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Bronsted - Lowry. Acids and Bases. This is our second conceptual definition. Acids – Substances that donate protons (H + ) Bases – Substances that accept proton(s). Compare with Arrhenius HCl (aq) H + (aq) + Cl - (aq) (Arrhenius).
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Bronsted - Lowry Acids and Bases
This is our second conceptual definition. • Acids – Substances that donate protons (H+) • Bases – Substances that accept proton(s) Compare with Arrhenius HCl(aq) H+(aq) + Cl-(aq) (Arrhenius) HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) (B-L) Acid Gives up protons Base Accepts protons
Other Examples NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Base: Accepts protons Acid: gives up protons Note: Protons always move from the acid to the base. Water is AMPHIPROTIC ( or AMPHOTERIC) meaning that it can function as either an acid or a base. Check the strength of acid chart in your data booklet to see which other substances are amphiprotic
Conjugate Acid / Base Pairs In any Bronsted-Lowry equation, there are 2 acids and 2 bases. HSO4-(aq) + HPO42-(aq) SO42-(aq) + H2PO4-(aq) Acid (Gives up H+) Base (accepts H+) Base (accepts H+) Acid (gives up H+) A conjugate acid/base pair consists of 2 substances that differ by the gain or loss of a single proton (H+ )
Therefore the conjugate pair will always be the acid from one side of the equation and the base from the other side. Each equation will have 2 conjugate acid/base pairs. In the above equation, the conjugate pairs are: HSO4-(aq) with SO42-(aq) and… HPO42-(aq) with H2PO4-(aq) Check your data booklet. Notice that the formula for each conjugate is the result of removing a single H+ from the formula for the acid.
Important notes The strongest acid is the one that gives up protons the easiest. That is: it functions very well as an acid. The strongest base is the one that has the greatest attraction for protons. That is: it functions very well as a base.
Table in your data booklet Acids Conjugate Bases Strongest Weakest Weakest Strongest
How to represent substances in Bronsted-Lowry reactions These are rules that you must know in order to be able to write these reactions. Learn them well.
I. Strong Electrolytes: Ionic salts Strong Acids Strong Bases These are written in dissociated (ion) form. That is: Ionic salts are written as individual ions. Example: NaCl(s)→ written as Na+(aq) and Cl-(aq) Strong Acids are written as the hydronium ion. Example: HCl (aq) → written as H3O+(aq) ALL STRONG ACIDS ARE WRITTEN THE SAME WAY Strong Bases are written as hydroxide ions. Example: NaOH(s)→ written as OH-(aq) ALL STRONG BASES (HYDROXY BASES) ARE WRITTEN THE SAME WAY
II. Weak Electrolytes: Weak Acids Weak Bases These are written in associated form. That is: they are written “as is”. Examples: CH3COOH(aq) NH3(aq) → written as CH3COOH (aq) → written as NH3(aq)
review time!! YEAH!!
Brønsted-Lowry Acids and Bases • Proton Transfer Reactions • Brønsted-Lowry: (conceptual definition) • acid donates proton (H+) • base accepts proton (H+) • Brønsted-Lowry base does not need to contain OH-.
Brønsted-Lowry Acids and Bases Proton Transfer Reactions acid base
Brønsted-Lowry Acids and Bases • Conjugate Acid-Base Pairs • Whatever is left of the acid after the proton is donated is called its conjugate base. • Similarly, whatever remains of the base after it accepts a proton is called a conjugate acid.
Brønsted-Lowry Acids and Bases • Conjugate Acid-Base Pairs • Consider: • After HA (acid) loses its proton it is converted into A- (base). Therefore HA and A- are conjugate acid-base pairs. • After H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs. • Conjugate acid-base pairs differ by only one proton.
Brønsted-Lowry Acids and Bases Proton Transfer Reactions acid base conjugate base conjugate acid
Strong Acids and Bases • Strong Acids • The strongest common acids are HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4. • Strong acids are strong electrolytes. • All strong acids ionize completely in solution:
back to the real world ( new stuff )
He always takes the credit for my work! They should be Lowry-Bronsted reactions. Here is the procedure for writing my reactions. It will always work. Lowry Bronsted
Procedure for writing Bronsted-Lowry reactions The reaction will always occur between thestrongestacid presentand thestrongest base present. 1. List all species present. (don’t forget the H2O (l) ) 2. Locate the strongest acid present using your data book. (start at the top left and work down. 3. Locate the strongest base present using your data book. (start at the bottom right and work up.) 4. Transfer a single proton from the acid to the base. 5. Fill in the products. (these will be the conjugate acids and bases of your reactants)
Examples. Species list 1. HBr(aq) is mixed with NH3(aq) H3O+(aq) NH3(aq) H2O(l) W.B. written “as is” S.A. written as H3O+ (aq) H3O+(aq) + NH3(aq) NH4+(aq) + H20(l) Acid Base Con - Acid Con - Base
Example 2CaSO3(aq) mixed with HF(aq) Species List W.A.→ “as is” HF(aq) Ca2+(aq) SO32-(aq) Ionic salt → Ca2+(aq) + SO32-(aq) HF(aq) H2O(l) HF(aq) + SO32- (aq) HSO3-(aq) + F-(aq) Acid Base Con - Acid Con - Base Note: All metallic ions are spectator ions and can be omitted. They cannot function as an acid or a base.
Which side is favored? • Strongest acid in the whole equation will dissociate the most. Therefore the side opposite the strongest acid is favored
Example 3KHCO3(aq) mixed with Na2HPO4(aq) Species List HCO3-(aq) HPO42-(aq) H2O(l) Na+ + HPO42- K+ + HCO3- HCO3-(aq) + acid HPO42-(aq) base H2PO4-(aq) + acid CO32-(aq) base
Example 4.NaOH(aq) is mixed with HCl(aq) Species List OH-(aq) H3O+(aq) H2O(l) S.B. → OH-(aq) S.A. → H3O+(aq) H3O+(aq) + acid OH-(aq) base H2O(l) + H2O(l) acid base Neutralization
Example 5.KCl(aq) is mixed with H2O(l) Species List Cl-(aq) K+(aq) + Cl-(aq) H2O(l) H2O(l) H2O(l) + H2O(l) No Reaction Or: H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Same as in water →like we said…no reaction
Kb An equilibrium constant (Keq) for a base
Where do we find Kb? • There are no Kb values in the data book. • We need to use the Ka values on the strength of acid chart in our data book to calculate Kb.
Relationship Between Ka and Kb • For a conjugate acid-base pair • KaKb = Kw • Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base.
Example 1 • Find Kb for CH3COO-. CH3COO-(aq)+ H2O(l)↔ CH3COOH(aq) + OH-(aq) Look up Ka for conjugate acid (CH3COOH) Ka = 1.8 x 10-5 Ka x Kb = Kw 1.8 x 10-5 x Kb = 1.0 x 10-14 Kb = 5.6 x 10-10
Example 2 Ionic salt • Find Kb for Na2CO3. Na2CO3(aq)↔ CO3-2(aq) + 2Na+(aq) CO3-2(aq) + H2O(l)↔ HCO3-(aq) + OH-(aq) Look up Ka for conjugate acid (HCO3-) Ka = 4.7 x 10-11 Ka x Kb = Kw 4.7 x 10-11 x Kb = 1.0 10-14 Kb = 2.1 x 10-4 Base
Example 3 A 0.100 mol/L propanoic acid solution (C2H5COOH(aq)), has a pH of 2.95. From these data, the Kb for the propanoate ion, C2H5COO-(aq) is________. C2H5COOH(aq) + H2O (l) C2H5COO–(aq) + H3O+(aq) Ka = (1.122 x 10 -3) (1.122 x 10-3) (0.100 – 1.122 x 10-3) = 1.273 x 10-5 Kb = Kw (1.0 x10-14 ) Ka (1.273 x 10-5 ) = 7.9 x 10-10
Example 4 Household ammonia is a cleaning product in which the concentration of ammonia is about 2 mol/L. The pOH of a 2 mol/L ammonia solution is ____________. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = [NH4+] [OH-] [NH3] Now solve : Kb = x2 2 Kb(NH3) = Kw Ka (NH4+) 1.785 x 10-5 = x2 2 = 1.0 x 10-14 5.6 x 10-10 X = 5.976 x 10-3 = 1.785 x 10-5 pOH = 2.2
Neutralization Basic Principle • A reaction that involves equal moles of H+ ions and OH-ions. moles H+ = moles OH-
Neutralization • Strong or weak acids/bases are of no concern in neutralization reactions since weak acids will totally dissociate in the presence of a base.
Recall Le Chatelier’s Principle: CH3COOH(aq)↔ CH3COO-(aq) + H+ (aq) Addition of base will remove H+ (H+ + OH- → H2O). A decrease in [H+] will cause a shift to the right to replace the H+. i.e. dissociation increases WS 18-2
ACIDS Monoprotic – an acid that has one ionizeable proton. (egs…HCl, HI, HF, CH3COOH) Diprotic (egs…H2SO4, H2S, H2CO3, H2PO4-) Triprotic (egs…H3PO4) BASES Monobasic species – a base that donates one OH-. (or a base that accepts one proton) (egs…KOH, LiOH, HCO3-) Dibasic species (egs…Ca(OH)2, Mg(OH)2, CO32-) Tribasic species (egs…Al(OH)3, PO43-) Terms
Sample Calculations • What volume of 0.800 mol/L HCl(aq) is required to neutralize 50.0 mL of 0.500 mol/L NaOH(aq)? 2. 0.800 L of 0.300 mol/L Ca(OH)2 (aq) requires what volume of 0.500 mol/L H3PO4 (aq) to neutralize it? 3. 150 mL of Ba(OH)2(aq) is used to neutralize 50.0 mL of 0.040 mol/L of H3PO4(aq) . What is the concentration of the base? 4. A 4.00 g sample of acetic acid requires 50.0 mL of NaOH(aq) for a neutralization. What is the concentration of the base?
A short cut formula a CAVA = b CBVB Volume of Base Volume of Acid [Acid] [Base] Number of OH- donated per mole of base (1,2 or 3) Number of ionizeable protons (1,2 or 3)
Titrations Volumetric Analysis
Titrations • Titration is a physical process that is used to find the concentration of an unknown solution. Titration involves adding one solution (titrant) from a buret to another solution (sample) in an Erlenmeyer flask. (titration flask)
(titrant) indicator is added known volume (sample)
Carry out the qualitative Titration Lab
Qualitative Titration LabWe will titrate HCl(aq) with NaOH(aq) CBL Settings: mode: time graph interval: 0.5s – 1.0s total time: 120s
Procedure • Place 10.0 mL of HCl(aq) in a 250 mL beaker. • Add about 85.0 mL of distilled H2O(l). • Fill the right hand side buret with the NaOH (aq). • Set up the apparatus as per the diagram. • Start the magnetic stirrer. • Push ‘start’ to begin collecting data. • Open the valve on the buret and allow the base to drain into the beaker. (keep adding base until the pH is above 12 for several readings) • Use graph link to print your graph.
buret pH probe 250 mL beaker Stirring bar top of stirrer
Notes • Watch speed of the magnetic stirrer • Be careful when handling the pH probe. Make sure that the probe is submerged in the buffer solution when you put it away. • You must wear your and your • Or you will be out! • Have your bench checked before you leave the lab. Student
Qualitative Titration # 2 • Follow the same procedure as lab 1 with the following changes. • Substitute CH3COOH(aq) for the HCl(aq) • That is CH3COOH (aq) with NaOH (aq) Qualitative Titration # 3 • Follow the same procedure as lab 1 with the following changes. • Substitute NaHCO3(aq) for the NaOH(aq) • That is: HCl(aq) with NaHCO3(aq)