PROBLEM 4: In the titration of 100 ml of weak acid, 27.63ml of 0.09381M of NAOH was needed to reach the equivalence point. Consequently, the pH at equivalence point was measured and the value was 10.99. Calculate the pH if the volume of the NaOH added is less than 8.16ml of the volume at equivalence point.

PROBLEM 4: In the titration of 100 ml of weak acid, 27.63ml of 0.09381M of NAOH was needed to reach the equivalence point. Consequently, the pH at equivalence point was measured and the value was 10.99. Calculate the pH if the volume of the NaOH added is less than 8.16ml of the volume at equivalence point.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

2, In chemistry lab you make a buffer solution by dissolving 0.825 g of potassium formate in 0.2000L of 0.350 M formic acid. What is the initial pH of this buffer solution, and what is the pH after the addition of 10.50 mL of 0.235 M hydrochloric acid?
Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)? The Ka for HOCl is 3.0 x 10-8 M.(THE ANSWER IS 4.96) I DONT NEED THIS ABOVE QUESTIONS DONE!!!! THE QUESTIONS BELOW RELATE TO IT THOUGH!!!! What is the pH after 4.80 mL of NaOH are added? What is the pH at the equivalence point? What is the pH after 9.80 mL of NaOH are added?
QUESTION 2 What is the calculated equivalence point volume when 21.2 mL of 0.250M Hl are titrated with 0.350M LIOH?
Question 3 of 5 How many moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 2.90? (Ka for HF is 6.8 x 10 4) mol 1 3. 6. 7. 8. 9. +/- 0. P Type here to search 2. 4- 近

For each combination of acid and base below, what is the relative pH at the equivalence point of the titration? 1) HCO₂H + NaOH: [Select] CIRC 2) HNO₂ + NH3: [Select] 3) H₂SO4 + LIOH: [Select] 4) HBr + KOH: [Select] <
In lab, you will be provided with a 1.8 M acetic acid solution (CH3COOH) and solid sodium acetate trihydrate (NaOOCCH3  3H2O). You want to make 100.0 mL of a 0.5 M acetic acid buffer at pH 5.00. How many grams of sodium acetate trihydrate, and what volume (mL) of the 1.8M acetic acid solution and deionized water should you combine? The pKa of acetic acid is 4.75. Show all work.
The half‑equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.500 moles of a monoprotic weak acid (?a=8.9×10−5) is titrated with NaOH, what is the pH of the solution at the half‑equivalence point? pH= ( Please type answer note write by hend )
Determine the pH at the point in the titration of 40.0 mL of 0.200 M HNNH2 with 0.100 M HNO: after 10.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. PREV 1 2 4 NEXT > Based on your ICE table and definition of Kb, set up the expression for Kb in order to determine the unknown. Do not combine or simplify terms. = 3.0 x 10-6 RESET [0] [0.200] [0.0200] [0.100] [0.140] [0.175] [x] [2x] [0.200 + x] [0.200 - x] [0.0200 + x] [0.0200 - x] [0.100 + x] [0.100 - x] [0.140 + x] [0.140 - x] [0.175 + x] [0.175 - x]
Learning Goal: To calculate the pH at the equivalence point for various types of titrations. The equivalence point in an acid-base titration is the point at which stoichiometrically equivalent quantities of acid and base have been mixed together. At this point the reaction is complete because all analyte has been consumed by titrant. On a titration curve, the equivalence point is represented by the point of inflection (where the curve changes concavity). The figure (Figure 1) shows the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. When 40.0 mL of the NaOH solution is added, the acid-base neutralization reaction is complete. When analyzing titrations involving weak acids or bases, consider how the neutralization reaction will impact the pH of the system. For example, when titrating a weak acid with a strong base, at the equivalence point all the base has been used to neutralize the acid forming its conjugate weak base Figure E 14 12 10 8 6 4 2 0 0 Equivalence point 20.0 40.0 mL…
Practice Question 19 part 2!Consider the titration of 20.0 mL of 0.20 M NaOH with 0.1 M HCl. A) Calculate the pH at the equivalence point B) Calculate the pH after adding 10 mL beyond the equivalence point.
(Correct!) When 0.300 g of a diprotic acid was titrated with 0.100 M LIOH, 40.0 mL of the LiOH solution was needed to reach the second equivalence point. Identify the formula of the diprotic acid. H₂C4H4O6 (Your correct answer) H₂C204 H₂Se H₂S H₂ Te
Part 1: A 42.50-mL sample of 0.0527 M HF (Ka = 6.6 10-4) was titrated with a 0.0600-M KOH solution. What would be the equivalence point of this titration? Part 2: (This is the follow-up question of part 1) What would be the pH of the titree solution at the equivalence point? Please solve part 2