1 / 11

SOLUTION STOICHIOMETRY LECTURE 3

SOLUTION STOICHIOMETRY LECTURE 3. ACIDS AND BASES. Arrhenius. Acids Give off H + Bases Give off OH - Substances that caused a problem??? Main example – NH 3. Br ønsted-Lowry. Came up with new definition to account for these Acids Give off (donate) protons (H + ) Bases

seanna
Télécharger la présentation

SOLUTION STOICHIOMETRY LECTURE 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SOLUTION STOICHIOMETRY LECTURE 3 ACIDS AND BASES

  2. Arrhenius • Acids • Give off H+ • Bases • Give off OH- • Substances that caused a problem??? • Main example – NH3

  3. Brønsted-Lowry • Came up with new definition to account for these • Acids • Give off (donate) protons (H+) • Bases • Accept protons (H+) • Remember conjugates??

  4. Neutralization • A special type of double replacement reaction • Reactants = • Acid • Base • Products • Water • A salt (ionic compound)

  5. Neutralization

  6. Strong Acid + Strong Base • Hydrochloric acid(aq) + sodium hydroxide(aq) • Demo • Molecular • HCl (aq) + NaOH (aq) --> NaCl (aq) + HOH (l) • Complete Ionic • H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) --> Na+(aq) + Cl-(aq) + HOH(l) • Net Ionic • H+(aq) + OH-(aq) --> HOH(l) Formation of water

  7. Weak Acid + Strong Base • Acetic acid (aq) + potassium hydroxide (aq) • Molecular • HC2H3O2 (aq) + KOH (aq) --> KC2H3O2 (aq) + HOH (l) • Complete Ionic • HC2H3O2 (aq) + K+(aq) + OH-(aq) --> K+(aq) + C2H3O2-(aq) + HOH(l) • Net Ionic • HC2H3O2 (aq) + OH-(aq) --> C2H3O2-(aq) + HOH(l)

  8. Stoichiometry • Write equations. • Balance net ionic equation. • Find moles (from original solutions). • Limiting reactant • Mole ratio • Get it out of moles • Example - • How many grams of water are formed when 50.0 mL of 0.356 M HI react with 28.0 mL of 0.488 M barium hydroxide?

  9. Titration • Delivery with a buret of a known solution (titrant) into an unknown solution (titrateor analyte) until the endpoint or equivalence point • Vocabulary • Titrant = solution of known concentration, in buret • Titrate = solution of unknown concentration, in beaker • Endpoint = point where an indicator first changes color, titrant exactly reacted with the titrate • Equivalence point = HUGE!!! • Demo = iron ions and permanganate (we’ll do it for real later in the year & simulated) http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redoxNew/redox.html

  10. HUGE POINT • Equivalence point • Occurs when MOLES = MOLES (stoichiometrically) Really important for titration calculations!!

  11. Example Titration Problem from AP exam • 0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate (from a buret) to titrate it and cause it to turn pink (the end point). The equation for this reaction is: • 5 Na2C2O4 (aq) + 2 KMnO4 (aq) + 8 H2SO4 (aq) 2 MnSO4 (aq) + K2SO4 (aq) + 5 Na2SO4 (aq) + 10 CO2 (g) + 8 H2O (l) (a) How many moles of sodium oxalate are present in the flask? (b) How many moles of potassium permanganate have been titrated into the flask to reach the end point? (c) What is the molarity of the potassium permanganate?

More Related