
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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If an unknown organic acid is titrated with 0.200 M NaOH and it takes 34.50 mL to reach the equivalence point. Show how to calculate the # moles of H3O+ at the equivalence point?
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- A 25. mL sample of a weak acid solution (KA= 4.3×10−4) is titrated with 0.20 M NaOH. 20.0 mL of NaOH are needed to reach the equivalence point What is the pH halfway to the equivalence point? 3.37 5.85 4.16 4.74arrow_forwardWhen titrating a 25 mL solution of 0.20M CH3COOH with 0.10M KOH, you realize you forgot to add an indicator initially. You've added 10 mL of KOH to the flask of acid. Ka of CH3COOH = 1.8 x 10^-5 1) Calculate the pH at this point 2) What color would your solution be if you add 40 mL KOH more. And a few drops of clorophenol blue indicator at this point. Why?arrow_forwardConsider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCI. Calculate the pH after 65.0 mL of HCI added. (K for CH3NH2=4.4 × 10-4) A. 10.96 B. 9.75 C. 7.00 D. 3.68 E. 2.54arrow_forward
- Bewis a titra on curve of a weak acid with NaOH. What region of the titration curve can the volume of NaOH at equivalence point be located? O с B A 2 A B Vol. NaOH (mL) 6 C 8 10arrow_forwardIf an unknown organic acid is titrated with 0.200 M NaOH and it takes 34.50 mL to reach the equivalence point. Show how to calculate the # moles of H3O+at the equivalence point? Answer: Moles of H30+= 0.0069 moles Suppose 0.345g of an unknown acid (with one ionizable H+ion) is titrated using the information in question above. Show how to calculate the Equivalent Mass of this acid.arrow_forwardConsider the titration of a 25.0mL sample of 0.195M acetic acid (CH3COOH) with 0.300M NaOH. How many mL of NaOH is required to reach the equivalence point? The Ka of acetic acid is 1.8x10-5.arrow_forward
- In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M NaOH, where K, HCOOH = 1.8x10-4 %3D the volume of NAOH added at one-half the equivalence point is about 27.4 mL O the volume of NaOH added at one-half the equivalence point is about 7.45 mL O the volume of NaOH added at one-half the equivalence point is about 12.6 mL O the volume of NaOH added at one-half the equivalence point is about 14.9 mLarrow_forwardHow do I approach this question?arrow_forwardCalculating the pH of a weak base titrated with a strong acid 0/5 Izabella An analytical chemist is titrating 165.0 mL of a 0.7400M solution of methylamine (CH3NH2) with a 0.4100M solution of HNO3. The pK of methylamine is 3.36. Calculate the pH of the base solution after the chemist has added 318.6 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO 3 solution added. Round your answer to 2 decimal places. pH = ☐ ☑ ? 18 Ararrow_forward
- HCN is a weak acid with a Ka = 4.90 x 10-10. A 50.00 mL sample of this HCN solution (0.250 M) is titrated with 0.500 M NAOH. What is the pH at the equivalence point?arrow_forwardIf an unknown organic acid is titrated with 0.200 M NaOH and it takes 34.50 mL to reach the equivalence point. Show how to calculate the # moles of H3O+at the equivalence point?arrow_forwardSuppose we titrate 20.0 mL of 0.100 M HCHO2 (Ka = 1.8 x 10-4) with 0.100 M NaOH. Calculate the pH. Before any base is added: pH = When half of the HCHO₂ has been neutralized: pH = After a total of 15.0 mL of base has been added: pH At the equivalence point: pH =arrow_forward
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