca = 40.078 amu).

Please write a paragraph of the objective stated using the two images attached below Please please answer as fast as possible thank you

Using the two images attached please write one paragraph explaining the analysis Also write another paragraph of the complete method in your own words Please please please answer as fast as possible

You are asked to determine the amounts of calcium ion in a 200 mL water sample using EDTA to titrate the sample. It took 45.5 mL of 0.3 M EDTA to reach endpoint. What is the mass (in grams) of calcium ion in the water sample? Hint: Find out moles of EDTA used to reach endpoint, relate to mol of calcium ion, convert mol of calcium to gram of calcium. The molar mass of Ca is 40.08 g/mol. Round and report your answer to the SECOND decimal place. Numeric value only, no unit.

A 50.00 mL aliquot of standard CaCO3 (0.3834g CaCO3 per liter) consumed 42.35 mL of EDTA solution. The same EDTA standard solution was used for the titration of a 3.00 mL water sample. This required 30.56 mL of the titrant. Calculate the molar concentration of EDTA standard solution. What is the total hardness of water sample (then after solving explain how it come up to the final answer.)

Please answer letter c

In a hardness of water determination like the one performed in this experiment, it was found that a 25.00-mL sample of water required 12.25mL of 0.0128 M EDTA solution for complete titration. What is the concentration (molarity) of Ca2+ ions in the water sample? What is concentration of CaCO3 (in mg/L) in the water sample? Describe the hardness level of the water using the U.S. government standards.

Concentration sodium thiosulfate solution used: 0.056 mol L-1 ACCURATE TITRATION VOLUME OF SODIUM THIOSULFATE SOLUTION USED Volume titration 1, (ml) Volume titration 2, (ml) Your Group 15.80 15.75 Group 1 16.05 15.70 Group 2 16.30 16.25 ANALYSIS Average volume used, (mL) Standard deviation Number of moles sodium thiosulfate used, (mol) Number of moles calcium iodate in solution, (mol) Volume of calcium iodate used, (mL) Concentration of saturated calcium iodate solution (mol L-¹) Unrounded value 15.9750 0.26220 Rounded value 15.97 0.262 10.0

10.0 pH 8.15, V= 8.6 mL 8.0 pH 6.0 pH= 3.88 D%3D4,3ml. 4.0 2.0 2.0 4.0 6.0 8.0 10.0 Volume (mL) The figure shows the data obtained in the pH titration of a biochemical substance that is a weak acid. Determine the pK, of this compound. exact number, no tolerance

Preparation and Standardization of KMnO4 solution Experimental data Complete the table below. Trial 1 0.2001 g Trial 2 0.2065 g Trial 3 Weight of sodium oxalate (Na2C2O4, MM= 134 g/mol) Titration data 0.2050 g Final reading Initial reading 29.86 mL 0.00 mL 30.66 mL 30.52 mL 0.00 mL 0.00 mL Total vol. of KMNO4 used Computed Molarity of KMNO4 solution Mean Molarity Computed Normality of KMNO4 Mean Normality of KMNO4 solution Reaction Involved: Calculations:

“This EDTA solution was used to titrate a 100 ml sample of water, adjusted to pH 10, using EBT as the indicator. The titration required 11.23 ml to reach a blue endpoint. Calculate the total concentration of Ca^2+ and Mg^2+ in the water. (The EDTA reacts i a 1:1 mole ratio with both Ca^2+ and Mg^2+.) Express the concentration in moles of metallic cation per liter.” I have found the molarity from part a.) to be 9.9788 x 10^-3. How do I solve part b?

I need help to calculate this data please thanks a lot

If you could help with the below pre-lab questions please. PRELABORATORY ASSIGNMENT Define the term "salt.'  The solubility of AgCl(s) in water is 1.3 x 10-5 M.  Calculate its Ksp.  The solubility of CsF2(s) in water is 2.1 x 10-4 M.  Calculate its Ksp. The of Pb(OH)2 is 4 x 10-15.  Calculate the molar solubility of this compound.

What is the mass of pure dry CaCO3 (MW: 100.0869 g/mol) standard used in the standardization of EDTA solution? Assume that 1.758 M EDTA solution was computed based on the titration of CaCO3 standard. The volume of titrant needed to reach the endpoint was 35.00 mL.   Ca2+ + Y4- →  CaY2-         0.1500 g                       5.630 g     6.158 g                                     0.6158 g

A student performed three KHP titrations in Part 1 in order to standardize their NaOH solution. They reported the following measurements: Mass of KHP + Vial Mass of 'Empty' Vial Final Burette Reading Initial Burette Reading Determination #1 Determination #2 Determination #3 14.1992 g 14.0219 g 14.0982 g 13.3149 g 13.2301 g 13.2779 g 39.21 mL 0.49 mL 35.46 mL 0.86 mL 37.42 mL 1.59 mL How many moles of NaOH did they use in their third titration? Report your answer to the correct number of significant figures and only report the numerical value (no units).

Redox Titration 2 KMnO4 + 5 (COOH)2 + 3 H2SO4 -------> 10 CO2 + 2 MnSO4 + K2SO4 + 8 H2O

Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass…