Equilibrium 225 in Visual C#.NET

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Equilibrium 225
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14 12 10 8 pH 6 4 2 0 10 20 30 40 50 60 Volume of NaOH added (mL) 70
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The Titration of a weak acid with a strong base
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Let s consider a typical titration problem A 1000 mL sample of 0150 M nitrous acid (pKa = 335) was titrated with 0300 M sodium hydroxide Determine the pH of the solution after the following quantities of base have been added to the acid solution: a b c d e f 000 mL 2500 mL 4950 mL 5000 mL 5500 mL 7500 mL
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a 000 mL Since no base has been added, only HNO2 is present HNO2 is a weak acid, so this can only be a Ka problem HNO2 H+ ( aq) + NO2 0 150 x x x K a = 10 335 = 4 5 10 4 = ( x ) (x ) 0 150 x
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Quadratic needed: x2 + 447 10 5x 670 10 5 = 0 (extra sig figs) x = [H+ ] = 80 10 3M pH = 210 b 2500 mL Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem Stoichiometry requires a balanced chemical equation and moles HNO2 + NaOH Na + + NO + H 2O 2
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Na+ + NO2 could be written as NaNO2, but the separated ions are more useful
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226 U Step 4 Review the Knowledge You Need to Score High
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Acid: 0150 mol 10000 mL = 00150 mol (This number will be used in all remaining steps) 1000 mL
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0300 mol 2500 mL = 000750 mol 1000 mL Based on the stoichiometry of the problem, and on the moles of acid and base, NaOH is the limiting reagent Base: HNO2 + init 0 0150 react 0 0148 final 000750 NaOH 0 00750 mol 0 00750 0000 Na + + 0 + 0 00750 NO + H 2O 2 0 + 0 00750 000750
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The stoichiometry part of the problem is finished The solution is no longer HNO2 and NaOH, but HNO2 and NO2 (a conjugate acid base pair) Since a CA/CB pair is present, this is now a buffer problem, and the Henderson Hasselbalch equation may be used pH + pKa + log (CB/CA) = 335 log (000750/000750) = 335 Note the simplification in the CB/CA concentrations Both moles are divided by exactly the same volume (since they are in the same solution), so the identical volumes cancel 000750 mol base 012500 L solution 000750 mol acid 012500 L solution c 4950 mL Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem again Stoichiometry requires a balanced chemical equation and moles
HNO2 + NaOH Na + + NO2 + H 2O
Base:
0300 mol 4950 mL = 00148 mole 1000 mL
Based on the stoichiometry of the problem, and on the moles of acid and base, NaOH is the limiting reagent HNO2 + 0 0150 00148 00002 NaOH Na + + NO + H2O 2 00148 mol 0 0 00148 +00148 +00 1 48 0000 00148
init react final
The stoichiometry part of the problem is finished The solution is no longer HNO2 and NaOH, but HNO2 and NO2 (a conjugate acid base pair)
Equilibrium 227
Since a CA/CB pair is present, this is now a buffer problem, and the Henderson Hasselbalch equation may be used pH = pKa + log (CB/CA) = 335 + log (00148/00002) = 52 d 5000 mL Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem Stoichiometry requires a balanced chemical equation and moles
HNO2 + NaOH Na + + NO2 + H 2O
Base:
0300 mol 5000 mL = 00150 mol 1000 mL
Based on the stoichiometry of the problem, and on the moles of acid and base, both are limiting reagents HNO2 + NaOH Na + + NO + H2O 2 0 0 0150 00150 mol 0 00150 00150 +00150 +00 1 50 00000 0000 00150 [ NO2 ] = 0 0150 mol / 0 150L = 0 100M L
init react final
The stoichiometry part of the problem is finished The solution is no longer HNO2 and NaOH, but an NO2 solution (a conjugate base of a weak acid) Since the CB of a weak acid is present, this is a Kb problem p K b = 14 000 p K a = 14 000 3 35 = 10 65
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