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BIOL499A (5)

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School
University of Alberta
Department
Biology (Biological Sciences)
Course
BIOL499A
Professor
Blaine Mullins
Semester
Winter

Description
1 2 Speed Bump by Dave Coverly Buffer solution A solution that contains appreciable amounts of both a weak acid (HA) and its conjugate base (A ). In other words we have a solution with appreciable amount of the acid and its common ion. It resists pH change when a strong acid or base are added. Optimized buffer is when the amount of acid is equal to the amount of the conjugate base. Read 19.31,9.4 Solution (underlined) can react Practice PS 4 is posted online. You can solve Q 1–7 with an acid A + H O 3 HA + H O 2   with a base HA + OH  A + H O 2 3 4 See figure 19.3 Demo 6 – Buffer solution – "The Race" + the "Rat Race" video clip UniversianldicatorpHcolor t e l o i v 01 og i dn i 9 eu l b8 7green wo l l ey6 5orange Equal amounts of acid and conjugate base give the der4 optimum buffer (greatest resistance to pH change). Typically buffer solutions are effective up to 1 pH unit either side of the listed buffer pH. 5 6 Solution 1 Buffer and buffer–like solutions NH  violet pH ~ 10.5 First a stoichiometry problem. 3 Basic solution We need to recalculate concentrations (or moles) based Solution 2 (a buffer) on a rxn. NH 4 and NH 3 violet pH ~ 10.5 Then an equilibrium problem. Basic solution Create an ICE table with the new concentrations and calculate the position of equilibrium. Dry ice (solid CO2) is added to both solutions. It dissolves and forms carbonic acid. Add the same amounts of the acid (carbonic acid) and solution 1 becomes acidic quicker than solution 2. In our demo solution 1 became yellow (pH ~ 6) while solution 2 turned light blue (pH ~ 8) 7 8 1 L of 0.15 M benzoic acid + 1 L of 0.080 M Sodium Add 1 mL of 1 M HCl (very strong acid), what  benzoate n HA = 0.150 mol nA = 0.080 mol happens ? +  First find new concentrations H + lC H 2O  H O3+ Cl 0.15 M x 1 L nH3O+ = (HCl)init 0.001 mol [HA] = = 0.075 M 2L Add 1 ml of 1 M HCl  0.001 mol of HA is formed.  0.080 M x 1 L [A  = = 0.040 M Why ? 2L  +  Rxn of acid and base H 3 + A  HA + H O 2 HA + H 2O  H 3 + A I 0.075 0 0.040 M This is the reverse reaction of an acid dissociation C – x + x + x
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