Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Sections
Buffer
Acid and Base Titration Curves
Weak Acid Strong Base Titrations
Weak Base Strong Acid Titrations
Strong Acid Strong Base Titrations
Titrations of Diprotic and Polyprotic Acids
Ksp
Additional Practice
Equivalence Point
Acid Base Indicators
Selective Precipitation
Formation Constant
Additional Guides
Henderson Hasselbalch Equation (IGNORE)
LearnAdditional Practice
Next SectionWeak Base Strong Acid Titrations

Whenever a WEAK ACID reacts with a STRONG BASE we use an ICF Chart to determine the pH of the solution. 

Weak Acid-Strong Base Reactions

Concept #1: Understanding Weak Acid–Strong Base titration reactions

Concept #2: If you use an ICF Chart and at the end you have remaining weak acid and conjugate base then you have a buffer so you use the Henderson-Hasselbalch Equation to find pH. 

Concept #3: If any excess moles of the strong base remain then we will use its concentration to find the pOH then pH of the solution. 

Concept #4: If the moles of both the weak acid and strong base are equal then we will have only conjugate base at the end of our ICF Chart Calculation. To find pH we would follow up with an ICE Chart. 

Weak Acid Strong Base Titration Calculations

Example #1:

Consider the titration of 75.0 mL of 0.0300 M H3C3O3 (Ka = 4.1 X 10-3) with 12.0 mL of 0.0450 M KOH. Calculate the pH.

 

Example #2:

In order to create a buffer, 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 175.0 mL of 0.300 M NaH?

 

Example #3:

Consider the titration of 75.0 mL of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The Ka of HNO2 is 4.6 x 10-4.

 

Previous SectionAcid and Base Titration Curves
Next SectionWeak Base Strong Acid Titrations
Additional Problems
What is the pH of a solution that contains 1.0 L of 0.10 M CH  3COOH and 0.080 M NaCH3COO after 0.03 moles of NaOH added? A. 4.33 B. 4.65 C. 4.74 D. 4.94 E. 12.48
What is the pH of a 1.00 L solution that is made by starting with 0.100 M HCOOH  (Ka = 1.8 X 10−4) with 0.035 moles of NaOH added?   A.  3.29      B.  3.48      C.  3.75      D.  4.00      E.  12.54
A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The K a of HF is 3.5 × 10 -4. A) 2.08 B) 3.15 C) 4.33 D) 3.46 E) 4.15
A solution is prepared by dissolving 0.23 moles of hydrazoic acid 0.27 moles of sodium azide in water sufficient to yield 1.00L of solution. The addition of 0.05 mol of sodium hydroxide to this buffer causes the pH to increase slightly. The pH does not increase drastically because the sodium hydroxide reacts with the ______ present in the buffer solution. The Ka of hydrazoic acid is 1.9x10 –5. Hydrazoic acid Azide H2O This is a buffer solution: the pH does not change upon the addition of an acid or base. All of the above
A 25.0mL sample of 0.150M of hydrazoic acid is titrated with 0.150M sodium hydroxide solution. What is the pH after 13.3mL of base is added? The Ka of hydrazoic acid is 1.9x10 –5. A. 4.45 B. 3.03 C. 1.34 D. 4.66 E. 4.78
A formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer? a) 3.67 b) 3.78 c) 3.81 d) 3.85 e) 3.95
A solution is made by mixing 500 mL of 0.167 M NaOH with 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of  H +, CH3COOH, CH3COO- , OH-, and Na+.
What is the pH of a one-liter solution that is 0.100 M in NH 3 and 0.100 M in NH 4Cl after 1.6 g of NaOH(s) has been added? Kb for NH3 is 1.8 × 10-5. (a) 4.74 (b) 9.26 (c) 9.46 (d) 9.62 (e) 11.12  
A 25.0 mL sample of 0.150 M hydrofluoric acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of hydrofluoric acid is 3.5 × 10-4. (a) 3.45 (b) 7.00 (c) 8.17 (d) 10.17 (e) 10.83
1.0-L of a buffer solution contains 0.800 moles of acid HA (K a = 2.00 x 10 -6 ) and 0.800 moles of its conjugate base, A− . What is the change in pH upon the addition of 0.0500 moles of a strong base to this solution (with no change in volume)? (a) +0.05 (b) +0.03 (c) +0.01 (d) -0.01 (e) -0.05  
In a titration, 25.00 mL of aqueous hydrofluoric acid requires the addition of 30.00 mL of 0.400 M NaOH to reach the equivalence (stoichiometric) point. What will be the corresponding pH? Ka (HF) = 6.76 × 10 -4 . (a) 9.12 (b) 8.25 (c) 5.75 (d) 7.00 (e) 1.74
Consider a buffer composed of a mixture of acetic acid and sodium acetate in equal amounts. Which of the following clearly indicates what happens to the buffer when a very small amount of concentrated LiOH is added to it.    [CH3COOH]      [CH 3COO – ]          pH A. increases          decreases        decreases B. decreases         increases          increases C. increases          decreases         increases D. decreases         increases          decreases E. increases          increases           increases
Find the pH of a solution made by titrating 60.0 mL of 0.125 M HCO  2H with 50.0 mL of 0.100 M LiH. The Ka of HCO2H is 1.8 x 10-4.
Calculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH. The Ka of HCN is 4.9 x 10-10.
What is the pH of a solution that contains 1.0 L of 0.10 M CH3COOH and 0.080 M NaCH3COO after 0.03 moles of NaOH added? A. 4.33 B. 4.65 C. 4.74 D. 4.94 E. 12.48
What is the pH of a 500 mL buffer containing 0.10 M Na 2HPO4/0.15 M KH2PO4 after adding 19 mL of 1 M NaOH?
A 30.0-mL volume of 0.50 M CH 3COOH (Ka = 1.8 x 10 -5) was titrated with 0.50 M NaOH. To reach the equivalence point, 30.0-mL of NaOH must be added. What will be the pH?
Calculate the pH of a solution after adding 28.522 mL of 0.285 M NaOH to 32.817 mL of a 0.318 M benzoic acid solution. Ka of benzoic acid is 6.5 x 10-5. A. 4.08B. 4.74C. 3.64D. 3.97E. 4.19
Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of a 0.10 M solution of aspirin (acetylsalicyclic acid, Ka = 3.0 × 10–4) solution.A)  3.5    B)  2.9    C)  4.1    D)  10.5    E)  1.8
What would be the final pH if 0.0100 moles of solid NaOH were added to 100mL of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10-4 ) and 0.300 M sodium formate?3.654.053.843.35
Determine the pH of a solution created by mixing 127.6 mL of 0.278 M formic acid, HCO2H, with 148.3 mL of 0.329 M sodium hydroxide, NaOH.A. 1.32B. 10.35C. 3.60D. 12.68E. 4.17
What is the pH at the equivalence point when 75.00 mL of a 0.150 M solution of acetic acid (CH3COOH) is titrated with 0.10 M NaOH to its end point?
A certain weak acid, HA, with a  Ka value of 5.61 × 10 −6, is titrated with NaOH.A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH?Express the pH numerically to two decimal places.
A 250.0 mL solution of 0.100 M HClO is titrated with 0.200 M NaOH. What is the expected pH of the resulting solution once 50.0 mL of the NaOH solution has been added to the HClO solution?A. 4.68B. 7.36C. 7.71D. 8.20E. 13.30
Calculate the pH at the equivalence point from the titration of 80.00 mL of 0.140 M hydrofluoric acid, HF, with 160.00 mL of 0.0700 M NaOH. Ka of HF is 3.5 x 10-4.A. 5.94B. 8.06C. 7.75D. 8.30E. 11.61