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The process of preparing for an acid-base titration experiment, focusing on the titration of hydrofluoric acid against sodium hydroxide. the objectives of the experiment, the concept of the equivalence point, and the calculation of the concentration of hydrofluoric acid and its initial pH using the molarity of sodium hydroxide and the Ka value of hydrofluoric acid. The document also explains the use of an ICE table and the quadratic equation to determine the initial pH.
Typology: Lecture notes
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You are encouraged to carefully read the following sections in Tro (3rd^ ed.) to prepare for this experiment: Sec 4.8, pp 168-174 (Acid/Base Titrations), Sec 16.4, pp 769-783 (Titrations and pH Curves).
Objectives: You will be able to: (1) determine the hydrogen ion concentration of a weak acid via titration against a strong base, (2) calculate the pH of a weak acid / strong base titration at the endpoint of a titration, (3) evaluate a pH titration curve and determine the pH of a weak acid / strong base titration at the equivalence point, (4) use the molar solubility, as determined from the hydrogen ion concentration, to determine the Ksp of an ionic compound.
Background: In an acid-base titration, a solution of unknown concentration is reacted with a standard solution of known concentration. The progress of the reaction is monitored using a pH meter or an indicator. When an indicator is used, the titration continues until the indicator changes color to signal the endpoint of the titration. The endpoint occurs in the first drops past the equivalence point. The equivalence point is the point at which the number of moles of base is stoichiometrically equivalent to the number of moles of acid.
To see how this data can be used, follow the titration of hydrofluoric acid against sodium hydroxide:
HF (aq) + NaOH (aq) H 2 O (l) + NaF (aq)
If 35 mL of HF required 25 mL of 0.250 M NaOH to reach the pink phenolphthalein endpoint, what is the concentration of HF? At this point, simply determine the concentration of the HF stoichiometrically.
[HF] = 0.0250 L NaOH x 0.250 mol NaOH x 1 mol HF x 1 = 0.179 M HF 1 L NaOH 1 mol NaOH 0.0350 L HF
Note that 0.179 M HF is equal to the concentration of the weak acid at the start of the titration. To determine the pH at the start of the titration, we will use this information, the Ka of HF, an ICE table for the dissociation of HF in water, and pH = -log [H 3 O+]
HF (aq) + H 2 O ( l ) H 3 O+^ (aq) + F-^ (aq) Ka = 3.5 x 10-
HF H 3 O+^ F- Initial Conc. 0.179 0 0 Change in Conc. - x + x + x Equilibrium Conc. 0.179 - x X x
3.5x10-4^ = Ka =
(x)(x) 0.179−x
Now solve for x using the quadratic equation and you get x = [H 3 O+] = 0.00774 M. Plugging that into the pH equation, you find the initial pH of the solution is 2.11.
At the equivalence point (endpoint in this experiment since the initial concentration of acid was determined experimentally from the endpoint) all of the weak acid will have reacted with the strong base to form the conjugate weak base as can be seen in the table below. The concentrations of the weak acid and the strong base that react are:
[HF] = (0.179 M HF)(0.0350 L) / (0.0600 L) = 0.104 M HF [OH-] = (0.250 M OH-)(0.0250 L) / (0.0600 L) = 0.104 M OH- Note: You must take into account the dilution from the presence of both solutions.
HF (aq) + OH-^ (aq) H 2 O (aq) + F-^ (aq)
HF OH-^ F- Before reaction 0.104 0.104 0 reaction changes -0.104 -0.104 +0. After reaction 0 0 0.
So, the reaction that determines the pH of the solution at the equivalence point is actually the interaction of the product weak base in water. An ICE table can be used to figure out the concentrations of the reactants and products at equilibrium.
F-^ (aq) + H 2 O (l) OH-^ (aq) + HF (aq) Kb = 2.9 x 10-
Initial Conc. 0.104 0 0 Change in Conc. - x + x + x Equilibrium Conc. 0.104- x (^) x x
2.9x10-11^ = Kb =
(x)(x) 0.104−x
Solve for x using the quadratic equation and you get x = [OH-] = 1.7 x 10-6^ M. Plugging that into the equation pOH = -log [OH-], you find the pOH of the solution is 5.77. Finally, using the equation 14.00 = pH + pOH, you find that the pH = 8.23.
Titration of a Saturated Potassium Hydrogen Tartrate (KHT) Solution Against Standardized NaOH
Preparation of the Saturated KHT Solution for Titration
Preparation of the Standard NaOH Solution for Titration Remember standardized NaOH is valuable and time consuming to make. Take only what you need from the bottle, and be certain the stock bottle is tightly sealed after you use it.
Preparation of the pH meter Be careful never to touch the pH-sensitive circuit of the pH meter with anything but water, aqueous solutions or blotting with a Kimwipe.
Titration
pH Curve for the Titration of Saturated Potassium Hydrogen Tartrate (KHT) Solution against Standardized NaOH
pH Curve Titration
Titration:
b. Run 2
HT-^ (aq) + H 2 O (l) H 3 O+^ (aq) + T2-^ (aq)
T2-^ (aq) + H 2 O (l) OH-^ (aq) + HT-^ (aq) Kb = 2.2 x 10-
Complete the following table by identifying the one of the four species (HA, A−, H 3 O+, OH−) that is most concentrated, the one that is least concentrated, and any pairs that have nearly equal concentration in the mixture at that point. The first entry is provided for you.
Point on curve
Highest concentration
Lowest concentration
Nearly Equal Concentrations
A HA
B
C
D
2
4
6
8
10
12
14
0 10 20 30 40 50
pH
Volume NaOH
A
B
C
D
b. What is the pH after 16 mL (equivalence point) of 0.150 M KOH has been added to the solution. (for C 2 H 3 O 2 - , Kb = 5.6 x 10-10)