letcure+f - Ch 10 Recap Acids/Bases Acids release[H in...

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Ch 10 – Recap Acids/Bases HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) original acid original base “new” acid CONJUGATE ACID “new” base CONJUGATE BASE - Acids release [H + ] in aqueous solution In the Arrhenius theory: - Bases release [OH ] in aqueous solution In Brønsted-Lowry theory: - Acids donate a proton in a chemical reaction - Bases accept a proton in a chemical reaction Acid/bases and their conjugates
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Ch 11 – Acids and bases HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) B(aq) + H 2 O(l) BH + (aq) + OH (aq) Let’s look at the general acid and general base reactions again: In the top reaction, water is the base (proton acceptor) and in the bottom reaction, water is the acid (proton donator). Water is AMPHOTERIC , meaning it can act both as an acid or base: H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Experiment has shown that for a sample of pure water at 25 °C: [H 3 O + ] eq = [OH ] eq = 1 x 10 7 M Amphoteric substances K eq = [H 3 O + ] [OH ] “ion-product constant” Which means K w = (10 7 ) 2 = 10 14 = K w (at 25 °C)
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Ch 11 – Acids and bases Thus whenever water is present, an acid/base equilibrium with an exceedingly low K eq is ALWAYS trying to be established. The low value of this equilibrium establishes a maximum amount of H + and OH which are allowed to exist in solution at the same time: - When an acid is added to water, H + goes up and OH goes down. Here, [H + ] eq > [OH ] eq and the solution will be called acidic . - When a base is added to water, OH goes up and H + goes down. Here, [OH ] eq > [H + ] eq and the solution will be called basic . In addition to this, K W helps us relate K a and K b for an acid/base pair: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) A (aq) + H 2 O(l) HA(aq) + OH (aq) K a K b H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) K w Overall, we can show that for an acid/base conjugate pair: K a ×K b = K w
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Practice problem #17 Rank the following species in order of acid strength, strongest to weakest. HNO 2 , HCH 3 COO, CH 3 CH 2 NH 3 + , HClO 4 , Na + . Ch 11 – Acids and bases The only way to do this is to determine the K a values: From the textbook we can get HNO 2 = 4.5 ×10 4 , HCH 3 COO = 1.8 × 10 5 , and HClO 4 = VERY LARGE (one of the strong acids). What about the other two? Na + is technically the conjugate acid of NaOH, a strong base. Na + cannot donate protons, and has no interest in reacting at all once formed (“spectator ion”). K a of Na + = 0 (does not exist) . CH 3 CH 2 NH 3 + is the conjugate acid of ethyl amine, which means: K a = 10 14 5.6 × 10 4 = 1.8 × 10 11 From the K a data, we can rank them as such: HClO 4 > HNO 2 HCH 3 COO 4 > CH 3 CH 2 NH 3 + > > Na + (stronger) (weaker) K a = K H hydrolysis of the acid from reaction with water
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Ch 11 – Acids and bases In most cases, [H 3 O + ] and [OH ] are very small numbers. For “simplicity,” chemists decided to define a scale based on the negative logarithm (
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This note was uploaded on 11/04/2009 for the course CHEM Chem 1C taught by Professor Farmer during the Spring '09 term at UCL.

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letcure+f - Ch 10 Recap Acids/Bases Acids release[H in...

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