Chem 162 2015 Chapter 18 Acid base solubility equilibria lecture d

Chem 162 2015 chapter 18 acid base solubility

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Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 11
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15.23d Calculate the pH of a mixture containing 0.100 M HC 3 H 5 O 2 and 0.100 M NaC 3 H 5 O 2 . (HC 3 H 5 O 2 , K a = 1.3 x 10 -5 ) *ET: Problem done previously. Note that there are fewer calculations when an uncharged substance reacts with H 2 O rather than a charged substance. Hence, I chose to react HP with water rather than NaP. However, either is OK. 0.100 M HP 0.100 M NaP (Get rid of Na + spectator ion). HP + H 2 O H 3 O + + P - HP(aq) + H 2 O( l ) H 3 O + (aq) + P - (aq) Initial 0.100 0 0.100 Change Equilibrium HP(aq) + H 2 O( l ) H 3 O + (aq) + P - (aq) Initial 0.100 0 0.100 Change -X +X +X Equilibrium 0.100 - X +X 0.100+X ([H 3 O + ][P - ])/[HP] = K a ((X) x (0.100 + X))/(0.100 - X) = 1.3 x 10 -5 Quadratic equation; simplify with small K rule: (X x 0.100)/(0.100) = 1.3 x 10 -5 X = 1.3 x 10 -5 pH = -log 1.3 x 10 -5 = 4.89 Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 12
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27d. Calculate the pH after 0.020 mol HCl is added to 1.00 L of a mixture containing 0.100 M HC3H5O2and 0.100 M NaC3H5O2. Kpropanoic acid= 1.3x10-5Note: In absence of a buffer, pH of 0.020M HCl = 1.70ET: This slide ties into the previous slide of an HP/P-buffer without a strong acid.ET: Given weak acid, conjugate base, and added strong acid concentrations; find pH.What equilibrium equation should be used for the ICE table? . So Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 13
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HENDERSON-HASSELBALCH EQUATION An equation for buffer solutions (students not responsible for this derivation) HA + H 2 O H 3 O + + A - ([H 3 O + ][A - ])/[HA] = K a [H 3 O + ] = K a x [HA]/[A - ] -log[H 3 O + ] = -logK a + -log([HA]/[A - ]) -log[H 3 O + ] = -logK a + log([A - ]/[HA]) pH 3 O + = pK a + log([A - ]/[HA]) pH = pK a + log([A - ]/[HA]) pH = pK a
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