1) 0.121 grams of oxalic acid dihydrate (MW=126.068 g/mole) is dissolved in wate
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1) 0.121 grams of oxalic acid dihydrate (MW=126.068 g/mole) is dissolved in water in an Erlenmeyer flask. Remember that each mole of oxalic acid dissolved in water results in 2 moles of hydronium ion so it takes 2 moles of NaOH to neutralize one mole of oxalic acid. It requires 14.22 mLs of your prepared sodium hydroxide solution to reach the endpoint of the titration. What is the molarity of your NaOH solution? 2) You are given an impure sample of oxalic acid dihydrate (MW=126.068 g/mole) and asked to determine the percent of oxalic acid dihydrate present in the sample. You place 0.110 grams of the unknown sample in an Erlenmeyer flask and dissolve it in distilled water. Titration with 0.140 M standardized NaOH solution requires 10.10 mLs to reach the titration endpoint. What is the percent of oxalic acid dihydrate in the sample? 3)A 20.00 mL of a sample of vinegar of unknown concentration was titrated with 21.63 mL of 0.102 M sodium hydroxide solution. How many moles of acetic acid are present in the 20.00 mL sample? 4) A 20.00 mL of a sample of vinegar of unknown concentration was titrated with 18.91 mL of 0.132 M sodium hydroxide solution. How many moles of acetic acid are present in the 20.00 mL sample?Explanation / Answer
1) Convert Oxalic acid into moles using formula 1 .121g/126.068 = 9.578X10^-4 moles of oxalic acid 2) We know from the info given that it takes 2 moles of NaOH to neutralize one mole of oxalic acid. So we know there is 9.578X10^-4 X 2 = 1.9196 X 10^-3 moles of NaOH in the 14.22 mLs. Using formula 3 convert 14.22 mLs into Liters 14.22mL/1000 = .01422 liters Molarity of the NaOH = moles/volume (in liters) 1.9196 X 10^-3 / .01422 = .13499289 #2) We need these formulas: 1) Molarity = moles/volume (volume in liters) 2) Moles = grams/molecular weight (molecular weight is calculated from periodic table) 3) 1000 mL = 1 L Given: 126.068 molecular weight of Oxalic acid .240g unknown sample (impure oxalic acid) 0.150M of NaOH 14.2 mLs to titration endpoint Calculations: 1) Convert 14.2 mLs to liters 14.2/1000 = .0142L 2) Calculate # moles of NaOH since Molarity = moles/volume we can say that Molarity * volume = moles, so .150M * .0142L = 2.13*10^-3 moles of NaOH 3) from the problem we know it takes 2 moles of NaOH to neutralize one mole of oxalic so, 2.13 * 10^-3 / 2 = 1.065*10^-3 moles of oxalic acid 4) Now we need to find our how many grams of oxalic acid we have. Since the molecular weight of the oxalic acid is given as 126.068 then 126.068 * 1.065*10^-3 = .13426242 grams of oxalic acid in the .240 grams of unknown 5) to find the percentage take .13426242 / .240 = .55942675 6) the percentage is .55942675 * 100 = 55.942675% #3) We need these formulas: 1) Molarity = moles/volume (volume in liters) 2) Moles = grams/molecular weight (molecular weight is calculated from periodic table) 3) 1000 mL = 1 L Given: 20.00 mL Vinegar with molarity of .780 Calculations: 1) Convert 20.00 mL to liters 20.00/1000 = .0200 L 2) calculate the # moles in the first solution since Molarity = moles/volume then Molarity * volume = # moles of vinegar .0200 * .780 = 0.0156 moles of vinegar in the 20.00 mL 3) Now the 20.00 mL of solution is diluted to a new volume of 100mL Molarity = moles/volume. Moles = 0.0156 Volume = 100ml/1000mL in a liter = .100L so, molarity = 0.0156/.100 = .156 molarity of the new solution 4) Now 10mL of the new solution is take to a flask, how many moles of vinegar in the flask? Simple, molarity = moles/volume, and molarity * volume = moles, so volume = 10mL/1000L/mL = .0100 L .156 * .0100 = 1.56*10-3 moles of vinegar in the flask.
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