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# lec18 - donation H 3 A – Three equilibria k a1 k a2 k a3...

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1 Last Time • Dilute solutions – autoprotolysis –[H 3 O + ] ~ 10 –6 –10 –7 • Strong acids and bases – know this • Weak acids and bases – be aware of Charge Balance • Total charge in the solution must be zero!! • For acids : [H 3 O + ] = [A ] + [OH ] K w = [H 3 O + ][OH ] = [H 3 O + ] ( [H 3 O + ] – [HA] ) • For bases : [OH ] = [H 3 O + ] + [M + ] K w = [H 3 O + ][OH ] = [H 3 O + ] ( [H 3 O + ] + [B ] ) Example • What is the pH of a 5.6 x 10 –8 M solution of HCl? Dilute Solutions – Weak Acids • Problem: Acid with K a and [H 3 O + ] ~ 10 –7 • Wrong answer: use “x” and find pH directly • Correct answer – solve for [H 3 O + ] using K w = [H 3 O + ][OH ] K a = [H 3 O + ] = [OH ] + [A ] A = [HA] + [A ] [H 3 O + ][A ] [HA]

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2 Example What is the pH of a 1 x 10 –6 M solution of phenol in water given that K a for phenol is 1.3 x 10 –10 ? Polyprotic Acids • Monoprotic – one H + donation : HA • Diprotic – two H + donation : H 2 A – Two equilibria : k a1 , k a2 – Example: H 2 S • Triprotic – three H
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Unformatted text preview: + donation : H 3 A – Three equilibria : k a1 , k a2 , k a3 – Example : H 3 PO 4 Diprotic Acids H 2 A(aq) + H 2 O(l) → H 3 O + (aq) + HA – (aq) K a1 HA – (aq) + H 2 O(l) → H 3 O + (aq) + A 2– (aq) K a2 H 2 A H 3 O + HA – A 2– [Initial] a 0 0 [Change] –x +x+y +x–y +y [Equilib.] a–x +x+y +x–y +y ← ← “Double” Equilibrium • Two equations two unknowns … K a1 = K a2 = • Can simplify when K a2 << K a1 • Use “weak acid approach” and proceed stepwise. x 2 – y 2 a – x y (x + y) x – y 3 Example Calculate the pH of a 0.1 M solution of H 2 S given that K a1 = 1.3 x 10 –7 and K a2 = 7.1 x 10 –15 . Example Calculate the concentration of all components of a 0.01 M H 3 PO 4 solution given that K a1 = 7.6 x 10 –3 , K a2 = 6.2 x 10 –8 , and K a3 = 2.1 x 10 –13 ....
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lec18 - donation H 3 A – Three equilibria k a1 k a2 k a3...

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