Exercise 15.4 - Titrations - Answers.pdf
Solving Titration Problems. A titration is a chemical process for finding the The pH of any strong acid Istrong base titration at the equivalence point.
ACID-BASE TITRATIONS (PROBLEMS)
being titrated with 0.50 M NaOH. Calculate the pH of this solution initially before any NaOH is added. At this point
Problem Solving
Apr 20 2016 Mole ratio of base to acid in titration reaction ? Volume of base solution titrated. 20.00 mL. Moles of base in solution titrated ? mol NaOH.
Acid-Base Titration and pH
PROBLEMS Write the answer on the line to the left. Show all your work in the space provided. 4. [H3O ] in an aqueous solution 2.3 10 3 M. 4.3
Acid – Base Titration Calculations (WA + SB)
Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid HC3H5O3 after it has been titrated with a total of 30.0 ml of 0.40 M KOH? Page 2. P G4 A (pg of ).
1 General Chemistry II Jasperse Buffers/Titrations/Solubility. Extra
What Kind of Solution/pH at End? p2 Titration Calculations p11. Preparation and Recognition of Buffers p4 pH Estimations/Calculations after acid/base are
Untitled
EXTRA PRACTICE: Titration Problems Practice. Titration Calculate the moles of base used the moles of acid used and the concentration of the original acid.
Acid – Base Titration Calculations (WB + SA)
1. Calculate the pH of 20.0 ml of a 0.060 M solution of ammonia NH3. 2. What volume of 0.040 M HI would be needed to
Test3 ch17b Buffer-Titration-Equilibrium Practice Problems
Answer: A buffer consists of a weak acid and its conjugate base in roughly equal amounts. If acid is added to the solution it is consumed by the conjugate
What is the M of NaOH if it takes 40.0 ml of NaOH to reach the
Worksheet - Titration Problems. 1. What is the M of NaOH if it takes 40.0 ml H₂Soy as Hel to neutralize the base. 30.3 ml of 0.305 M NaOH are required to ...
w336-titrations-worksheet.pdf
calculations for the concentration of the base? 3). It takes 38 mL of 0.75 M NaOH solution to completely neutralize 155 mL of a sulfuric acid solution
Exercise 15.4 - Titrations - Answers.pdf
Solving Titration Problems While the volumes of acid and base should probably be converted to liters as long as they are the same unit
Acid – Base Titration Calculations (WA + SB)
Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid HC3H5O3 after it has been titrated with a total of 30.0 ml of 0.40 M KOH? Page 2. P G4 A (pg of ).
Titration Calculations Strong Acid/Strong Base Calculations (1) Use
added during a titration to 25.0 mL of a 0.12M HCl solution with 0.15M NaOH solution? For strong acid/base titration perform stoichiometry calculation
Bookmark File PDF Acid Base Titration Lab 39 Answers ? - covid19
easy to acquire as without difficulty as download lead Acid Base Titration. Lab 39 Answers. It will not say yes many grow old as we accustom before.
1 General Chemistry II Jasperse Buffers/Titrations/Solubility. Extra
What Kind of Solution/pH at End? p2 Titration Calculations p11. Preparation and Recognition of Buffers p4 pH Estimations/Calculations after acid/base.
Where To Download Redox Practice Problems With Answers Copy
with states phases
A Dating Analogy for Acid-Base Titration Problems
alleviate this confusion. The use of other analogies to help clarify other confusing concepts related to solution concen- trations and stoichiometric ratios
Worksheet 22 – Weak Acid/Strong Base Titrations A. Initial pH This
This can only be used after the initial point and before the equivalence point when the acid and its conjugate base are both major species in the solution. At
Test3 ch17b Buffer-Titration-Equilibrium Practice Problems
What Kind of Solution/pH at End? p2 Titration Calculations p11. Preparation and Recognition of Buffers p4 pH Estimations/Calculations after acid/base.
![Titration Calculations Strong Acid/Strong Base Calculations (1) Use Titration Calculations Strong Acid/Strong Base Calculations (1) Use](https://pdfprof.com/Listes/27/6236-27TitrationCalculationAnswerKey.pdf.pdf.jpg)
Titration Calculations Strong Acid/Strong Base Calculations (1) Use balanced equation to do stoichiometric calculation. (2) Determine pH from amount of strong acid/base that is in excess. Note: At stoichiometry point of equal acid and base, pH =7. Example: What is pH after 0.0 mL, 10.0mL, at equivalence point, and 50.0 mL of base has been added during a titration to 25.0 mL of a 0.12M HCl solution with 0.15M NaOH solution? For strong acid/base titration, perform stoichiometry calculation first; then calculation resulting concentration with total volume; finally, calculate pH directly. (A) 0.0 mL base: Solution is 0.12M HCl pH = -log[H+] = -log(0.12) = 0.92 (B) 10.0mL added base: HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) (0.0250L)(0.12M) (0.0100L)(0.15M) 0.0030 mol 0.0015 mol 0 -0.0015 mol -0.0015 mol +0.0015 mol 0.0015 mol 0 0.0015 mol [HCl] = 0.0015 mol/0.0350L = Therefore since strong acid: [H+] = so pH = -log (0.043) = 1.37 (C) At Equivalence Point: Volume of base added = (0.0030mol HCl)(1mol NaOH/1mol HCl)(1L/0.15mol NaOH) = 0.020 L = 20. mL added base Since NaCl does not hydrolyze water, pH is neutral 7.00. (D) 50.0mL added base: HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) (0.0250L)(0.12M) (0.0500L)(0.15M) 0.0030 mol 0.0075 mol 0 -0.0030 mol -0.0030 mol +0.0030 mol 0 mol 0.0045 mol 0.0030 mol [NaOH] = 0.0045 mol/0.0750L = Therefore since strong base left: [OH-] = so pOH = -log (0.060) = 1.22 pH = 12.78
Weak Acid/Strong Base Calculations What is pH after 0.0 mL, 10.0mL, at equivalence point, and 50.0 mL of base has been added during a titration to 25.0 mL of a 0.12M HF solution with 0.15M NaOH solution? Ka = 6.8 x 10-4 (1) Use balanced equation to do stoichiometric calculation. (2) Determine new concentrations by dividing by total volume. (3) Use appropriate equilibrium reaction and ICE chart to determine pH. Stoichiometric Reaction: HF(aq) + NaOH(aq) -> H2O(l) + NaF(aq) Equilibrium Reaction: HF(aq) + H2O(l) -> H3O+(aq) + F-(aq) (A) Addition of 0.0 mL of base: Only weak acid present. HF (aq) + H2O - H3O+(aq) + F- (aq) I 0.12 M 0 0 C - x + x + x E 0.12 - x x x €
][F [HF]6.8x10
(0.12-x) 8.7 10-3 -log (8.7 10-3) 2.06(B) What is pH after 10.0mL of 0.15M NaOH solution has been added to 25.0 mL of 0.12M HF solution? Ka = 6.8 x 10-4 (1) Use balanced equation to do stoichiometric calculation. (2) Determine new concentrations by dividing by total volume. (3) Use appropriate equilibrium reaction and ICE chart to determine pH. (1) Stoichiometric Reaction: HF(aq) + NaOH(aq) -> H2O(l) + NaF(aq) (0.0250L)(0.12M) (0.0100L)(0.15M) 0.0030 mol 0.0015 mol 0 -0.0015 mol -0.0015 mol +0.0015 mol 0.0015 mol 0 mol 0.0015 mol (2) New concentrations: [HF] = 0.0015 mol/ 0.0350L = 0.043M [F-] = 0.0015 mol/ 0.0350L = 0.043M (3) Equilibrium Reaction:
HF (aq) + H2O - H3O+(aq) + F- (aq) I 0.043M 0 0.043M C - x + x + x E 0.043 - x x 0.043 + x
][F [HF]6.8x10
(x (0.43+x)) (0.43-x) 6.8 10-4 -log (6.8 10-4) 3.17 Note: Could also useHenderson-Hasselbalch
equation since this buffer region titration curve. (C) What is pH at equivalence point? First need to determine volume at equivalence point. €0.0250 L
0.12mol HF
1mol NaOH
1mol HF
0.15mol NaOH
=0.0200 L or 20.0 mL(1) Use balanced equation to do stoichiometric calculation. (2) Determine new concentrations by dividing by total volume. (3) Use appropriate equilibrium reaction and ICE chart to determine pH. (1) Stoichiometric Reaction: HF(aq) + NaOH(aq) -> H2O(l) + NaF(aq) (0.0250L)(0.12M) (0.0200L)(0.15M) 0.0030 mol 0.0030 mol 0 - 0.0030 mol - 0.0030 mol + 0.0030 mol 0 0 0.0030 mol (2) New concentrations: [HF] = 0 mol/0.0450L = 0 M [F-] = 0.0030 mol/0.0450L = 0.067 M €
1x10 -146.8x10
=1.5x10 -11(3) Equilibrium Reaction: Only conjugate base now left. So must use equilibrium reaction for conjugate base and calculate Kb. F- (aq) + H2O - OH-(aq) + HF (aq) I 0.067 M 0 0 C - x + x + x E 0.067 - x x x
[OH ][HF]1.5x10
-11 (0.067-x) 1.0 10-6 pOH -log (1.0 10-6) 6.00 8.00(D) What is pH after 50.0mL of 0.15M NaOH solution has been added to 25.0 mL of 0.12M HF solution? Ka = 6.8 x 10-4 (1) Use balanced equation to do stoichiometric calculation. (2) Determine new concentrations by dividing by total volume. (3) Use appropriate equilibrium reaction and ICE chart to determine pH. (1) Stoichiometric Reaction: HF(aq) + NaOH(aq) -> H2O(l) + NaF(aq) (0.0250L)(0.12M) (0.0500L)(0.15M) 0.0030 mol 0.0075 mol 0 - 0.0030 mol - 0.0030 mol + 0.0030 mol 0 0.0045 mol 0.0030 mol (2) New concentrations: [OH-] = 0.0045 mol/0.0750L = 0.060 M [F-] = 0.0030 mol/0.0750L = 0.040 M €
1x10 -146.8x10
=1.5x10 -11(3) Equilibrium Reaction: F- (aq) + H2O - OH-(aq) + HF (aq) I 0.040 M 0.060 M 0 C - x + x + x E 0.040 - x 0.060 + x x €
[OH ][HF]1.5x10
-11 (x (0.060+x)) (0.040-x) 1.0 10-11 pOH -log (1.010-11)
11.00 3.00NOTE: "x" is NOT the OH- concentration. The OH- concentration is 0.060M +x. Since there is excess strong base in this last addition, the pH is determined by the strong base concentration. The weak conjugate base F- adds an insignificant amount.
[OH-] 0.060 0.060 1.0 10-11 0.060 pOH -log (0.060) 1.22 12.78quotesdbs_dbs2.pdfusesText_4[PDF] acid base test review answers
[PDF] acid/base stoichiometry practice problems answers
[PDF] acide acétique
[PDF] acide base ph cours
[PDF] acide base ph exercice
[PDF] acide base ph terminale s
[PDF] acide base physique chimie
[PDF] acide base physique terminale s
[PDF] acide base physique ts
[PDF] acide et base conjuguée
[PDF] acide et base de bronsted
[PDF] acide et base de lewis
[PDF] acide et base exercices corrigés pdf
[PDF] acide et base pdf