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Titration Calculations

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An example of titration problem: I have a 25.00 mL sample of a strong acid at an unknown concentration. After adding 13.62 mL of a 0.096 M NaOH solution, equivalence was reached. Determine the pH of the acid. What is the goal of a titration problem? What is significant about equivalence and why is that important?

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Equivalence At equivalence: Moles of H + = Moles of OH - added If we know the [OH-] added, then at equivalence it will be equal to the [H+] of the original acid. …After addition of 13.62 mL of a 0.096 M NaOH solution, equivalence was reached. How can we get the [OH-]?

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Be part of the solution, not the problem! 13.62 mL NaOH 1 L = 0.01362 L NaOH added 1000 mL 0.096 M NaOH = X moles NaOH = 1.308 x10 -3 mol OH - 0.01362 L NaOH What does that number tell us? 1.308 x 10 -3 mol OH - added = 1.308 x 10 -3 mol H + in original acid

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An example of titration. We calculated that there were 1.308 x 10 -3 moles H + in original sample. Are we done? We need the pH of the acid. How do we do that? pH = - log [H + ] – We need the Molarity of H + Molarity = 1.308x10 -3 mol / 0.025L= 0.052 M pH = - log [0.052] = pH 1.28 (2 Dec. Places) Problem: I have a 25.00 mL sample of a strong acid at an unknown concentration. After adding 13.62 mL of a 0.096 M NaOH solution, equivalence was reached. Determine the pH of the acid.

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Titrations not to Equilibrium Titrations are not required to go all the way until the equilibrium point. Some problems ask how the pH of a solution will change as titration occurs.

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Problem type #2 Determine the pH when 49.0 mL of NaOH is added to a 50.0 mL HCl solution. Notice that equivalence or complete neutralization is not mentioned, nor obtained. In these types of problems, we need to determine how many H + ions and OH - ions are neutralized and then determine what ions are left over to affect the pH of the resulting solution.

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Start with the Equation NaOH + HCl H 2 O + NaCl If we remove the aqueous spectator ions, the equation looks like this: OH - + H + H 2 O To solve this question, we need to know how many of the ions are used to make the water.

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Need to determine moles not M Another tricky part about this question is that the volume will change when the two solutions combine. When volume changes, Molarity changes so we need to convert everything to MOLES: Molarity = Moles/Liters

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Converting to Moles NaOH = 0.100 M = X mol 4.90 x10 -3 mol 0.049 L HCl = 0.100 M = X mol 5.00 x10 -3 mol 0.050 L

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Repurposed ICE Chart We need to calculate the concentrations at equilibrium. What did we have initially? What did we Change? and what is left over? OH - + H + H 2 O Initial Change Equilibrium

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What did we begin with? We started with the hydrochloric acid. How much did we have? Where will that go in the ICE Box? OH - + H + H 2 O Initial:05.00x10 -3 0

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What did we change? We added the sodium hydroxide. Where will that go in the ICE Box? How much did we add? OH - + H + H 2 O Change4.90x10 -3 00

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What did we end up with? How do we determine the concentrations at equilibrium. (Initial + Change = Equilibrium) But remember a neutralization reaction is taking place so we must make water. Initial 05.00x10 -3 0 Change 4.90x10 -3 00 Eq. 4.90x10 -3 5.00x10 -3 0 OH - + H + H 2 O

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What is left over at equilibrium? The question remains, how much water is made? In this reaction all possible reactants are used up. Equilibrium 4.90x10 -3 5.00x10 -3 4.90x10 -3 OH - + H + H 2 O So we are limited by the amount of the [OH - ] ion. That means we have excess [H + ] ions that will affect the pH of the solution!!

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How many ions are left over? So how many moles of [H + ] do not react? 5.00 x 10 -3 - 4.90 x 10 -3 = 1.0 x 10 -4 mol H + These ions remain in the solution and will affect the pH of the solution. So we have the moles of H +, but what do we need to determine the pH? Eq. 4.90x10 -3 5.00x10 -3 4.90x10 -3 OH - + H + H 2 O

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Tricky Volume! We need the Molarity! But this is tricky because the volume of the solution changed when we added the acid to the base. (Molarity = Moles / Liters) What is the final volume of the solution? 49.0 mL of NaOH + 50.0 mL of HCl = 99.0 mL 1.0 x 10 -4 / 0.099 L = 1.01 x10 -4 [H + ]

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And Finally the pH part! pH = -log (1.0 x10 -3 [H + ]) pH = 3.00 This makes sense (hopefully) because we’re not at equilibrium so the pH isn’t at 7.

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