Q: 4. The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL…
A: Acid-base titration: Titration is a method to determine the concentration of unknown acid by using…
Q: Titration of a 20.0 mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point.…
A: Molarity is defined as moles of solute present per liter kf a solution. molarity=moles of…
Q: A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence…
A: Molarity = Moles of solute/Volume of Solution(in L) Moles of Solute = Molarity x Volume of…
Q: Calcium oxalate, CaCO4, is very insoluble in water. What mass of sodium oxalate,Na2C2O4, is required…
A: Given: Volume of CaCl2 = 37.5 ml Molarity of CaCl2 = 0.104M
Q: A hydrochloric acid solution is standardized by titrating 0.4541 g of primary standard…
A: Given:The mass of primary standard tris(hydroxymethyl)aminomethane = 0.4541 g. Volume = 35.37 mL =…
Q: What is the concentration of NaCl in a solution if titration of 50.00 mL of the solution with 0.1500…
A:
Q: a 50.0mL sample of phosphoric acid, H3PO4, is titrated with 42.50 mL of .250 M NaOH. What is the…
A:
Q: Sodium hydroxide is a commonly used base in neutralization titrations. It is a normal procedure to…
A: Primary standards are reagents that are used for determining the concentration of the unknown. A…
Q: Example: Titration of o.2121 g of pure Na CO, (134.00 g/mol) required 43-31 mL of KMNO What is the…
A:
Q: What was the concentration of the weak acid?
A: Titration is the process of adding a measured volume of an acid or base of known molarity to an acid…
Q: Titration of a 25.00 mL sample of acid rain required 12.4 mL of 0.009879 M NaOH to reach the end…
A: Volume of sulphuric acid,V1 = 25.00mL Volume of NaOH,V2 = 12.4mL Concentration of NaOH,M2 =…
Q: 25.00 mL of a diprotic acid solution, whose concentration is unknown, is titrated with 12.54 mL of…
A: Given data: Volume of diprotic acid = 25.00 mL Volume of NaOH = 12.54 mL Concentration or molarity…
Q: Titration of a 10.0 mL sample of acid rain required 2.1 mL of 0.0824 M KOH to reach the end point.…
A: Given: Volume of sample is 10 mL. Volume of base is 2.1 mL. Molarity of base solution is 0.0824 M.…
Q: In a titration of KOH, 0.0725 g of KHP is weighed into a flask and dissolved in deionized water. The…
A: The molar mass of KHP is 204.22 g/mol. The number of moles of KHP can be calculated by dividing the…
Q: What is the percent of CaCO3 in an antacid given that a tablet weighed 1.3198 g reacted with 50.00…
A:
Q: Suppose that 4 g of solid Fe2O3 is added to 25 mL of 0.5 M HClsolution. What mass of Fe2O3 will…
A: Reaction of Fe2O3 with HCl take place as Fe2O3 + 6HCl -----> 2FeCl3 + 3H2O Given mass of…
Q: 2) A 50.0 mL sample of phosphoric acid, H3PO4, is titrated with 42.50 mL of 0.250 M NAOH. What is…
A: We define concentration as the amount of solute present in the solution. There are various ways in…
Q: 2. What weight of AgNO, is required to precipitate the chloride in 750 mg of 14.0 % BaCl,?
A:
Q: Suppose in a titration of a monoprotic acid a student needed 31.36 mL of 0.1105 M NaOH to reach the…
A: Answer :- 60.87 g/mol The Molar Mass of unknow monoprotic acid = 60.87 g/mol…
Q: 0.3043 g of pure KHP was weighed out and titrated to an end point with 15.12 mL of a NaOH solution…
A: Neutralization reaction between KHP (C8H5KO4) and NaOH is given below,
Q: You titrated a 21.00 mL solution of 0.0500 M oxalic acid with a freshly prepared solution of KMnO4.…
A: Given, Volume of oxalic acid = 21.00mL. Molarity of oxalic acid = 0.0500M. Volume of KMnO4 solution…
Q: What is the percent of Caco, in an antacid given that a tablet weighed 1.3198 g reacted with 50.00…
A: From given Percentage of CaCO3 and weight of CaCO3 in milligrams is calculated analytically with…
Q: Calculate the molar concentration for H2 SO4 when 41 mL of it was completely titrated by 10.9 mL of…
A:
Q: In an acid-base titration, you were given with a 0.1050 M sodium hydroxide solution placed in a…
A:
Q: During the standardization of NaOH , a drop of titrant was present on the tip of the buret when…
A: A titration is a procedure that uses a known concentration of a solution to calculate the…
Q: The titration of 0.9297 g of KHP (molar mass=204.227 g/mol) requires 41.02 mL of NaOH solution, What…
A: Given: Mass of KHP = 0.9297 g Volume of NaOH = 41.02 mL Molar mass of KHP = 204.227 g/mol
Q: In a titration reaction 20.00 mL of 0.500 M KOH reacts completely with 12.55 mL of sulfuric acid,…
A: Concentration of KOH = 0.500 M Volume of KOH = 20.00 mL (0.02L) Volume of sulfuric acid =…
Q: A 5.0 g impure sample of sodium carbonate was titrated with an acid. The sample utilized 31.55 ml of…
A: The relationship between normality and molarity of a substance follows: Normality = n × Molarity…
Q: 50.00 mL of a sulfuric acid solution, whose concentration is unknown, is titrated with 12.54 mL of…
A: Given: The Volume of sulfuric acid=50.00 mL=0.050 L Molarity of sulfuric acid=M mol/L The Volume of…
Q: Lactic acid, HC3H5O3, is a monoprotic acid that forms when milk sours. An 18.5 mL sample of a…
A: Molarity of the solution is equal to the number of moles of solute that can be dissolved in per…
Q: (4) If the end point of the titration was calculated to have occurred at a volume of 17.32 mL and…
A:
Q: Suppose 198.2 mg of a triprotic acid is titrated with 42.67 mL of 0.1126 M NaOH. What is the molar…
A:
Q: To do a titration (neutralization), a student started with 18.25 ml NaOH in a buret with a…
A:
Q: How many mL of the standard AgNO3 solution were used in the titration?
A: Solution: We have filled standard AgNO3 solution in burette and used for titration. In the given…
Q: What is the percent of CaCO3 in an antacid given that a tablet that weighed 1.3198 g reacted with…
A:
Q: A particular solution of 10 mL NaOH is supposed to be approximately 0.1000 M. To determine the…
A: This problem can be solved by using molarity equation which may be expressed as ; M1V1=M2V2
Q: An aqueous solution of barium hydroxide is standardized by titration with a 0.180 M solution of…
A: Given : Molarity of HCl = 0.180M Volume of base = 28ml Volume of acid = 26.4ml
Q: What is the titer(expressed in milligrams per milliliters) of a solution of oxalic acid dihydrate…
A: Given that - Volume of oxalic acid dihydrate Solution = 21.49 mL Mass of calcium oxide = 0.2203 g…
Q: A) Suppose you have a compound which has an empirical formula of C4H5N2O and molar massof 194 g/mol.…
A: “Since you have asked multiple questions, we will solve the first question for you. If you want any…
Q: In the titration of a volume of 50 mL of a hydrochloric acid solution with Solution density 1.045 g…
A: Density is measure of mass per unit volume. Density =MassVolume Molarity is the ratio of moles of…
Q: Suppose in a titration of a diprotic acid a student needed 31.84 mL of 0.1105 M NaOH to reach the…
A:
Q: An aqueous solution of hydroiodic acid is standardized by titration with a 0.108 M solution of…
A: given :- molarity of Ba(OH)2 solution = 0.108 M volume of Ba(OH)2 solution = 21.8 mL volume of HI…
Q: Adic acid (H2C6H8O4) is an important carboxylic acid compound for the industry in which nylon is…
A: The answer to both the questions can be given as:
Q: What is the concentration of NaCl in a solution if titration of 50.00 mL of the solution with 0.1500…
A:
Q: What is the mass-volume percent of acetic acid (CH3COOH) in a vinegar sample if 5.00 mL of the…
A: Given that - Volume of vinegar = 5.00 mL Molarity of NaOH = 0.1150 M Volume of NaOH = 30.50 mL…
Q: What is the molarity of the acid solution?
A:
Q: Suppose in a titration of a diprotic acid a student needed 25.64 mL of 0.1105 M NaOH to reach the…
A:
Q: I have trouble solving this becaue I don't know how to convert the 3g of a random chloride to mL.
A: The relation between chloride and AgNO3 in the form of balanced equation .
Q: A 15.00 mL solution of phosphoric acid (H3PO) is titrated with 11.20 mL of 0.155 M sodium hydroxide.…
A: Given data: Volume of acid = 15 mL Volume of base = 11.20 mL Molarity of base = 0.155 M Formula…
Q: Titration of a 25.00 mL sample of acid rain required 13.23 mL of 0.009567 M KOH to reach the end…
A:
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
- 1. Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, CH3COOH. For 20.0 cm3 of the vinegar, 26.7 cm3of 0.60 mol dm-3 NaOH solution was required.What was the concentration of acetic acid in the vinegar if no other acid was present? (Thanks for your help) A. 0.45 mol dm-3B. 0.60 mol dm-3C. 0.80 mol dm-3D. 1.60 mol dm-3Fast pls i will give u like for sure solve this question correctly in 5 min pls A student analyzed a CaCO3 antacid tablet by neutralizing the base with an excess of .60 M HCL and back-titrating the excess with .50M NaOH. He recorded the following data: mass of tablet = 1.3g amount of active ingredient CaCO3 per label = 500 mg volume of HCl added: 25 mL initial buret reading : 34 mL final buret reading : 47.4 mL If the concentration of the provided HCl is actually higher than .60M, would the determined mass of CaCO3 in the antacid tablet be too high or too low?A carefully weighed 280mg Calcium carbonate was used in the standardization of an EDTA solution. Initially, 25mL of the titrant was consumed but only after the addition of another 10mL of the titrant did an endpoint was visible.What is the MW of calcium carbonate?102 g/mol98 g/mL100 g/mL100 g/molWhat is the molar concentration of the standardized EDTA solution?0.08m0.8M0.08N0.08MHow many grams of EDTA (MW: 292 g/mole) is required to prepare 250mL of a 0.025M solution?0.18250.18521.8251.1852
- Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours. CH3COOC2H5 + OH- → CH3COO- + C2H5OH After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4. Answer the following: Calculate the number of moles of OH- that reacted with ethyl acetate. Calculate the number of moles of ethyl acetate in the 20.00 mL solution. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?A Medical Technology student was given a capsule of a multivitamins and she was asked to determine the % by mass (w/w) of ascorbic acid present in the capsule. The student analyzed the 2.001 g sample using volumetric titration. The following data was generated in the analysis: KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O C6H8O6 + I2 → C6H6O6 + 2I- + 2H+ Table 1. Standardization of KIO3 Molarity of Ascorbic Acid Standard Solution 0.03542 M Volume of Ascorbic Acid 25.00 mL Volume of KIO3 8.70 mL Molarity of KIO3 ______________M Table 2. Determination of Ascorbic Acid Concnetration Initial burette reading, KIO3 0.00 mL Final burette reading, KIO3 33.60 mL Volume consumed, KIO3 33.60 mL MM of Ascorbic Acid 176.12 g/mole choices 7.30% 30.1% 33.6% 32.5%
- 3I- + OCl- + 2H+ ----> I3- + Cl- + H2: A 25.00 ml sample of liquid bleach was diluted to 1000 ml in a graduated flask. A 25 ml portion of the diluted sample was pipetted into an Erlenmeyer flask and treated with excess KI to oxidize OCl- to Cl- and I3- was produced at the end of the reaction. The released I3 was titrated with 0.09892 M Na2S2O3 and 8.96 ml was spent to reach the turning point besides the starch indicator. What is the percent by weight/volume of NaOCl in the bleach sample? (NaOCl:74.44 g/mol)Titration of the I2 produced from 0.2645g of primary standard KIO3 required 45.36mL of sodium thiosulfate. Calculate the concentration of the Na2S2O3 in M. IO3- + 5I- + 6H+ --> 3I2 + 3H2O I2 + 2S2O32- --> 2I- + S4O62- KIO3 mw = 214.001g/molA sodium thiosulfate solution is standardized using pure copper as the primary standard. A sample of copper weighing 0.2624 g is dissolved in acid, excess KI is added, and the liberated iodine was titrated with 42.18 mL of sodium thiosulfate solution. Calculate the molarity of sodium thiosulfate solution. 2Cu2+ + 4I- → 2CuI + I2 I2 + 2S2O32- → 2I- + S4O62-
- The 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%A 25.0025.00 mL solution of 0.088300.08830 M NaINaI is titrated with 0.051500.05150 M AgNO3AgNO3. Calculate pAg+pAg+ following the addition of the given volumes of AgNO3AgNO3. The KspKsp of AgIAgI is 8.3×10−178.3×10−17. 37.40 mLpAg+=37.40 mLpAg+= VepAg+=VepAg+= 47.30 mLpAg+=47.30 mLpAg+=The thiourea (NH2)2CS in a 1.600 g sample of an unknown organic material was extracted into a dilute acid solution and was titrated with 42.50 mL of 0.0010 M Hg2+. Equation: 4 (NH2)2CS + Hg2+ → [(NH2)2CS]4Hg2+ Find the percentage thiourea present in the sample.