A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralizedusing 18.1 mL of a standard HCI.Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standardbase is equivalent to 0.1005 g KHP.MW: KHP = 204.22 NazCO3 = 106.0 ASA = 180.16ASA+ NaOH+H20ÓHNaOH+ NaOH+ CH;COONNa acetateNaNaNa AcetylsalicylateNa salicylateх'ss NaOH + HCІNaCl + H,OThe Normality of the STD. Base isThe Normality of the STD. Acid isThe weight of pure ASA in grams isThe H of aspirin isThe % ASA in the sample is???

A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralized using 18.1 mL of a standard HCI. Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standard base is equivalent to 0.1005 g KHP. MW: KHP = 204.22 Na2CO3 = 106.0 ASA = 180.16 ASA +NaOH + H,O `Na HO + NaOH + CH;COONa Na acetate `Na `Na Na Acetylsalicylate Na salicylate x'ss NaOH + HCІ NaCl + H2O The Normality of the STD. Base is The Normality of the STD. Acid is The weight of pure ASA in grams is ? ?

A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralized using 18.1 mL of a standard HCI. Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standard base is equivalent to 0.1005 g KHP. MW: KHP = 204.22 Na2CO3 = 106.0 ASA = 180.16 ASA +NaOH + H,O `Na HO + NaOH + CH;COONa Na acetate `Na `Na Na Acetylsalicylate Na salicylate x'ss NaOH + HCІ NaCl + H2O The Normality of the STD. Base is The Normality of the STD. Acid is The weight of pure ASA in grams is ? ?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter14: Equilibria In Acid-base Solutions

Section: Chapter Questions

Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...

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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).
A buffer is prepared using the butyric acid/butyrate (HC4H7O2/C4H7O2)acid-base pair. The ratio of acid to base is 2.2 and Ka for butyric acid is1.54105. (a) What is the pH of this buffer? (b) Enough strong base is added to convert 15% of butyric acid to the butyrate ion. What is the pH of the resulting solution? (c) Strong acid is added to the buffer to increase its pH. What must the acid/base ratio be so that the pH increases by exactly one unit (e.g., from 2 to 3) from the answer in (a)?
Calculate the pH change when 10.0 mL of 0.100-M NaOH is added to 90.0 mL pure water, and compare the pH change with that when the same amount of NaOH solution is added to 90.0 mL of a buffer consisting of 1.00-M NH3 and 1.00-M NH4Cl. Assume that the volumes are additive. Kb of NH3 = 1.8 × 10-5.
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