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Acid-Base Reactions and Definitions: Hydrogen Ion and Hydronium Ion Chemistry

Learn about definitions of acids and bases, including the hydrogen ion (H+) and hydronium ion (H3O+). Explore Arrhenius and Bronsted-Lowry definitions, monoprotic and diprotic molecules, strong and weak acids, conjugate acid-base pairs, acid ionization constant (Ka), and more!

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Acid-Base Reactions and Definitions: Hydrogen Ion and Hydronium Ion Chemistry

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  1. Acid-Base reactions

  2. Definitions of acids and bases • Read pages 637-640 to fill in your note template

  3. Hydrogen Ion/Hydronium ion • In acid base chemistry we will often write the Hydrogen ion H+ as the Hydronium ion H3O+ • These two ions are considered to be the same thing as H+ will combine with water to form H3O+ • H+ will often be referred to as a proton (it has just one proton, no electrons).

  4. Arrhenius acid/base

  5. Bronsted-Lowry Acid/Base

  6. Conjugate Acid/Base pairs(Bronsted Lowry) • NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) • HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq) • H2SO4(aq) + H2O(l) HSO4-(aq) + H3O+(aq) • HSO4-(aq)+H2O(l)  SO42- (aq) + H3O+(aq) • HCO3-(aq) + H2O(l)  H2CO3(aq) + OH-(aq)

  7. Monoproticand diprotic molecules • Monoprotic • Mono = 1 • Protic = proton • Monoprotic molecules have a single proton they can donate ex. HCl • Diprotic • Di = 2 • Protic = proton • Diprotic molecules have two protons which can be donated ex. H2SO4

  8. Strong and weak acids • Strong acids – acids which dissociate completely in solution • HCl is a strong acid, the dissociation of HCl can be written: HCl(aq) H+(aq) + Cl-(aq) or:HCl(aq)+H2O(l) H3O+(aq) + Cl-(aq) • Both equations yield the same Keq equation as H3O+ = H+ • If we were to look at a Keq equation: • Keq=[H+][Cl-]Keq for this reaction is 3.2x109. It strongly [HCl] favours the products, we can assume that all of the HCldissociates.

  9. Strong acids • There are 6 strong acids: • HCl – Hydrochloric acid • HBr – Hydrobromic acid • HI – Hydriodic acid • HNO3 – Nitric acid • HClO4 – Perchloric acid • H2SO4 - Sulfuric acid • All of these acids have a Keq greater than 1. • Note: H2SO4 is diprotic, only the first H+ dissociates completely to form HSO4- which is a weak acid.

  10. Naming Acids quick review • Binary acids (two elements only) • Receive the prefix (at the start) “hydro” • Receive the suffix (at the end) “ic acid” • Ex. H2S is hydrosulfuric acid • Acids containing polyatomic ions • If the polyatomic ion ends in “ate” change the suffix to “ic acid” • Ex. HNO3 contains the nitrate ion, so it is nitric acid • If the polyatomic ion ends in “ite” change the suffix to “ous acid” – nitric acid • Ex. HNO2 contains the nitrite ion, change the suffix to “ous acid” • Nitrous acid

  11. Acid Ionization constant Ka • For acids we do not use Keq, instead it is denoted as Ka. • Ka = Acid ionization constant • The principles used are the same as Keq Ka= [products]coefficients [reactants]coefficients • The acid ionization constant can be used to find the concentration of H+/H3O+ in a solution.

  12. Weak acids • Weak acids do not dissociate completely in solution • Their Ka is less than 1. • We can use Ka to determine the concentration of ions in solution. • Example problem: • The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ • Step 1: Balanced equation • HF(aq)  H+(aq) + F-(aq)

  13. Example problem The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ • Step 2: Create an ICE table H2O(l) + [HF](aq) [H3O+](aq) + [F-] (aq) • Step 3: Fill in your Ice table • Step 4: Acid ionization equation. • Ka= [H3O+][F-] [HF]

  14. Example problem The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ H2O(l) + [HF](aq) [H3O+] (aq) + [F-] (aq) • Step 5: Substitute Equilibrium values • Ka= (x)(x) (1.0-x) • Step 6: you may be able to assume “x” is small, dependant on the size of Ka. Generally 1.0 x10-4 is small enough but not always!

  15. Example problem The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ H2O(l) + [HF](aq) [H3O+] (aq) + [F-] (aq) • Step 7: Solve for x • Ka= (x)(x) where Ka= 3.5x10-4 (1.0)

  16. Example problem The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ H2O(l) + [HF](aq) [H3O+] (aq) + [F-] (aq) • Step 8: Check to see that x is actually small:

  17. Example problem The Ka for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentrations of HF, F- and H3O+ [HF](aq) [H+] (aq) + [F-] (aq) • Step 9: Use x to find the equilibrium concentrations. • This will give you your equilibrium concentrations.

  18. Example number 2: • Find the equilibrium concentration of [H+] for a 0.250 M solution of acetic acid. Ka for acetic acid is 1.8x10-5 • Step 1: write out the dissociation equation. • Step 2: Create an ICE table • Step 3: Fill in the Ice table • Step 4: Write out the acid ionization equation • Step 5: Substitute equilibrium values into Ka equation • Step 6: use “x” is small where appropriate • Step 7: solve for x • Step 8: check to see if “x” is small is appropriate. (if not re-solve without x is small) • Step 9: Substitute calculated x to find [H+]

  19. Strong bases • Much like acids, bases can be either weak, or strong depending on whether they dissociate completely in solution or not. • A strong Base completely dissociates into its ions in solution. • Strong bases are: Lithium hydroxide - LiOH Calcium hydroxide – Ca(OH)2 Sodium hydroxide – NaOHStrontium hydroxide - Sr(OH)2 Potassium hydroxide – KOH Barium Hydroxide – Ba(OH)2 Ex. LiOH(aq)  Li+(aq) + OH-(aq) Sr(OH)2(aq)  Sr2+(aq) + 2OH-(aq) Note that bases with two hydroxide ions (OH-) dissociate in one step unlike diprotic acids.

  20. Weak bases • Weak bases, like weak acids, do not completely dissociate into their ions • Weak bases react with water to produce a hydroxide ion, typically by accepting a proton from water molecules. • Ex. • NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) • Note that a compound does not have to contain an OH- group in order to be a base. • For bases we do not use the acid ionization constant, instead we use the base ionization constant, Kb

  21. Example problem: • You have a 0.5 M solution of the weak base ammonia. Find the concentration of hydroxide ions at equilibrium. The base ionization constant is 2.5 x 10-6. • Step 1: write out the dissociation equation. • Step 2: Create an ICE table • Step 3: Fill in the Ice table • Step 4: Write out the base ionization equation • Step 5: Substitute equilibrium values into Kb equation • Step 6: use “x” is small where appropriate • Step 7: solve for x • Step 8: check to see if “x” is small is appropriate. • Step 9: Substitute calculated x to find [OH-]

  22. Water, acid or base?

  23. Water: • Acts as an acid or base • Can donate a proton to become OH- • Can receive a proton to become H3O+ • Pure water acts as both an acid and a base H2O(l) + H2O(l)  OH-(aq) + H3O+(aq) • This is called the Autoionization of water: • The ability of water to act as both an acid and a base even in pure water

  24. Ion product constant for water (Kw) • The ion product constant for water is exactly the same as all the other K’s we have discussed (Keq, Ksp, Ka, Kb) • H2O(l) + H2O(l)  OH-(aq) + H3O+(aq) • Because water is a pure liquid it is not involved in the Kw equation • Kw= [OH-][H3O+] • Kw at 25⁰C is 1.0x10-14 • To find [OH-] and [H3O+] we can simply set up an ICE table

  25. Ion product constant for water (Kw) [OH-] [H3O+] • Kw=1.0x10-14=[OH-][H3O+] = (x)(x)

  26. Ion product constant for water (Kw) • Used to find [OH-] for problems involving acids, or [H+] for problems involving bases. • Once you know the concentration of either [OH-] or [H3O+] you can easily find the concentration of the other using the ion product constant for water (Kw) • Ex. The concentration of [OH-] in a solution is 1.2x10-4 M, find the concentration of [H+] in this solution.

  27. The pH scale • The pH scale is used to quickly convey information whether a solution is basic or acidic • Basic solutions have higher [OH-] than [H+]. • Acidic solutions have higher [H+] than [OH-]. • Kw can be used to determine [OH-] and [H+]. • pH= - log[H+]

  28. Example problem • A 0.50M solution of hydrochloric acid is made. What is the pH of this solution? • Step 1: what do you know and what do you need to find? • [HCl]=0.50M • pH=? • We also know HCl is a strong acid • Step 2: write the dissociation equation • HCl(aq)H+(aq)+Cl-(aq) • Since HCl is a strong acid and will completely dissociate, the initial concentration of HCl=[H+] • pH= -log[H+] • pH= -log(0.50)

  29. Example problem. • Find the pH of a 1.0 M solution of Nitrous acid. Ka for nitrous acid is 4.6x10-4. • Step 1: write out the dissociation equation. • Step 2: Create an ICE table • Step 3: Fill in the Ice table • Step 4: Write out the acid ionization equation • Step 5: Substitute equilibrium values into Ka equation • Step 6: use “x” is small where appropriate • Step 7: solve for x • Step 8: check to see if “x” is small is appropriate. (if not re-solve without x is small) • Step 9: Substitute calculated x to find [H+] • Step 10: Use [H+] to find pH using pH=-log[H+]

  30. Finding [H+] from pH • We can find our equilibrium [H+] by using the equation for pH • pH=-log[H+] • Log[H+]=-pH • [H+]=10-pH • Ex. Find the [H+] at equilibrium for a solution with pH = 3.8 • [H+]=10-3.8 how you solve this depends on your calculator • Either: • Punch in -3.8 then the 10x button (2nd function log typically) • Or hit 10x then -3.8

  31. Example problem. • The pH of a solution of hydrofluoric acid is 5.2. Ka for hydrofluoric acid is 3.5x10-4. What was the initial concentration of hydrofluoric acid in solution? • 1. Identify what we have and what we must find out. pH = 5.2 Ka=3.5x10-4 Initial [HF]=? • So how can we get our equilibrium concentration of hydrofluoric acid?

  32. Example problem. • The pH of a solution of hydrofluoric acid is 5.2. Ka for hydrofluoric acid is 3.5x10-4. What was the initial concentration of hydrofluoric acid in solution? • 2. Write a balanced equation • HF(aq) + H2O(l)  F-(aq) + H3O+(aq) • 3. Write the acid ionization equation • Ka= [F-][H3O+] We only have one known at this point, we [HF] also know how to find [H3O+] from pH • 4. Find the concentration of [H3O+] • pH= -log[H3O+] • -pH = log[5.2] remember to use the 10x on your calculator.

  33. Example problem. • The pH of a solution of hydrofluoric acid is 5.2. Ka for hydrofluoric acid is 3.5x10-4. What was the initial concentration of hydrofluoric acid in solution? • HF(aq) + H2O(l)  F-(aq) + H3O+(aq) • Once you have your concentration of H3O+ you can look at the balanced equation. For every mol of H3O+ produced you will also get one mol of F- so we really have three values in our Ka expression • 5. Use the coefficients to find the concentration of the anion. • 6. Solve for [HF] • Ka= [H3O+][F-] [HF]

  34. Example problem. • The pH of a solution of hydrofluoric acid is 5.2. Ka for hydrofluoric acid is 3.5x10-4. What was the initial concentration of hydrofluoric acid in solution? • 7. Set up an ICE table the initial concentrations of ions will be 0. HF(aq) + H2O(l)  F-(aq) + H3O+(aq) • 8. Fill in the ICE table, the values you obtained from the acid ionization expression are the equilibrium values • 9. work backwards to find the initial concentration of hydrofluoric acid. • Check your answer does it make sense? Is it what you were looking for?

  35. Find the initial concentration of acetic acid for a solution with pH 5.8. The acid ionization constant for acetic acid is 1.8x10-5 Finding initial concentrations from pH • 1. Identify what we have and what we must find out. • 2. Write a balanced equation • 3. Write the acid ionization equation • 4. Find the concentration of [H3O+] using pH=-log[H3O+] • 5. Use the coefficients to find the concentration of the anion. • 6. Solve Ka for [Acid] • 7. Set up an ICE table the initial [ions] will be 0. • 8. Fill in the ICE table • 9. work backwards to find initial [acid] • 10. Check answer

  36. Finding Ka from pH and initial [acid] Find the initial concentration of acetic acid for a solution with pH 5.8. The acid ionization constant for acetic acid is 1.8x10-5 • 1. Identify what we have and what we must find out. • 2. Write a balanced equation • 3. Write the acid ionization equation • 4. Find the concentration of [H3O+] • 5. Use the coefficients to find the concentration of the anion. • 6. Set up an ICE table the initial [ions] will be 0. • 8. Fill in the ICE table, the values you obtained from the acid ionization expression are the equilibrium values • 9. work backwards to find the initial concentration of hydrofluoric acid. • 10. Look back, is your answer what you were looking for?

  37. Example problem. • Find the pH of a 0.5 M solution of acetic acid. Ka for acetic acid is 1.8x10-5.

  38. What can we do? • We can find Ka given initial [acid] and an equilibrium [H+] • We can find initial [acid] given Ka and [H+] • We can find [H+] given Ka and initial [acid] • We can find [H+] given equilibrium [anion], pH or [OH-] • We can find [anion],pH and [OH-] from [H+] • We can find [OH-] using Kw when given [H+]

  39. [OH-] Equilibrium [anion] pH

  40. Identify how we will get to the answer • Find the initial concentration of acetic acid for a solution with pH 3.4. The acid ionization constant for acetic acid is 1.8x10-5 • The acid ionization constant for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the concentration of hydroxide ions in this solution. • Find the concentration of hydroxide ions for a solution with pH 5.8. • Find the pH of a 1.0 M solution of Nitrous acid. Ka for nitrous acid is 4.6x10-4. • The acid ionization constant for Hydrofluoric acid (HF) is 3.5x10-4 a 1.0 M solution of HF is created, find the pH of this solution.

  41. What will we do next? • Just like there is a pH scale representing the concentration of H+ there is a pOH scale for concentration of OH-. • So we can do all the same calculations with acids as we can with bases. • We can find Kb given initial [base] and an equilibrium [OH-] • We can find initial [base] given Kb and [OH-] • We can find [OH-] given Kb and initial [base] • We can find [OH-] given equilibrium [cation], pOH or [H+] • We can find [cation], pOH and [H+] from [OH-] • We can find [H+] using Kw when given [OH-]

  42. One more thing! • pH + pOH always equals 14. So given pOH or pH we can find the other value very quickly.

  43. For Acids For Bases pOH [OH-] pH [OH-] Equilibrium [anion] pH Equilibrium [cation] pOH

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