A 100-mL sample of hard water is titrated with the EDTA solution in Problem 2. The same amount of Mg2+ is added as previously, and the volume of EDTA required is 31.84 mL.
a. What volume of EDTA is used in titrating the Ca2+ in the hard water? _____________mL
b. How many moles of EDTA are there in that volume? _____________ moles
c. How many moles of Ca2+ are there in the 100 mL of water? _____________ moles
d. If the Ca2+ comes from CaCO3, how many moles of CaCO3 are there in one liter of the water? How many grams of CaCO3 are present per liter of the water? _____________ mol / L _____________ g / L
e. If 1 ppm CaCO3 = 1 mg per liter, what is the water hardness in ppm CaCO3? _____________ ppm CaCO3
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Chapter 28 Solutions
Chemical Principles in the Laboratory
- A 15.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 300 mL with water. A 20 mL aliquot was titrated with 5.02 mL of 0.5352 M HCl to reach PHP endpoint. Another 20 mL aliquot was titrated to the BCG endpoint, using up 18.87 mL of titrant in the process. Note: Answer in two decimal places only. If there is no answer, type in 0.00. Use the molar masses 105.989 ?/??l ??2cO3 and 84.007 g/mol NaHCO3 indicated. The volume of titrant needed to neutralize NaOH is _________ mL.The volume of titrant needed to neutralize Na2CO3 is _____________ mL.The volume of titrant needed to neutralize NaHCO3 is _____________ mL. The mass of NaOH is ___________ g.The mass of Na2CO3 is __________ g.The mass of NaHCO3 is __________ g. The percent weight of NaOH is __________ %.The percent weight of Na2CO3 is __________ %.The percent weight of NaHCO3 is __________ %.arrow_forwardA 50.00-mL aliquot of solution containing 0.310 of MgSO4 (FM 120.37) in 0.500 L required 37.77 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution? Answer in 3 significant figures. Do not include the unit. need asaparrow_forwardA solution contains 1.694 mg of CoSO4 per milliliter. Calculate the volume of 0.08640M EDTA needed to titrate a 25.00-mL aliquot of this solution.Molar Mass: CoSO4 = 155.0 with solution please :))arrow_forward
- You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. A) How many milliliters of EDTA solution were used to titrate hardness that actually came from the unknown? B) How many moles of EDTA reacted with hardness-causing ions from the unknown sample? C) How many moles of hardness-causing ions were present in the unknown sample? D) Assuming that the total hardness of water is due to CaCO3, how many grams CaCO3 does it correspond to? E) What is the total hardness of the unknown water in ppm CaCO3?arrow_forwardAsking for lab help Fe3+ (aq) + SCN- (aq) = FeSCN2+ (aq) + heat pale yellow redWe have a solution at equilibrium after mixing 10 drops of 0.10M FeCl3 (aq), 10 drops 0.10M KSCN (aq), and then adding enough deionized water to bring the whole thing up to 25mL. We then added 2mL of that solution to a test tube. If I add 2.0mL of deionized water to that test tube, following Le Chateiler's Principles, how should the equilibrium be expected to change? Please give a detailed explination as to why.arrow_forward100 mL of mineral water containing Mg2+ and Ca2+ is taken and titrated with 86.65 mL of 0.06120 M EDTA. NH4F was added to the second 100 mL portion taken from the mineral water to mask the magnesium in the sample as MgF2, and when this sample was titrated with the same EDTA solution, 38.56 mL was spent. Find the concentration of CaCO3 and MgCO3 in the sample in ppm.arrow_forward
- The water hardness of tap water has been determined to be 120 (= 120 mg CaCO3 per 1 L of water). After the tap water has run through the ion exchange column, titration of a 20.00 mL sample used only 0.70 mL of 0.0100 M EDTA, to reach the endpoint. Calculate the effectiveness of the ion exchange column: what is the % of the calcium, removed by the ion exchange column? Provide your answer in % and without decimal. ( Please type answer note write by hend)arrow_forwardA solid sample containing chloride and with a mass equal to 6.000 g was dissolved in water and transferred to a 100.0 ml flask. A 50.0 mL aliquot of this solution was diluted to 250.0 mL. In the titration of 20.0 mL of the diluted solution, 32.0 mL of 0.2000 mol L-1 AgNO3 standard solution were used. Calculate the % of impurities in the analyzed sample. Data: Cl = 35.45arrow_forwardAt 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 = 0.35arrow_forward
- At 40.0 mL 0.0050 M Sr2 + pH 10, 0.0100 M EDTA is titrated. Calculate the pSr for the equivalence point. Arm (SrY) = 4.3x108, α4 = 0.35 A. 5.37 B. 4.69 C. 5.33 D. 5.55arrow_forwardFollowing the monograph procedure, determine the weight in grams of sodium carbonate (MW-106 g/mol) used to standardize a 0.987 N sulfuric acid solution. 1. What specific type of titration is involved? A. Direct Acidemetry B. Direct Alkalimetry C. Residual Acidemetry D. Residual Alkalimetry 2. What is the analyte- titrant chemical reaction involved? A. Strong Base+Weak Acid B. Weak Acid+Strong Base C. Strong Acid+Weak Base D. Weak Base+Strong Acid 3. What is the preferred indicator for the chosen type of reaction above? A. Methyl Orange B. Penolphtalein C. Methyl Red 4. Consider that the burette was completely filled to the 0mp mark before titrating. What is the volume of titrant consumed based from the image below? A. 22.9 mL B. 22.2 mL C. 21.3 mL D. 22.7 mL 5. What is the unknown weight (grams) the problem? Your Answer:arrow_forwardThe zinc contained in a 0.7555-g sample of foot powder was titrated with 21.27 mL 0.01645 M EDTA. Find the percent Zn (MM=65.41) in the sample. Answer: 3.029%arrow_forward