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
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Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
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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
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
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
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 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 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
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
A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 0 mL of added base.
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?
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
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
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 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?
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.
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.
Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(a) no KOH added
A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M  NaOH solution. Calculate the pH after the following volumes of base have been added.0 mL of the base
A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M  NaOH solution. Calculate the pH after the following volumes of base have been added.34.5 mL of the base
Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10  -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.b. 50.0 mL
Lactic acid is a common by-product of cellular respiration and is often said to cause the “burn” associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.
Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. (K  a for HCN = 6.2 x 10-10.)c. Calculate the pH after 75.0 mL of KOH has been added.
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (c) 15.00 mL
A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M  NaOH solution. Calculate the pH after the following volumes of base have been added.35.5 mL of the base
Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42.2 mL of 0.0520 M CH 3COOH
A 10.0 mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH.What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.
Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.a. 100.0 mL of 0.10 M HC7H5O2 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH
A weak monoprotic acid is titrated with 0.100 M NaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of base is added, the pH of the solution is 3.62.Estimate the pKa of the weak acid.
1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.Write the governing equilibrium.A.  CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)B. CH3NH3Cl(aq) ⇌ CH3NH3+(aq) + Cl-(aq)C. Ba(OH)2(aq) ⇌ Ba2+(aq) + OH-(aq) D. 2 CH3NH3Cl(aq) + Ba(OH)2(aq) ⇌ 2 CH3NH2(aq) + BaCl2 + 2 H2O(l)E. CH3NH3+(aq) + OH-(aq)  ⇌ CH3NH2(aq) + H2O(l)
A 1.0L buffer solution contains 0.100 mol of HC2H3O2 and 0.100 mol of NaC2H3O. The value of Ka for HC2H3O2  is 1.8×10−5. Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer.
During the titration of 10.0 mL of a 0.100 M aqueous solution of acetic acid (Ka = 1.8 x 10-5), what is the pH following the addition of 20.0 mL of a 0.100 M aqueous solution of NaOH?A. 4.76B. 7.00C. 8.72D. 12.5E. 13.5
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.(a) before addition of any KOH(b) after addition of 25.0 mL of KOH(c) after addition of 30.0 mL of KOH(d) after addition of 50.0 mL of KOH(e) after addition of 60.0 mL of KOH
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.(a) before addition of any KOH(b) after addition of 25.0 mL of KOH(c) after addition of 30.0 mL of KOH(d) after addition of 50.0 mL of KOH(e) after addition of 60.0 mL of KOH
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
A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.What will the pH be at the equivalence point of the above titration?A. 7.60B. 6.03C. 7.97D.  2.83E. 11.17
Fifty milliliters of a 0.10 M aqueous NaOH solution is needed to reach the equivalence point in a titration of a 50. mL sample of benzoic acid (Ka = 6.3 x 10-5). What is the pH of the solution at the equivalence point?A. 5.6B. 7.0C. 8.4D. 10.2E. none of the above
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
1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.Calculate the pH of the solution (pKa(CH3NH3+) = 10.62).A. 11.00B. 10.49C. 10.00D. 11.17E. 10.75
Calculate the pH of a solution formed by adding 0.20 moles of NaOH to 1.00 liter of 0.50 M acetic acid. Ka = 1.8 x 10-5 for acetic acid.A. 4.24B. 3.74C. 4.92D. 4.57E. 2.63
245 mL of a 0.143 M solution of Ba(OH)2 was mixed with 400 mL of 0.53 M acetic acid. What is the final pH?A. 4.0B. 4.4C. 9.6D. 12.1E. 13.8
1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.Is this an acidic or basic buffer? Justify your answer.A. neutralB. acidic neutralC. basic neutral
Paracresol (C7H8OH) has a Ka = 5.5 x 10-11. If 100.0 mL of a 0.5000 M aqueous paracresol solution is mixed with 100.0 mL of 0.5000 M aqueous sodium hydroxide, the resulting solution will have a pH
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).