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 Volume of acid solution. 24.09 mL. Molarity of acid solution. 1.605 M. Mole ratio of base to acid in titration reaction. 2 mol base: 1 mol acid.
Titration Curve Practice Problems
Prior to the half-equivalence point there will be more weak base in the solution with increasing amounts of conjugate acid ion being created as the strong acid
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
solution with 0.15M NaOH solution? For strong acid/base titration perform stoichiometry calculation first; then calculation resulting concentration with ...
Titrations Practice Worksheet
If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4) what is the concentration of the H2SO4 solution?
Exercise 15.4 - Titrations.pdf
NaOH? Solving Titration Problems. A titration is a chemical Sketch the titration curve for the titration of a generic weak base B
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
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EXTRA PRACTICE: Titration Problems Practice. Titration Calculate the moles of base used the moles of acid used and the concentration of the original acid.
Exercise 15.4 - Titration
When the acid or base produces multiple H or OH respectively the molarity of the solution must be multiplied by number of ions produced.
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
Titrations Practice Worksheet
Find the requested quantities in the following problems: sulfuric acid solution (H2SO4) what is the concentration of the H2SO4 solution? A".o^-.
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
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
p4 pH Estimations/Calculations after acid/base Titration-Related Problems ... Answer: A buffer consists of a weak acid and its conjugate base in roughly ...
Section 19.1. Acid-Base Buffer Solutions
Example: Calculate [H3O+] in a solution that is 0.10 M in HF and 0.20 M in NaF. Also calculate % ionization. Problem: Use HF(aq) ? H+ (aq) + F- (aq) ?
Questions with Answers- Amino Acids & Peptides A. Two of the
_____. Which must happen if amino acid Y is titrated with NaOH from pH=1 to. pH=14? a). Y must exist entirely in the fully protonated form at the lowest half-.
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.
ACID-BASE TITRATIONS
Calculate the pH of solution at the following volumes of. NaOH added: 0 10.00
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 €
K a [H 3 O ][F [HF]6.8x10
-4 x 2 (0.12-x) x 8.7 X 10-3 M pH -log (8.7 X 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
K a [H 3 O ][F [HF]6.8x10
-4 (x (0.43+x)) (0.43-x) x 6.8 x 10-4 M pH -log (6.8 x 10-4) 3.17 Note: Could also useHenderson-Hasselbalch
equation since this is buffer region of titration curve. (C) What is pH at equivalence point? First need to determine volume at equivalence point. €0.0250 L
0.12mol HF
1L1mol NaOH
1mol HF
1L0.15mol 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 €
K b 1x10 -146.8x10
-4 =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
K b [OH ][HF] [F1.5x10
-11 x 2 (0.067-x) x 1.0 x 10-6 M pOH -log (1.0 x 10-6) 6.00 so pH 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 €
K b 1x10 -146.8x10
-4 =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 €
K b [OH ][HF] [F1.5x10
-11 (x (0.060+x)) (0.040-x) x 1.0 x 10-11 M pOH -log (1.0 x10-11)
11.00 so pH 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 x 0.060 M 1.0 x 10-11 M 0.060 M pOH -log (0.060) 1.22 so pH 12.78quotesdbs_dbs19.pdfusesText_25[PDF] acid/base stoichiometry practice problems answers
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