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For question 5 could you find the best value and explain how you arrive to the best value please
thank you!
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- Figure 15-3 outlines the classic scheme for separating a mixture of insoluble chloride salts from one another. Explain the chemistry involved in the various steps of the figure.Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).Which of the indicators in Fig. 14-8 could be used for the titrations in Exercises 61 and 63?
- The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSeO4, 0.0118 g/100 mL. (b) Ba(BrO3)2H2O, 0.30 g/100 mL. (c) NH4MgAsO46H2O, 0.033 g/100 mL. (d) La2(MoO4)3, 0.00179 g/100 mLMalonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH, stoichiometric points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (Sec Exercise 113.). A 4.59 mL sample of HCl, specific gravity 1.3, required 50.5 mL of 0.9544N NaOH in a titration. Calculate the % w/w HCl.
- Titration of Acids and Bases Part B. Determination of the molar concentration of an unknown acid. Molarity of N₂OH = .09M ☆ Volume of unknown acid used (mL) Final buret reading (mL NaOH) Initial buret reading (mL NaOH) Volume of NaOH used (mL) Molarity of unknown acid Trial 1 10mb Moles 9.6 mL 0.5mL 9.1mL (OH) Sample calculations for molarity of unknown acid solution: Trial 2 10mL 18,9 Average molarity of unknown acid solution: Trial 3 10mL 28.3mL 9.6 ml 18.9ml 9,3mL 9.4mL mol NaOH used? M N₂OH = M acid = .1 Gril To multiple decimals1. A 4.59 mL sample of HCl, specific gravity 1.3, required 50.5 mL of 0.9544N NaOH in a titration. Calculate the %w/w HCl. 2. A 10 mL sample of sulfuric acid solution required 16.85 mL of NaOH solution in a titration. Each mL of the NaOH solution was equivalent to 0.2477 g of potassium hydrogen phthalate. Calculate the sulfuric acid content in %w/v. MW of sulfuric acid, 98; MW of potassium hydrogen phtalate(KHC8H4O4 ) = 204.22Calculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.
- This is about determinition of acid content in vinegar through titration, please answer the questions. Please answer only the items nummber 4-5. i only put other questions for guide. 1. What is the purpose of standardization? 2. What is the primary standard used (name and formula)? ANS: KHP 3. Suppose that the KHP is not completely dry. Will the reported molar concentration of the sodium hydroxide solution be higher, lower, or unaffected? Explain. 4. In preparing the buret for titration the final rinse is with the NaOH titrant rather than with distilled water. Explain. 5. The procedure suggests the addition of only 2 drops of phenolphthalein. What will be the effect to the analysis if larger amount of phenolphthalein is added?Vinegar? required to neutralize the acetic acid in each sample of vinegar, 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. Convert the average volume of NaOH required to reach endpoint from mL to L Average volume of NaOH required to reach endpoint in mL 15. Subtract the initial volume from the final volume to calculate the volume of NaOH I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH MI TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (NaOH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.39 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 -27.54 27.54ML (Show calculations here.) a7.54ML X IL 0-02754L :0.02754L nd bon 1600 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of…RESULTS AND CALCULATIONS: Standardization of NaOH solution 1.trial 2.trial 15.0 mL of 0.1015 M HCl + 15.3 mL NaOH 15.0 mL of 0.1015 M HCl + 15.2 mL NaOH Calculation of the average molarity of NaOH solution; Titration of unknown acid 1.trial 2.trial 15.0 mL of unknown acid + 12.2 mL NaOH 15.0 mL of unknown acid + 12.4 mL NaOH Calculation of the moles and molarity of the unknown acid solution;