A 25.0 mL solution of 0.0460 M EDTA was added to a 49.0 mL sample containing an unknown...
Question:
A {eq}25.0 \ mL {/eq} solution of {eq}0.0460 \ M \ EDTA {/eq} was added to a {eq}49.0 \ mL {/eq} sample containing an unknown concentration of {eq}V^{3+} {/eq}. All {eq}V^{3+} {/eq} present formed a complex, leaving excess {eq}EDTA {/eq} in solution. This solution was back-titrated with a {eq}0.0490 \ M \ Ga^{3+} {/eq} solution until all the {eq}EDTA {/eq} reacted, requiring {eq}10.0 \ mL {/eq} of the {eq}Ga^{3+} {/eq} solution. What was the original concentration of the {eq}V^{3+} {/eq} solution?
Back Titration:
Complexometric titration involves complexing agents like EDTA. There are different types of EDTA titrations. In case of a back titration, an additional amount of EDTA is added to the solution. After the reaction, this additional EDTA is then titrated with another solution with a known concentration.
Answer and Explanation: 1
Become a Study.com member to unlock this answer! Create your account
View this answerGiven Data
- The volume of EDTA solution is 25.0 mL.
- The molarity of EDTA solution is 0.0460 M.
- The volume of {eq}{{\rm{V}}^{3 + }} {/eq} solution is...
See full answer below.
Ask a question
Our experts can answer your tough homework and study questions.
Ask a question Ask a questionSearch Answers
Learn more about this topic:

from
Chapter 24 / Lesson 7Learn about the titration equation and its uses in titration calculations. Find titration calculation examples covering how to calculate molarity from titration.
Related to this Question
- A 25.0-mL solution of 0.0670 M EDTA was added to a 34.0-mL sample containing an unknown concentration of V3. All V3 present formed a complex, leaving excess EDTA in solution. This solution was back-titrated with a 0.0440 M Ga3 solution until all the EDTA
- A 25.0 ml solution of .0600 M EDTA was added to a 57.0 ml sample containing an unknown concentration of V 3 + . All V 3 + present formed a complex, leaving excess EDTA in solution. This solution was back-titrated with a .0440 M G a 3 + solution until
- A 50 mL sample containing Ni2+ was treated with 25 mL of 0.05 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5 mL of 0.05 M Zn2+. What was the concentration of Ni2+ in the original solut
- The concentration of Ca2+ in an unknown solution was determined by titration with the EDTA solution standardized at 5.231 x 10-4 M. In the determination, 25.00 mL of the unknown solution was diluted to 250.00 mL with deionized water. A 25.00 mL aliquot of
- Before performing an analysis of calcium in a powdered cement sample with an EDTA titration, the EDTA must be standardized. A 50.0 mL known solution of 5.67 x 10-3 M Ca2+ required 38.10 mL of an unknown EDTA solution to reach its end point. Based on t
- Assume that 50 ml of 0.01 M EDTA solution were added to 25 ml of Co solution and the excess EDTA required 20 ml of 0.005 M Mg solution.
- A 15.00 mL sample of a standard solution containing 1 g of CaCO3/L required 8.45 mL of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaC
- 25.00 mL aliquots of the solution are titrated with EDTA to the Eriochrome Black T endpoint. A blank containing a small measured amount of Mg^{2+} requires 2.60 mL of the EDTA to reach the endpoint. An aliquot to which the same amount of Mg^{2+} is added
- A titration was performed to standardize an EDTA solution. Use the following data to calculate the molarity of the EDTA solution: 20.00 mL of a standard solution containing 0.02000 M Ca2+ required 22.
- You have been tasked with determining the concentration of Ca2+ in an unknown solution of calcium carbonate (CaCO3). You titrate 50.0 ml of the unknown solution with 0.0250 M EDTA. It requires 16.85 m
- A .600 g sample of CaCO_3 is dissolve in concentrated HCl (12M). The solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant. 1. What is the molar
- 1. A graduate measured out a 100 mL hard water sample and titrated it with 37.6 mL of 0.0100 M EDTA solution for a total hardness endpoint, and 29.3 mL of 0.0100 M EDTA solution for a calcium endpoint
- A 50.0 mL sample containing Cd^2+ and Mn^2+ was treated with 68.8 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 12.7 mL of 0.0170 M Ca^2+. The Cd^2+ was displaced from EDTA by t
- A 20.0 mL solution containing 0.250 M Ni^{2+} ion was titrated with 0.0500 M EDTA while the system was buffered at pH=4.00. Determine the volume of EDTA needed to reach the equivalence point.
- In the laboratory, you have stock solutions of TRIS (1.000 M) and EDTA (0.500 M). If you mix 30.0 mL of TRIS with 50.0 mL of EDTA and 20.0 mL H_20, what are the concentrations of TRIS and EDTA in the final solution?
- A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04978 M EDTA solution. The solution is then back titrated with 0.02398 M Zn2 solution at a pH of 5. A volume of 15.45 mL of the Zn2 solution was needed to reach the xy
- 0.4250 g of the dried calcium carbonate was diluted to 500 ml and 50 ml of this solution required 46.83 ml of an unknown EDTA solution.
- A 50.0 mL solution containing Ni^{2+} and Zn^{2+} was treated with 25.0 mL of 0.452 M EDTA to bind the metals. The excess EDTA required 12.4 mL of 0.0123 M Mg^{2+} for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to
- In the titration of 25.00 mL of a water sample, it took 19.840 mL of 3.525 x 10-3 M EDTA solution to reach the endpoint. The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express this hardness as
- A 1.000-mL aliquot of a solution containing Cu_2 and Ni_2 is treated with 25.00 mL of a 0.03146 M EDTA solution. The solution is then back titrated with 0.02115 M Zn_2 solution at a pH of 5. A volume
- In the titration of 25.00 mL of a water sample, it took 18.240 mL of 3.210x10-3 M EDTA solution to reach the endpoint. Calculate the number of moles of EDTA required to titrate the water sample.
- A 0.4505 g sample of CaCO3 was dissolved in the HCl and the resulting solution diluted to 25.00 mL in a volumetric flask. A 25.00 mL aliquot of the solution required 29.25 mL of EDTA solution for titr
- Titration of 50 mL of a wastewater sample with 0.0737 M EDTA yielded a calmagite endpoint after the addition of 12.13 mL of the EDTA. What is the hardness expressed in mg/L of CaCO_3?
- In the titration of 25.00 mL of a water sample, it took 19.290 mL of 3.765x 10-3 M EDTA solution to reach the endpoint. Calculate the number of moles of EDTA required to titrate the water sample.
- 25 mL of tap-water is titrated with 0.005 M EDTA. 7.95 mL of EDTA is required for the complete reaction. What is the molarity of that total water hardness in the 25 mL sample?
- A 25.00-mL aliquot of a standard calcium solution (M = 0.0068; 0.3412 grams of CaCO3 was added to a 500 mL flask) was titrated with a previously prepared EDTA solution. In the titration, 32.55 mL of the EDTA solution was required to reach the end point. W
- A 48.30 mL aliquot from a 0.480 L solution that contains 0.360 g of MnSO4 (Fw = 151.00 g/mol) required 41.8 mL of an EDTA solution to reach the endpoint in a titration. What mass (in milligrams) of CaCO3 (FW = 100.09 g/mol) will react with 1.85 mL of the
- A 0.4505g sample of CaCO_3 was dissolved in HCl and the resulting solution was diluted to 250ml in a volumetric flask. A 25 ml aliquot of the solution required 24.25 ml of an EDTA solution for the tit
- A 0.5118 g sample of CaCO_3 is dissolved in concentrated HCl (12 M) the solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant. How much molarity of the Ca^{2+} in the 250 mL of the solution?
- If the ratio of water drops to EDTA is 6.7 for the known sample and 10 for the unknown sample what is the concentration of the unknown sample? (The concentration of the known sample is 315 ppm) a. 15
- A 0.5118 g sample of CaCO_3 is dissolved in concentrated HCl (12 M) the solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant. How many moles of Ca^{2+} is in a 25 mL aliquot of the solution?
- Calculate the total amount (in moles) of Ca and Mg presented in a sample of water that required 12.86 ml of 0.0005 M EDTA solution to completely complex them.
- A corrosion technologist pipetted a 100.00 mL hard water sample and titrated it with 37.64 mL of 0.01 M EDTA solution for a total hardness endpoint and 29.32 mL of 0.01 M EDTA solution for a calcium endpoint. Calculate the concentration of magnesium and c
- A 0.3205g sample of CaCO3 was dissolved in HCl and the resulting solution diluted to 250.ml in a volumetric flask. A 25 mL sample of the solution required 18.75mL of EDTA solution for titration to the
- From standardization, you determine your EDTA solution is 0.0115 M. It takes 25.11 mL of your EDTA solution to reach the endpoint when titrating 50.00 mL of your unknown. What is the ppm of calcium carbonate in the unknown? The molar mass of calcium ca
- If the endpoint is reached when 50.00 mL of a tap water sample is titrated with 12.00 mL of 5.00 x 10-3 M EDTA solution, what is the hardness of the water sample in ppm?
- Before performing an analysis of calcium in a powdered cement sample with an EDTA titration, the EDTA must be standardized. A 50.0 mL known solution of 5.67 times 10^{-3} M Ca^{2+} (aq) required 38.10
- You use 1.0093 g of calcium carbonate to make your standard solution. It takes 22.33 mL of your EDTA solution to reach the endpoint when you titrate 25.00 mL of your standard calcium solution. What is the molarity of the EDTA solution? The molar mass of
- A 25.00 mL sample of seawater was titrated for total hardness, the combined concentrations of magnesium and calcium ion, using 12.60 mL of an EDTA solution that was 0.1000 M. Assuming that the stoichiometry of the reaction between EDTA and both the ions i
- A corrosion technologist pipetted a 100.00 ml hard water sample and titrated it with 37.64 ml of 0.01 M EDTA solution for a total hardness endpoint, and 29.32 ml of 0.01 M EDTA solution for a calcium endpoint. Calculate the concentration of magnesium ion
- A solution was prepared by dissolving 0.450 g of MgSO4 in 0.500 L of water. A 50.0 mL aliquot of this solution consumed 37.60 mL of EDTA solution in a titration. How many mg of CaCO3 will react with 1.00 mL of the EDTA solution?
- How would you make 250 mL of a 15% EDTA solution? How much of the solution should be used to make 500 mL of a 0.1% EDTA solution?
- The 25.00 mL sample of tap water required 12.82 mL of 0.0100 M solution of EDTA. Calculate the hardness
- A student titrates a 50.0 mL sample of water with a 0.005920M EDTA solution. If the titration requires 10.05 mL of EDTA to reach the endpoint, what is the total concentration of Mg^{2+} and Ca^{2+} in the water sample in moles per liter?
- Calculate the Concentration of Ni 2 + in a solution which prepared as mixing 50.0ml 0.0300 M Ni 2 ? and 50.0mlml 0.0500 M EDTA at pH= 3, kA= 2.5x10-11 KNiY=4.2x1018
- In a titration of a 31-mL water sample, it took 22.5 mL of 0.200 M EDTA to reach the endpoint. How many moles of EDTA is in 22.5 mL of 0.200 M EDTA?
- A 48.60 mL aliquot from a 0.510 L solution that contains 0.420 g of MnSO_4 (FW = 151.00 g/mole) required 39.1 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO_3 (FW = 100.09 g/mole) will react with 1.01 mL o
- 100mL of the water sample is titrated with 0.0200M EDTA solution. It takes 9.0mL to reach the endpoint. How many moles of EDTA are used in the titration?
- A 45.50-mL sample of a phosphoric acid solution of unknown concentration is titrated with a 0.3161 M calcium hydroxide solution. A volume of 34.99 mL of the base was required to reach the endpoint. What is the concentration of the unknown acid solution? W
- Calculate the concentration of 17.0 ml of EDTA if you use 25.0 ml of 0.05 M Zn^2+.
- How many grams of EDTA is needed to prepare a solution of 250mL of 0.010M (mol/L) EDTA? The molar mass of EDTA is 372.244 g/mol.
- Calculate the amount of disodium EDTA dihydrate (Na2H2Y . 2H2O), MM = 372.2 g/ mol,required to prepare 250 mL of 0.016 M EDTA solution. Show your calculation.
- In the titration of 25.00 ml of a water sample, it took 20.740 ml of 4.305 x 10^{-3}M EDTA solution to reach the endpoint. Convert the number of ml of bottled water used in each sample titration to Liters. (enter your answer with 3 significant figures)
- A solution was prepared by dissolving 0.450 g MgSO_4 (FW = 120.363 g / mol) in 0.500 L of water. A 50.0 mL aliquot of this solution consumed 37.60 mL of EDTA solution in a titration. How many milligrams of CaCO_3 (FW = 100.089 g / mol) will react with 1.0
- Tris-acetate EDTA (TAE) buffer is sold at 50X concentration. You will need 25 L of this solution at a concentration of 1X for use during agarose gel electrophoresis in the lab. How would you make this solution?
- A 0.5118 g sample of CaCO_3 is dissolved in concentrated HCl (12 M) the solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant. How many moles of CaCO_3 are used (MW=100.1)?
- Determine the total water hardness in a 25.00 mL sample of water that required 12.45 mL of 0.01111 M EDTA for titration.
- Calculate the amount of , required to prepare 500 mL of 0.01 M EDTA solution.
- 25.0 mL tap water is titrated with 0.005 M EDTA. 7.95 mL EDTA is required for a complete reaction with water hardness. What is the molarity of the total water hardness?
- 10.0 ml of a Cu^{2+} solution of unknown concentration was placed in a 250 ml Erlenmeyer flask. An excess of KI solution was added. Indicator was added and the solution was diluted with H_2O to a total volume of 75 ml. The solution was titrated with 0.20
- Suppose you titrated a sample of distilled water with EDTA, approximately what volume of EDTA would be required?
- If 75 mL of a hard water sample required 23.00 mL of 0.200 M EDTA, calculate the total hardness.
- 10.0 mL of a Cu^{2+} solution of unknown concentration was placed in a 250 mL Erlenmeyer flask. An excess of KI solution was added. Indicator was added and the solution was diluted with H_2O to a total volume of 75 mL. The solution was titrated with 0.25
- Convert the number of mL of bottled water used in each sample titration to liters. The titration of 25.00 mL took 19.290 mL of 3.765x10-3 M EDTA solution to reach the endpoint.
- A 50.00 mL sample of groundwater is titrated with 0.0900 M EDTA. Assume that Ca_2 accounts for all of the hardness in the groundwater. If 11.40 mL of EDTA is required to titrate the 50.00 mL sample, w
- A 0.2481 g sample of marble was dissolved in 100 mL, and a 10.00 mL aliquot of the solution was titrated to an end point with 23.56 mL of 0.01052 M EDTA solution. What is the molecular weight?
- Tris-Borate-EDTA, or TBE, is commonly made as a 5x solution. What volumes of 5x TBE and water are required to make 500 ml of a 0.5x solution, which is often used in electrophoresis?
- If you are titrating 25.00 mL of a 1.00 mg/mL CaCO3 solution, how many mL of 0.010 M EDTA solution will be required to reach the equivalence point?
- Calculate the concentration of EDTA. -25 mL of 0.01004M Zn^2+ -Titrated with 12.15mL of EDTA (1:1 ratio) Why isn't M1 0.01004 M so that 0.01004M * 25mL = M2 * 12.15mL?
- In this experiment an EDTA titration was performed. Another common type of titration is an acid-base titration. Specifically, a neutralization titration can be used to determine the concentration of a
- A 50.00 mL sample of groundwater is titrated with 0.0350 M EDTA. Assume that Ca2+ accounts for all of the hardness in the groundwater. If 14.50 mL of EDTA is required to titrate the 50.00 mL sample, what is the hardness of the groundwater in molarity and
- Provide a step-by-step procedure for preparing a 500 mL stock solution of 2.0 M Tris and a 500 mL stock solution of 0.5 M EDTA. Then, explain how you would use these stock solutions to prepare a 100 mL Tris-EDTA buffer (TE buffer) with a final concentrati
- Suppose you have completed a titration of a water sample. Your average titre value is 20 mL,concentration of EDTA was 0.01 M and your aliquot of sample was 50mL. Calculate the total water hardness (as equivalent CaCO3 in ppm) of the water sample. HINT: re
- A 490 ml sample of an H_3 PO_4 solution of unknown concentration is titrated with a 1.100 times 10^{-2} M Na OH solution. A volume of 7.32 ml of the Na OH solution was required to reach the equivalence point. What is the concentration of the unknown H_3 P
- A titration is performed as follows: *4.004 mL of Fe2+ solution of unknown concentration is charged into a 100 mL beaker *45 mL of deionized water is added *the solution is titrated with 0.695 M Ce4+ solution is added *the equivalence point is reached aft
- For rxn 1, 10.0 mL of a solution of unknown concentration was placed in a 250 mL Erlenmeyer flask. An excess of KI solution was added. An indicator was added and the solution was diluted with to
- Calculate the weight of the disodium EDTA necessary to create 100 mL of your standard EDTA solution (0.01 M EDTA).
- 10.0 mL of a solution of H 3 P O 4 of unknown concentration are titrated with 0.100 M N a O H . 45.0 mL of the 0.100 M N a O H are required to reach the endpoint of this titration. What is the molar concentration of H 3 P O 4 in the original solut
- Provide the step by step solution. 1. How many milliliters of 0.0440 M EDTA is required to react with 50.0 mL of 0.0200 M Cu^{2+}? 2. With 50.0 mL of 0.0200 M Sc^3+?
- A titration is performed as follows: 8.367 mL of Fe2+ solution of unknown concentration is charged into a 100 mL beaker. 45 mL of deionized water is added. The solution is titrated with a 0.803 M Ce4+ solution. The equivalence point is reached after 23.58
- In an experiment, a 1.000 mL sample of an unknown solution is diluted with 50 mL of distilled water and about 3 mL of pH 10 buffer solution. Given the number of moles of the unknown solution used, what is the volume of the original unknown solution after
- A 5 mL sample of a solution of K_2CO_3 was diluted with water to 25 mL. A 15 mL sample of the dilute solution was found to contain 5M CO_3^{2-}. What was the concentration of K_2CO_3 in the original undiluted solution?
- 05 mL of sample water needs 10.9 mL .01M EDTA solution. What is the accurate Hardness of the sample in CaCO_3 (mg/L)?
- A sample of 0.250 g of nickel complex was titrated with 0.150 M EDTA. If the titration of the nickel sample required 7.5 mL of EDTA to reach the equivalence point, what is the percent nickel in the sa
- In a titration of a 31-mL water sample, it took 22.5 mL of 0.200 M EDTA to reach the end point. Approximately how many total moles of Ca2+ and Mg2+ are present in the water sample?
- You weigh out 0.966 g of an unknown sample and dissolve this in 50 mL of deionized water. Then, you titrate this solution with a standardized HCl solution whose concentration is 0.1125 M. 34.42 mL of the HCl titrant solution is required to reach the secon
- A 10 mL of unknown solution to be analyzed is diluted to 100.0mL, and then 10mL of that solution is titrated with the permanganate solution. Why not skip all the dilution process, and just pipet out 1ml of the original unknown solution and titrate it dire
- The experiment is about the determination of water hardness. Please answer as many as you can. Define hard water. What is EDTA? Is it a primary or secondary standard? Why was ammoniacal buffer used in this titration? List the concentration units: ppm,
- Dissolved 0.7500 g calcium chloride dihydrate and impure, in distilled water and the volume was elevated to 200.00 ml in a volumetric flask. A rate of 10.00 ml of the same consumed the titration 18.9 ml 0.0125 M EDTA using CALCON as indicator. What is the
- How much Tris-Acetate-EDTA (TAE) stock buffer (3.25 M) is required to make 400 ml of solution that has a concentration of 250 mM? How much water?
- 20.0 ml of solution containing 10.0 ml of 1.0 M NaX (unknown base) and 10.0 ml of 1.0 M HX (unknown acid) was diluted with 50.0 ml water. The pH is 4.92. (a) What is the concentration of HX after dilution and X- after dilution? (b) What is the equilibriu
- Calculate the water hardness as ppm CaCO_3 unknown sample is 50 mL it is titrated with 25.30 mL 0.01 M EDTA
- A 6.06 g sample of a solid containing Ni is dissolved in 20.0 mL water. A 5.00 mL aliquot of this solution is diluted to 100.0 mL and analyzed in the lab. The solution was determined to contain 5.41 ppm Ni. 1) Determine the molarity of Ni in the 20.0 mL s
- How do I find the edta molarity?
- 23.02 mL of a 0.1930 \dfrac{mol}{L} EDTA solution is required to titrate a sample to the equivalence point. How many moles of divalent metal cations were in your sample?
- A 0.18 g sample of the unknown was placed in a flask and dissolved in 125 mL of deionized water. A solution of HCl (0.13 M) was used to titrate the sample, and 16 mL was required to reach the endpoint
- A 4.58 g sample of a solid containing Ni is dissolved in 20.0 mL water. A 5.00 mL aliquot of this solution is diluted to 100.0 mL and analyzed in the lab. The solution was determined to contain 5.83 ppm Ni. Determine the molarity of Ni in the 20.0 mL solu
- A solution of 0.0500 M HCI is used to titrate 28.0 mL of an ammonia solution of unknown concentration. The equivalence point is reached when 15.5 mL HCI solution have been added. (Assume K_w = 1.01 ti
- The chromium (density = 7.10 g/cm 3 ) plated on a 9.75 cm 2 surface was dissolved with HCl and diluted to 100.00 mL. A 25.00 mL aliquot was buffered to pH 5, and 50.00 mL of 0.00862 M EDTA was added.Titration of the excess chelating reagent required 7.36
- Suppose you dissolved 0.7500 g calcium chloride, dihydrate and impure, in distilled water, and the volume was elevated to 200.00 mL in a volumetric flask. A rate of 10.00 mL of the same consumed the titration 18.9 mL 0.0125 M EDTA using CALCON as an indic