Chem_215_Lecture_11_Color - 8.2: Buffer Solutions Buffers...

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Unformatted text preview: 8.2: Buffer Solutions Buffers resist large pH changes upon addition of an acid or base. Contains: Weak acid + conjugate base or Weak base + conjugate acid One component neutralizes acids; the other neutralizes bases. Unbuffered: 1.00 L of Pure Water: pH = 7.00 Add 10.0 ml of 2.0M HCl pH = 1.69 Buffer: 1.00 L of 1.50M Acetic Acid/ 1.20M Sodium Acetate: pH = 4.65 Add 10.0 ml of 2.0M HCl pH = 4.63 1 Buffer solutions consume added acid or base 2 For a buffer solution containing a weak acid and its conjugate base: Henderson-Hasselbalch Equation pH = pK a + log10 [HA] [A ] - Notes: Equation was derived by assuming M - x M. [conjugage base]/[acid] should lie between 0.10 and 10.0 Molarity of each component should exceed Ka by factor of 100. 1.50M Acetic Acid / 1.20 M Sodium Acetate: pH = 4.74 + log10 [1.2] [1.5] Ka = 1.8 x 10-5; pKa = 4.74 = 4.74 - 0.09 = 4.65 3 Preparing Buffers pH = pK a + log10 [HA] [A ] - 4 8.3 Exact Treatment of Buffer Solutions When the pH of a buffer is near 7.0, it may be necessary to consider the autoionization of water. Use charge balance and mass balance equations. [H 3O + ]2 - K w [H 3O + ] [ A ]0 + [H 3O + ] [H 3O + ][A - ] = Ka = [H 3O + ]2 - K w [ HA] [HA]0 [H 3O + ] When [A-]o = 0, the above equation reduces to that derived earlier for a dilute weak acid dissolved in water. 5 8.1 The Common Ion Effect Recall that for polyprotic acids, the production of [H3O+] from the first ionization greatly inhibits the second ionization. This general phenomenon is quite general, and is known as the "common ion effect". Calculate the pH of 1.0 M HF (Ka = 7.2 x 10-4) in: a) Pure water b) 1.0 M NaF 6 Indicators - a weak acid with a conjugate base of different color Color 1 Phenophthalein: HIn + H2O H3O+ + InColor 2 Colorless Pink HIn: In-: [H 3O + ][In - ] Ka = [HIn] Ka [H 3 O ] + [In - ] = + [H 3 O ] [HIn] Ka Acidic- Large Basic- Small [H3O+] [H3O+] << 1 [ In - ] << 1 [ HIn] [In-] << [HIn] Colorless [In-] >> [HIn] 7 Pink Ka [H 3 O + ] >> 1 [ In - ] >> 1 [ HIn] Indicator Color and pH HIn + H2O H3O+ + In[H 3O + ][In - ] K HIn = [HIn] K HIn [In - ] = [HIn] [H 3 O + ] pH = pK HIn [In - ] + log10 [HIn] Hi [H3O+] Low pH Low [H3O+] Hi pH When: pH = pKHIn: Intermediate color pH < pKHIn 1 : Acid color pH < pKHIn + 1 : Base color 8 HIn + H2O H3O+ + In[H 3O + ][In - ] K HIn = [HIn] [In - ] + log10 [HIn] (High pH) pH = pK HIn (Low pH) 9 Colors of Indicators depend upon pH 10 Titration: Measurement of the number of moles of acid or base in a sample. Involves neutralization using known quantities of base or acid. Equivalence point: moles acid = moles base. Endpoint: point at which indicator undergoes color change. Titration of 25.0ml of 0.10 M HCl Strong acid + strong base: pH = 7 at equivalence pt. 11 Volume of 0.100M NaOH, mL Equivalence point: moles acid = moles base Endpoint : Point at which color of indicator changes. Titration of 25.0ml of 0.10 M HCl Best Indicator choice (Equiv. pt. = Endpoint) Poor Indicator choice (Equiv. pt. Endpoint) Volume of 0.100M NaOH, mL Choose indicator such that endpoint coincides with equivalence point. 12 Titration of a weak acid using NaOH pH > 7 at equivalence point (moles base = moles acid) Titration of 25.0 ml of 0.100 M Acetic Acid Good indicator choice (Equiv. pt. = Endpoint) Poor indicator choice (Endpoint reached BEFORE equiv. pt) 13 ...
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