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Chemical reactions & Solutions & ACIDS BASES

Chemical reactions & Solutions & ACIDS BASES. Acid  Increases the concentration of H + ions in an aqueous solution Base  Increases the concentration of OH - ions in an aqueous solution H + + H2O  H3O + Hydronium

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Chemical reactions & Solutions & ACIDS BASES

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  1. Chemical reactions&Solutions&ACIDS BASES

  2. GSCI 163 Spring 2010

  3. Acid  Increases the concentration of H+ ions in an aqueous solution Base  Increases the concentration of OH- ions in an aqueous solution H+ + H2O  H3O+ Hydronium No distinction is made as to whether or not the H+ or the H3O+ is actually present in the solution. We will consider treat these two entities as the same thing. OH- Hydroxide Acid  Molecule that transfers a proton Base  Molecule that accepts a proton In examining a reaction where a proton is transferred from one molecule to another an acid base pair can be identified GSCI 163 Spring 2010

  4. HCl + H2O HCl + H2O  H3O+ + Cl- + H2O HCl is an acid because it increase H3O+ HCl is an acid because it donates a proton to H2O H3O+ [Here H2O would be the base that accepts the proton] GSCI 163 Spring 2010

  5. NaOH + H2O NaOH + H2O  Na+ + OH- + H20 NaOH is a base because it increase OH- NaOH is base because the OH- can accept a proton [When considering this as a base we would need to consider a reaction with the NaOH NaOH + HCl  Na+ + OH- + HCl  Na+ +H2O + Cl- Here the OH- accepts a proton and HCl donates a proton. So we have an acid base reaction. ] GSCI 163 Spring 2010

  6. Keq= [H3O+] [OH-] Water reaches an equilibrium state when the product of the concentration of H3O+ [H3O+] and the concentration of OH- [OH-] [H3O+] [OH-] = 1x10-14moles/liter Concentration =# moles per liter In one liter of pure water you will have [H3O+]=1x10-7molar concentration of dissolved hydrogen ions (H+). [OH-]=1x10-7molar concentration of dissolved hydroxide ions (OH-). GSCI 163 Spring 2010

  7. One might guess that the rate at which a reaction occurs must be proportional to the amount of stuff there is which interacts. In a one step simple process Aproducts Rate = K[A] Extrapolating to a two reactant process A+B products Rate= K[A][B] A+B products  A+B in equilibrium The rate to produce products is the same as the rate of the products to produce the reactants. For salt in water at some point the rate salt is dissolved is the same as the rate that salt precipitates. This is an equilibrium and determines the solubility of the salt. GSCI 163 Spring 2010

  8. H2O H3O+ + OH- H2O If you increase the number of H+ then they find OH- and form H2O until the numbers are depleted so that the likelihood of two connecting matches the rate that H2O disassociates. Because the amount of water is large we can assume it is constant and so the disassociation rate is always the same. Equilibrium is then established when the combination rate is the same. Since this rate depends on the product we establish this product. [H3O+] [OH-] = 1x10-14 moles/liter GSCI 163 Spring 2010

  9. Simple Model Dance for couples (MW only to enter) Couple concentration [MW] Allow couples to argue or separate. Single men concentration [M] Single women concentration [W] If the number of separations is small we will assume that [WM] remains fixed. GSCI 163 Spring 2010

  10. Eventually all couples will separate due to boredom or disagreements ect. and we can characterize their separation with a rate. Every minute one couple separates for every 1000 couples at the dance. The process would gradually disassociate all the couples. Let us assume that when a single man meets a single women they might recognize each other or simply want to meet and talk. They therefore e-associate into a couple. GSCI 163 Spring 2010

  11. The process of forming couples must depend on the number of single men and single women in an area. Consider: • Equal numbers • Lots of men few women • Lots of women few men It is plausible that the rate of forming couples depends on the product. rate = [M][W] This at least has the correct behavior. • Lowering either number increase rate • Increasing either increases the rate • Lowering one can be compensated by increasing the other. Might feel confident that the meeting probability has to depend on this product. GSCI 163 Spring 2010

  12. Therefore at some point the room stabilizes with a certain number of couples separating and and equal number forming. The number of singel men and women is then fixed. GSCI 163 Spring 2010

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