159_lect_18_acid_base_titrations-

159_lect_18_acid_base_titrations- - Lecture 18 Topics 1...

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3/31/2008 1 Lecture 18 Topics 1. Acid-Base Titrations 19.2 1. Indicators 2. Equivalence Points vs End Point 2. Solubility Products 19.3 1. What are they? 2. How to determine solubility? 3. What are they used for? 1 All Rights reserved Gene Hall Why titrations? L In reactions monitoring, need to know how much acid or base are produced. L In environmental analysis need to know buffering capacity of soil for acid rain damage. L For an acid or base spill, need to know how much acid or base is present and potentially what is the acid or base. 2 All Rights reserved Gene Hall
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3/31/2008 2 Two methods to monitor titrations. L Use of a pH meter is the most accurate. L Use indicators that change based on the chemical form of the indicator as a function of the pH. L Indicators are weak organic acids noted as HIn. L HIn(aq) + H 2 O(l) = H 3 O + (aq) + In - (aq) L Color 1 color 2 3 All Rights reserved Gene Hall The useful pH ranges for several common indicators. Note that most indicators have a useful range of about two pH units, as predicted by the expression p K a +/- 1. 4 All Rights reserved Gene Hall
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3/31/2008 3 Titration curves L Equivalence point is when the moles of acid or base have been neutralized by the base or acid in the respective titrations. L End point is when the indicator shows a change in the pH that is close to the equivalence point. Often the end point is higher or lower than the equivalence point and contributes to a small titration error. 5 All Rights reserved Gene Hall Five Titration Cases L Case 1: strong acid with strong base « pH = 7 at equivalence point L Case 2: weak acid with strong base « pH >7 (basic) at equivalence point. L Case 3: weak base with a strong acid « pH < 7 (acidic) at equivalence point. L Case 4: weak base with a weak acid « pH depends on the titrants, not often done. L Case 5: weak polyprotic acid with strong base « pH depends on the pK a’ s of the polyprotic acid. 6 All Rights reserved Gene Hall
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3/31/2008 4 Case 1 SA with SB L Case 1: HCl(aq) + NaOH(aq) = L H + (aq) + OH - (aq) = H 2 O (l) L Suppose we have 40-ml of 0.1 M HCl (SA) and titrate that with 0.1 M NaOH (SB). L 40 mL * 0.1M = 4 mili moles of acid = x mL * 0.1M where x=40 ml to neutralize the acid. Can calculate pH at different steps in titration. Titrate base into acid. pH at start = 1.0. 7
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This note was uploaded on 04/03/2008 for the course CHEM 159-160 taught by Professor Zbaida during the Fall '07 term at Rutgers.

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159_lect_18_acid_base_titrations- - Lecture 18 Topics 1...

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