If given 10ml of vinegar and told that 23ml of .4M NaOH was used to reach the end point during a titration: CH3COOH + NaOH → Na+CH3COO- +H2O A) Calculate the moles of Acetic Acid. (GMW: 60.0g/mol) B) Calculate the percent Acetic Acid (note: vinegar has a density of 1.005g)
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If given 10ml of vinegar and told that 23ml of .4M NaOH was used to reach the end point during a titration: CH3COOH + NaOH → Na+CH3COO- +H2O
A) Calculate the moles of Acetic Acid. (GMW: 60.0g/mol)
B) Calculate the percent Acetic Acid (note: vinegar has a density of 1.005g)
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- Answer the following three questions based on the information given below. Consider a buffer made with 34.0 mL of 0.25 M butanoic acid (HC4H-O2) and 16.0 mL of 0.46 M sodium butanoate (NAC4H7O2). The K, for butanoic acid (HC H-O,) is 1.5 X 10 5. Round all numerical answers to the 100en place (2 decimal places). Assume 298 K. What is the pH, if 20.0 mL of 1.0 M NAOH is added to this buffer? Type your answer..In the experiment conducted to determine the amount of acetic acid in vinegar, 8.2 mL of NaOH standard solution adjusted to 0.098 M was spent besides the phenolphthaleley indicator. Accordingly, which of the following is the amount of acetic acid in the vinegar sample in grams? (Ma (CH3COOH): 60.05 g/mol) A. 0.048 B. 0.024 C. 0.096 D.0.072. 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.
- 10.) 0.085 g of an unknown diprotic acid is titrated with 0.115 M NaOH. The first equivalence point occurred in the titration at a volume of 9.00 mL of NaOH added; The second equivalence point occurred at a volume of 18.00 mL of NaOH added. How many moles of NaOH were used to reach the second equivalence point in this diprotic acid titration? mol NaOH molThis 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?10:57 H₂O 0.1 NaCl Solution A Solution B Initial pH 7.0 5.D 3.0 10.0 Acid-Base Titration % acidity from vinegar label Molarity of NaOH 2030 Volume of Vinegar (ml) Initial Volume of NaOH (ml) Final Volume of NaOH (mL) Report Table AB.3: Effect of Adding Base pH after 5 Drops NaOH 11.0 9.0 3.0 11.0 Acid/Base Titration, Buffers, and PH Report Sheet pH after 10 Drops NaOH 5% 0.101 Trial 1 Report Table AB.4: Concentration of Acetic Acid in Vinegar 11.0 12.0 3.0 11.0** 5.00 ML 0.00 ML 41.20 ML Total pH Change Trial 2 5,00 ML 000 ML 38.40mL Buffer? (Y or N) Trial 3 Show your calculations for calculating the %(m/v) for acetic acid in vinegar for Trial 1 in Report Table AB.4. X 5.00 ML 0.01 ML 41.30mL
- If you have a 74.0 mL solution of 0.160 M benzoic acid, C6H5COOH, and it is being titrated with 0.270 M NaOH, answer each of the questions below. Note that the K, value of benzoic acid is 6.40 × 10-5. a) How many moles of benzoic acid are present in the original solution? C6H;COOH : mol b) What is the volume of the equivalence point, Veg, for the titration in litres, L? Veq : c) What is the pH at the equivalence point? pH :Now you use the above standardization NaOH (0.0702mol/L) to titrate a 25.0mL aliquot of H2SO4. It takes 12.0mL of your standardized NaOH to come to the equivalent point. Calculate the molarity of H2SO4Malic acid (C4H6O5) is one of the alpha hydroxy acid which is naturally found in apples. The reaction of HCl with sodium maleate to form malic acid and sodium chloride is according to the given equation in the pictureYou have been asked to prepare a maleate buffer(Kc=3.908 x 10-4 ) using sodium maleate (M.W=178.05g/mol) and 7.29 x 103 ppm HCl. How many milllilitres of HCl should you add to 250.00mL of 13.4 x 105 ppm sodium maleate to produce a buffer solution with pH of 3.37.
- Answer the following three questions based on the information given below. Consider a buffer made with 34.0 mL of 0.25 M butanoic acid (HC4H702) and 16.0 mL of 0.46 M sodium butanoate (NaC4H7O2). The K, for butanoic acid (HC4H-02) is 1.5 X 105. Round all numerical answers to the 100n place (2 decimal places). Assume 298 K. What is the pH, if 2.0 mL of 1.0 M NAOH is added to this buffer? Type your answer..0.915g of KHP (204.22g/mol) was dissolved in 100ml of distilled water and titrated euth sodium hydroxide solution. 27.55ml titrant was needed to reach the equivalenec point. There is 0.00448 moles of KHP present in the sample. a) How many moles of sodium hydroxide are needed to neutralize each mole of KHP? b) Therefore, how many moles of NaOH were needed to neutralize the KHP sample? c)What is the molar concentration (mol/L) of the sodium hydroxide solution?Question 14 of 15 Si A solution of phosphoric acid (H,PO,) with a known concentration of 0.250 M H.PO, 4 is titrated with a 0.800 | M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H,PO, according to the following balanced chemical equation: Н.РО, + NaOH → Na PO, + 3 H,0 STARTING AMOUNT