Following a standardization reaction, the concentration of an aqueous solution of NaOH is determined to be 0.4510 M. a. What is the mass of NaOH in 50.0 mL of this solution? Show your work. b. What mass of KHP would be needed to completely react with 50.0 mL of this NAOH solution? Show your work.
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- Ff a solid, unknown acid were contaminated with pure, absorbed water prior to weighing out the sample, what effect (higher, lower, or none) would this have on the determined molar mass?Using the experimental data at 1/2 the equivalence point volume, calculate the Ka of acetic acid. Volume of NaOH at Equivalence point for HCl Trial = 5.083ml Equivalence point pH for HCl Trial = 6.57 pH 1/2 Equivalence point for HCl Trial = 3.285 Volume of NaOH at Equivalence point for HC2H3O2 Trial= 4.025ml Equivalence point pH for HC2H3O2 Trial = 4.13 pH 1/2 Equivalence point for HC2H3O2 Trial = 2.065A 10.00-mL aliquot of a known HCl solution (3.800 g/L) was titrated with a 4.100 g/L NaOH solution. Calculate the range of the blank corrected titration volumes using the titration data below: Blank Trial 1 Trial 2 Trial 3 Initial Volume (mL) 12.50 0.50 12.30 34.75 Final Volume (mL) 13.10 10.45 22.35 45.27
- 1. Recreate the graph in the potentiometric titration of a diprotic acid (FW=116.07) given the following data. Show necessary computations.Mass of acid = 0.1032 gKa1 = 3.21 × 10-5Ka2 = 8.85 × 10-8pH at EP1 = 5.77pH at EP2 = 7.69NNaOH = 0.05 NNote: Label EP1 and EP2. The graph should show the actual volume of titrant (mL) and pH level at ½ EP1, EP1, ½ EP2, and EP2.A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point. If the multivitamin was manufactured on March 2021, when is its most likely expiration date? June 2021 September 2021 December 2021 March 2022A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point. 1. How many mg of ascorbic acid is present in the sample after production? 2. How many mg of ascorbic acid are present in the 3 month old sample? 3. How many mg of ascorbic acid are present in the 6 month old sample? 4. If the multivitamin was manufactured on March 2021, when is its most likely expiration date? a. June 2021 b. September 2021 c. December 2021…
- A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point.A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point. A. How many mg of ascorbic acid is present in the sample after production? 521.0 mg 506.7 mg 492.53 mg 488.12 m B. How many mg of ascorbic acid are present in the 3 month old sample? 492.53 mg 488.12 mg 452.58 mg 435.91 mg C. How many mg of ascorbic acid are present in the…An automatic titration was performed with the following results: Standardized NaOH Concentration: 0.1191 M Volume of HA- product examined in each trial: 25.00 mL Volume of Standardized NaOH Titrant used to achieve the first equivalence point: Trial #1: 1.300 mL Trial #2: 1.137 mL Trial #3: 1.140 mL Calculate the Molarity for each trial, Average Concetration (M), Deviation (M), Standard Deviation (M), and RSD (M)
- In this experiment, it is assumed that No. of moles of (Ca2+ + Mg2+) = moles (Ca2+). Calculate the number of moles of Ca2+ in the hard water sample. please show working out, thank you Sample No. 1 2 3 4 Final Burette reading (mL) 24.40 46.90 24.70 47.20 Initial Burette reading (mL) 0.10 24.40 2.10 24.70 Final-Initial Titre (mL) 24.30 22.50 22.60 22.50Trial 1: Initial pH= 3.92mL Mass of KHP and Paper= 0.868g Mass of paper= 0.357g Mass of KHP= 0.511g Trial 2: Initial pH= 4.09ml Mass of KHP and Paper= 0.870g Mass of paper= 0.359g Mass of KHP= 0.511g 1)Use Kb, the number of moles of C8H4O4^2- at the equivalence point, and the total volume at that point to calculate the pH for each sample at the equivalence point. Compare these calculated results with the experimental results.Construct a curve for the titration of 20.00 mL 0.0500 N succinic acid (HO2CCH2CH2CO2H) by 0.1000 N NaOH solution. MS Excel® can be used. Titrant volumes: 15.00 mL, 20.00 mL, and 25.00 mL. (CH2)2(CO2H)2 ⇌ (CH2)2(CO2H)(CO2)^− + H^+ Ka1 = 6.21 × 10^-5; pKa1 = 4.207 (CH2)2(CO2H)(CO2)− ⇌ (CH2)2(CO2)2^2− + H^+ Ka2 = 2.31 × 10^-6; pKa2 = 5.636