A solution of citric acid (H3CgH;O,) with a known concentration of 0.200 M H;C,H;O, is titrated with a 0.750 M N2OH solution. How many ml of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H;CGH;O, , according to the following balanced chemical equation: H;CgH;O, + 3 NaOH – Na;CgH;0, + 3 H20

A solution of citric acid (H3CgH;O,) with a known concentration of 0.200 M H;C,H;O, is titrated with a 0.750 M N2OH solution. How many ml of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H;CGH;O, , according to the following balanced chemical equation: H;CgH;O, + 3 NaOH – Na;CgH;0, + 3 H20

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

The hydroxide ion concentration of an aqueous solution of 0.490 M acetylsalicylic acid (aspirin), HC,H;04. is [OH] = |M.
6.) It takes 22.5 mL of a 1.25 M KOH stock solution to reach the equivalence point in the titration of 10.0 mL, of hydrochloric acid (HCI) of unknown concentration. a) What is the balanced formula equation? b) What is the net ionic equation, and if you had to classify this reactipp, what war, gou classify it as? c) What is the concentration of the hydrochloric acid solution you start with?
A solution of citric acid (H.CeHsO,) with a known concentration of 0.200 M H.CcHsO, is titrated with a 0,750 | M NAOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H:CCH:O, according to the following balanced chemical equation: H.CcHsO, + 3 3 H:0 NaOH – Na:CeHsO, + STARTING AMOUNT ADD FACTOR ANSWER RESET 1 3 1000 6.67 15.0 20.0 0,020 25.0 11.25 2.22 0.001 0.750 2.00 x 10 0.200 + g H.CcHsO, M H.CcHsO, mL NaOH mol NaOH L NaOH mol H.CHSO, LH.CeHsO, 11:33 PM 68°F Mostly cloudy 10/5/2021 P Type here to search prt sc delete tg tvo insert 144 トト」米 esc home @ 23 backspace 3. 4 6. 7 8 Y P Q tab
4. A solution prepared to be initially 1 M in NH3 and 0.5 M in HCl is (Kb for NH3 = 1.8 x 10¯5): (A) a solution with a pH less than 7 that is not a buffer solution (B) a buffer solution with a pH between 4 and 7 (C) a buffer solution with a pH between 7 and 10 (D) a solution with a pH greater than 7 that is not a buffer solution (E) a solution with a pH of 7
2. Using the procedure described in this module, a student determined the percent KHP in an impure sample of KHP. A 3.150-g sample of impure KHP required 41.50 mL of 0.1352M NaOH solution for titration. (a) Calculate the number of moles of NaOH required for the titration. (b) Calculate the number of moles of KHP present in the impure sample of KHP. (c) Calculate the number of grams of KHP present in the impure sample. (d) Calculate the percent of KHP in the impure sample, using Equation 8. Equation 8: percent KHP in the impure sample, % = ( mass of KHP in the sample,g/ mass of sample analyzed, g) (100%)
7. The separate samples of nitric and acetic acids are both titrated with a 0.100 M solution of NaOH(aq). (Y) 25.0mL of 1.0 M HNO3(aq) (Z) 25.0 mL of 1.0 M CH3COOH(aq) Determine whether each of the following statements concerning this titration is true or false. (A) A larger volume of NaOH(aq) is needed to reach the equivalence point in the titration of HNO3 (B) The pH at the equivalence point in the HNO3 titration will be lower than the pH at the equivalence point in the CH3COOH titration (C) Phenolphthalein would be a suitable indicator for both titrations Group of answer choices a.) A) False B) True C) True b.) A) False B) False C) True c.) A) False B) True C) False d.) A) True B)True C)True
5 6 7 8 Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What is the initial solution pH (i.e., when 0.00 mL of titrant have been added)? (Two decimal places) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What volume (in mL) of titrant will have been added at the half-equivalence point? (One decimal place) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K₂ of hypobromous acid is 2.8 x 10.⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide…
4) A 25.00-mL sample of vinegar is diluted with deionized to 100.0 mL. Then a 10.00-mL aliquot (portion) of the dilute vinegar solutionis titrated with 0.1055 M NaOH solution using phenolphthalein as theindicator. (a) If 19.60 mL of the base solution is required to reach the end-point, calculate the molar concentration of acetic in dilute vinegar solution. (b) What is the molar concentration of acetic acid in the original (undiluted) vinegar? The acid-base reaction occurs as follows: HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)