Chapter_17 - Chapter 17: AcidBase Equilibria Solutions of a...

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CHM 1220/1225 Chapter 17 1 of 6 Chapter 17: Acid–Base Equilibria Solutions of a Weak Acid or Base 1. Acid–Ionization Equilibria When a weak acid is dissolved in water, an equilibrium is established between the molecular form of the acid (generic form: HA) and the hydrogen ion (hydronium ion, H 3 O + ) and anion (generic form: A ). The K C for an acid equilibrium is K A . To solve problems with weak acids, we combine the skills from Chapter 15 (equilibrium) and Chapter 16 (acids and bases). The equilibrium can be written in generic terms: HA( aq ) + H 2 O( l ) <===> H 3 O + ( aq ) + A ( aq ) The expression for Ka = [H 3 O + ][A ] [HA] Example What is the pH of a 0.12 M solution of formic acid, HCOOH? K A for formic acid is 1.7 x 10 –4 at 25ºC. First: the initial amount is given in Molarity; no conversion is required Second: complete the chart H 2 O + HCOOH <===> H 3 O + + HCOO Initial 0.12 M 0 0 Change –x +x +x Equilibrium (0.12 – x) M X M X M Third: write the equilibrium constant expression and substitute known values and expressions K A = [H + ][HCOO ] 1.7 x 10 –4 = x 2 [HCOOH] (0.12 – x) Fourth: Solve the equation: Assume that 0.12 – x = x 2.04 x 10 –5 = x 2 x = 4.5 x 10 –3 Test assumption: 0.12 – 4.5 x 10 –3 = 0.12 Fifth: Substitute the value of x into the expressions for equilibrium concentrations to answer the
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Chapter_17 - Chapter 17: AcidBase Equilibria Solutions of a...

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