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The following titration curve is expected from which acid-base pair?
See the picture below.
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- Express the concentration of acetic acid in both samples as % by mass of acetic per 100mL of solution given : %= mass/ volumex100% Given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of…NaOH + HCl ---> NaCl + H2O 123.60 mL of 3.90 M aqueous NaOH is required to titrate 346.00 mL of an aqueous solution of HCl to the equivalence point. The molarity of the analyte is 1.39 M. True or FalseIn 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)
- Calculate the percent difference in the two values for the molarity of the NaOH solution by: %Difference =[M1-M2]\Mavg x100% Given: 1st value for the molarity of NaOH = 0.0647 M 2nd value for the molarity of NaOH= 0.2627 M NaOH vs H2SO4 Burette solution is NaOH and the pipette solution is 10.0 mL of 0.205 M H2SO4 Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 31.8 mL 31.8 mL 31.7 mL Titration 1 0.0 mL 31.5 mL 31.5 mL Titration 2 0.0 mL 31.7 mL 31.7 mL To find the average volume Average volume = 31.8 mL + 31.5 mL + 31.7 mL331.8 mL + 31.5 mL + 31.7 mL3 = 31.7 mL NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL…What is the pH of the solution after mixing 0.171 g of Mg(OH)2 (MW=58.321 g/mol) with 68.9 mL of 0.0569 M HCl? The resulting solution was diluted to 100 mL. Round your calculated value for pH to two figures to the right of the decimal point.The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. A. What is the pH at the beginning of the titration, Vbase = 0.00 mL? B. What is the pH at the equivalence point? C. What is the pH of the titration when 5.00 mL of base have been added? D. What is the pH when the volume of base added equals half the volume of the equivalence point? E. What is the pH of the titration when 20.00 mL of base have been added? F. What is the pH of the titration when 30.00 mL of base have been added?
- 5 SO2 + 2 MnO4- + H2O ---> 5 SO42- + 2 Mn2+ + 4 H+ The MnO4- solution is poured in a burette and then titrated till the equivalent point is aquired. How many L are needed of MnO4? (the burette starts at 5,7 and is 37,3 after titration)A hydrochloric acid solution is standardized by titrating 0.5008 g of primary standard tris(hydroxymethyl) aminomethane, C4H11NO3, with 99.9780% purity. If 39.68 mL isrequired for the titration, what is the molarity of the acid?In the tiration of what base with KHP (endpoint) can we use "moles of base= 2 x moles of KHP" equivalence ?
- The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. a) Sketch the pH vs volume of added base titration curve for this reaction. pH vertically, volume of base horizontally. Label your axes with correct pH and volumes. (Attach more space, if needed) b) what is the pH at the beginning of the titration, Vbase = 0.00 mL ? c) what is the volume of the base needed to reach the equivalence point ? (label the equiv. point) d) what is the pH at the equivalence point?e) what is the pH of the titration when 5.00 mL of base have been added ? f) what is the pH when the volume of base added equals half the volume of the equivalence point? g) what is the pH of the titration when 20.00 mL of base have been added? h) what is the pH of the titration…Molarity of (NH3) solution (M) from bottle- 5.0 Initial reading of buret (NH3) (mL)- .27 Final reading of buret (NH3) (mL)- 8.25 Volume of Cd(NO3)2 solution (mL)- 10.00 Volume of Na2C2O4 solution (mL)- 10,00 please find Total volume of solution after titration (mL) Total moles of C2O42- (mol) Molarity of C2O42- (M) Total moles of Cd2+ (mol) Moles of [Cd(NH3)4]2+ (mol) Molarity of [Cd(NH3)4]2+ (M) Moles of NH3 added by titration (mol) Moles of NH3 that did not react with Cd2+ (mol) Molarity of NH3 that did not react with Cd2+ (M) Kf for [Cd(NH3)4]2+16.The titration of an aliquot of 6 mL of acetic acid of commercial use, with NaOH 0.1 , was made by titration, so that to reach the equivalence point 40 ml of NaOH were consumed.Determine the percentage concentration (m/m) of the acetic acid, if its density is 1.05 g/ml. HC2H3O2 acetic acid