Ch.16 - Aqueous Equilibrium See all chapters
All Chapters
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

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.

 

Additional Problems
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. 5.94 B. 8.06 C. 7.75 D. 8.30 E. 11.61
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. 1.32 B. 10.35 C. 3.60 D. 12.68 E. 4.17
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 the titration of 15.0 mL of 0.200 M H 3PO4(aq) with 0.200 M NaOH(aq). What is/are the major species in solution after the addition of 15.0 mL of base? 1. H3PO4(aq) and H2PO4− (aq) 2. H2PO4− (aq) and HPO4− (aq) 3. PO43− (aq) 4. H2PO4− (aq)
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
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
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
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
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. 4.68 B. 7.36 C. 7.71 D. 8.20 E. 13.30
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 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
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.
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
In a strong base with weak acid titration, which of the following is correct? A. the equivalence point is greater than 7 because only the conjugate base remains B. the equivalence point is greater than 7 because the base is stronger than the acid C. the equivalence point equals 7 because [H3O + ] = [OH - ] D. the equivalence point is less than 7 because only the conjugate acid remains E. the equivalence point is less than 7 because the acid is stronger than the base
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
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
In a titration a student used 16.60 mL of 0.100 M NaOH to titrate a 0.2000-g sample of an unknown acid. Which of the following acids would the unknown most probably be? Assume that only the hydrogens bonded to oxygen (in BOLD) are titrated by the sodium hydroxide, and that, due to experimental error, the student's value may not be identical with the theoretical value. A. succinic acid. HOOCCH2CH2COOH,molar mass = 118 B. benzoic acid. C6H5COOH. molar mass = 122 C. phthalic acid. C6H4(COOH)2. molar mass = 166 D. oxalic acid. HOOC—COOH. molar mass = 90. E. Not enough information is provided to narrow the selection to one.
In the titration of acetic acid (CH 3COOH) with NaOH, when one-half of the acetic acid has been titrated (to the half-equivalence point), the two (2) species present at equal concentrations are _________.a. H+ and OH-b. Na+ and OH-c. CH3COOH and H+d. CH3COO- and CH3COOHe. CH3COO- and H+
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
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
A student titrates an unknown monoprotic acid with NaOH solution from a buret. After the addition of 12.35 mL of NaOH, the pH of the solution is 5.22. The equivalence volume is 24.70 mL. What is the Ka of the acid?
Given the concentration of a strong base solution (assume a Group IA metal with hydroxide), what is the correct procedure to determine the pH of the solution?A) Set concentration of the base equal to [OH -], convert to [H3O+] and solve for pHB) Find Kb and convert to Ka and use a table to find the equilibrium concentration of [H 3O+] and solve for pHC) Set concentration of the base equal to [H 3O+] and solve for pHD) Use a table to find the equilibrium concentration of the conjugate acid which is equal to [H  3O+] and solve for pHE) Find Kb , and set-up a table to find the equilibrium concentration of [OH -], convert to [H3O+] and solve for pH
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?
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?
Gas A enters the tube, passes through the contents of B, and emerges as gas C. If gas A is a mixture of carbon monoxide and carbon dioxide, and gas C is carbon monoxide, what is the most likely substance in the tube?(A) concentrated HCl(aq)(B) pure water(C) phosphorous pentoxide(D) sodium hydroxide solution
Whenever we react a WEAK ACID with a STRONG BASE or a STRONG ACID with a WEAK BASE we use an ICF CHART.In this case the units must be in_______________. 
Nitrous acid, HNO2 has a Ka of 4.6 x 10-4. If the 100 mL of a 0.2 M solution of nitrous acid is mixed with a 200 mL of a 0.1 M solution of NaOH which of the following is NOT true of the resulting solution? 1. [Na+] > [H 3O +] 2. [NO2-] = [OH -] 3. they are all true 4. [NO2-] > [HNO 2] 5. [OH-] > [H3O+]
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.
When a 0.25 M HIO3 solution reacts with 10.0 mL of a 0.30 M K 2O solution the pH at the equivalence point will be:a) Less than 7.b) More than 7.c) Equal to 7.
When a 0.25 M H 3PO4 solution reacts with 10.0 mL of a 0.30 M BaH 2 solution the pH at the equivalence point will be:a) Less than 7.b) More than 7.c) Equal to 7.
Consider the titration of 50.0 mL of 0.125 M H  3PO4 with 0.160 M Ba(NH2)2. How many milliliters of the 0.160 M Ba(NH2)2 are needed to reach the equivalence point? 
An aqueous solution contains dissolved NH4Cl and NH3. The concentration of NH3 is 0.50 M and the pH is 8.95. The Kb of NH3 is 1.8x10-5a. What is the molar concentration of NH4 + in the buffer?      b. What is the pH after adding 0.10 moles of NaOH to 1.00 L of this solution? Assume no volume change upon addition of the NaOH?   
A 30.0 ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. a. Calculate the pH at 0 mL of added base. b. Calculate the pH at 5 mL of added base. c. Calculate the pH at 10 mL of added base. d. Calculate the pH at the equivalent point. e. Calculate the pH at one-half of the equivalence point. f. Calculate the pH at 20 mL of added base. g. Calculate the pH at 25 mL of added base.
An analytical chemist is titrating 202.5 mL of a 0.2200 M solution of propionic acid (HC2H5CO2) with a 0.5400 M solution of KOH. The pK a of propionic acid is 4.89. Calculate the pH of the acid solution after the chemist has added 89.35 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.
If 0.3705 g of pure KHP solid were NaoH and the endpoint was reached at 24.80 mL, what is the concentration of the NaOH solution? (NOte: MW of KHP = 204.223 g/mol).