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lecture+12sf

# lecture+12sf - Titration A titration is a procedure where a...

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Titration A titration is a procedure where a base is added to an acid, or an acid is added to a base until the stoichiometric amount of base or acid is added. This is called the equivalence point . The equivalence point is detected by a color change in an indicator. Strong Acid/Strong Base Titration Examples: HCl + NaOH HNO 3 + KOH Equation: H 3 O + (aq) + OH - (aq) → 2H 2 O( ) K = w K 1 When a reaction has a high K (as in a titration reaction) first react to completion before doing any equilibrium calculations. K for above reaction = 14 10 x 1 1 - = 1 x 10 14 Assume we add NaOH (source of OH - ions) to a solution initially containing 0.10 mol HCl in 1.00 L of solution. Make a table and then a plot of pH vs. mol OH - added. H 3 O + (aq) + OH - (aq) → 2H 2 O( ) K = w K 1 = 10 14 (large) Showing a calculation in detail (before the equivalence point): Adding 0.010 mol OH - to the initial 0.10 mol H 3 O + , all in 1.00 L. We react to completion since K is large, then calculate the pH based on the excess [H 3 O + ] concentration H 3 O + (aq) + OH - (aq) → 2H 2 O( ) Start 0.10 0.010 RTC -0.010 -0.010 RTC = “react to completion” 0.090 “0” Excess [H 3 O + ] = 0.090 M pH = -log (0.090) = 1.05

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Adding 0.090 mol OH - to the initial 0.10 mol H 3 O + , all in 1.00 L. We react to completion since K is large, then calculate the pH based on the excess [H 3 O + ] concentration H 3 O + (aq) + OH - (aq) → 2H 2 O( ) Start 0.10 0.090 RTC -0.090 -0.090 RTC = “react to completion” 0.010 “0” Excess [H 3 O + ] = 0.010 M pH = -log (0.010) = 2.00 Showing another calculation in detail (after the equivalence point): Adding 0.11 mol OH - to the initial 0.10 mol H 3 O + , all in 1.00 L. We react to completion since K is large, then calculate the pH based on the excess [OH - ] concentration H 3 O + (aq) + OH - (aq) → 2H 2 O( ) Start 0.10 0.11 RTC -0.10 -0.10 RTC = “react to completion” “0” 0.01 Excess [OH - ] = 0.010 M pOH = -log (0.010) = 2.00 pH = 14.00 - 2.00 = 12.00 Summary: Start with 0.10 mol HCl in 1.00 Liter mol OH - added pH 0 1.00 initial 0.10 M HCl 0.010 1.05 xs [H 3 O + ]: 0.10 - 0.010 = 0.090 M 0.050 1.30 xs [H 3 O + ]: 0.10 - 0.050 = 0.050 M 0.090 2.00 xs [H 3 O + ]: 0.10 - 0.090 = 0.010 M 0.099 3.00 xs [H 3 O + ]: 0.10 - 0.099 = 0.001 M 0.0999 4.00 xs [H 3 O + ]: 0.10 - 0.0999 = 0.0001 M 0.10 7.00 equivalence point, no xs [H 3 O + ] or [OH - ] 0.1001 10.00 xs [OH - ]: 0.1001 - 0.10 = 0.0001 M 0.101 11.00 xs [OH - ]: 0.101 - 0.10 = 0.001 M 0.110 12.00 xs [OH - ]: 0.11 - 0.10 = 0.01 M 0.150 12.70 xs [OH - ]: 0.15 - 0.10 = 0.05 M 0.200 13.00 xs [OH - ]: 0.20 - 0.10 = 0.10 M The pH graph of titrating 1.00 L of 0.10 M HCl with NaOH is shown below.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH 0 0.05 0.1 0.15 0.2 0.25 mol NaOH added 0.10 mol HCl titrated with NaOH (1.00 Liter total) pH Note the sharp change of pH near the equivalence point. A tiny amount of NaOH added near the equivalence point changes the pH from 3 to 10. The equivalence point (pH 7) is the

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lecture+12sf - Titration A titration is a procedure where a...

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