Buffers and Titrations, Part 2 Ch. 21 Suggested HW: 14, 18, 30, 42, 44

Weak Acid/Base Chemistry Recap For a weak acid dissociation For the protonation of the conjugate base: Therefore: HA aq + H 2 O L ↔ H 3 O + aq + A − (aq) K a = H 3 O + [ A − ] [HA] A − aq + H 2 O L ↔ OH − aq +HA (aq) K b = OH − [HA] [ A − ] K a x K b = H 3 O + [ A − ] [HA] x OH − [HA] [ A − ] = H 3 O + OH − = K w =𝟏 𝟎 −𝟏𝟒

Buffer Recap For a buffer (HA/A-) where pKa = -log Ka As the base concentration goes up, the logarithm of the ratio of base to acid also goes up, resulting in an increase in pH of the buffer. The opposite is then true for the addition of acid. pH=pKa+ log base acid

Buffer Capacity and pH Range Two important characteristics of a buffer are its capacity and pH range. Buffer capacity is the amount of acid or base that a buffer can neutralize before the pH begins to change vastly. ex. A buffer of 0.1M HF/ 0.1M NaF has the same pH as 1.0 M HF/ 1.0 M NaF, but less buffer capacity pH range is the range over which a buffer works effectively. In real practice, the maximum buffer range is reached when one component is 10 times more concentrated than the other. 𝑝𝐻=𝑝𝑘𝑎+ log 𝐵 𝐴 for B:A ratio of 10:1 , 𝑝𝐻=𝑝𝐾𝑎+1 for B:A ratio of 1: 10, 𝑝𝐻=𝑝𝐾𝑎−1 When designing a buffer, the pKa of the acid should be as close as possible to the desired pH max buffer range = pKa ± 1

Example: Selecting and Preparing a Buffer You work for a major pharmaceutical company that has developed a new drug to combat sickle cell anemia. You have a fresh batch of live cells to test this drug. The cells must be maintained in 100 mL of a buffered solution at pH=7.4. The buffering capacity should be high, so it is recommended that the concentration of the acid component be 0.5M. Assuming that you have access to any acid and base (non-ionic forms as well as sodium salts), what conjugate acid/base pair would you use to create your buffer? What masses of each would you dissolve into solution? Use the pka table on the next slide.

Example contd. Potential options for a pH=7.4 buffer would be (acid/base): HClO/NaClO pKa =7.40 (buffer range of 6.40 to 8.40) NaH2PO4/Na2HPO4 pKa = 7.21 (buffer range of 6.21 to 8.21)

Example contd. Using the hypochloric acid/ hypochlorite buffer with [HClO]=0.5 M : 7.4=7.4+ log [𝑁𝑎𝐶𝑙𝑂] [𝐻𝐶𝑙𝑂] 0= log [𝑁𝑎𝐶𝑙𝑂] [𝐻𝐶𝑙𝑂] 10 0 = [𝑁𝑎𝐶𝑙𝑂] [𝐻𝐶𝑙𝑂] 1= [𝑁𝑎𝐶𝑙𝑂] [𝐻𝐶𝑙𝑂] 𝐻𝐶𝑙𝑂 =0.5 M 𝑁𝑎𝐶𝑙𝑂 =0.5 M Acid: 0.100 L x 0.5 mol H 2 CO 3 L x 52.45 g 𝐻𝐶𝑙𝑂 mol 𝐻𝐶𝑙𝑂 =𝟐.𝟔𝟐 𝐠 Dissolve in 100 mL of solution. Base: 0.100 L x 0.5 mol Na𝐶𝑙𝑂 L x 74.45 g NaH CO 3 mol NaH CO 3 =𝟑.𝟕𝟐 𝐠

Titration Curves In a titration, one substance of known concentration (e.g. base) is added to another (e.g. acid). A colored indicator or pH meter can be used to monitor the reaction progress. Using a pH meter, a titration curve can be generated, showing pH as a function of the volume of titrant added

Titration Curves The shape of a titration curve makes it possible to determine the equivalence point (point at which the moles of acid and base are stoichiometrically equivalent) Titration curves can also be used to determine values of Ka or Kb if weak acids/bases are being titrated To understand the characteristics of titration curves, we will examine three types of acid/base titrations: Strong Base into Strong Acid Strong Base into Weak Acid Strong Base into Polyprotic acid

Strong Acid/Strong Base Titration Curves (e. g. 1M NaOH added to 0 Strong Acid/Strong Base Titration Curves (e.g. 0.1M NaOH added to 0.1M HCl) The general shape of this type of titration curve is shown to the right. The initial pH is simply the pH of the HCl(aq) before addition of NaOH(aq) At first, the pH rises slowly due to the presence of un-neutralized acid. At the equivalence point, pH=7 and the moles of acid and base are stoichiometrically equal Beyond equivalence, pH rises rapidly due to presence of free base

Group Example Calculate the pH when 49 mL of 0.1M NaOH is added to 50 mL of 0.1M HCl? (ans: pH = 3) Calculate the pH when 51 mL of 0.1M NaOH is added to 50 mL of 0.1M HCl? (ans: pH = 11)

Weak Acid/Strong Base (e. g. 1M NaOH added to 0 Weak Acid/Strong Base (e.g. 0.1M NaOH added to 0.1M CH3COOH with Indicator) As the acid is neutralized, its conjugate base forms, resulting in the formation of a buffer when [B]/[A] > 1/10. At the midpoint of the buffer region, pH = pKa; [CH3COOH] = [CH3COO-] At the equivalence point, all of the acetic acid is now acetate, which is a weak base. Thus, the pH at the equivalence point is greater than 7.

Group Example Given that Ka of CH3COOH is 1.8 x 10-5 M, calculate the pH when 49 mL of 0.1 M NaOH is added to 50 mL of 0.1M CH3COOH. First, calculate the moles of acetic acid remaining Now, plug the concentrations into the Ka expression to find [H3O+]. The total volume is 99 mL. Refer to slide 2 for clarification. 𝐶 𝐻 3 𝐶𝑂𝑂𝐻 𝑎𝑞 +𝑂 𝐻 − (𝑎𝑞) ↔𝐶 𝐻 3 𝐶 𝑂𝑂 − (aq) + H2O I 0.005 mol 0.0049 mol 0 C -.0049 mol -.0049 mol +.0049 mol E .0001 mol 0 mol .0049 mol 1.8 𝑥 10 −5 = 𝐻 3 𝑂 + [ 𝐶𝐻 3 𝐶𝑂𝑂−] [𝐶𝐻 3 𝐶𝑂𝑂𝐻] = 𝐻 3 𝑂 + [.0495𝑀] [.001𝑀] 𝐻 3 𝑂 + =3.64 𝑥 1 0 −7 M pH = 6.44

Example Calculate the pH at the equivalence point from the titration of 50 mL of 0.1M acetic acid by 50 mL of 0.1M NaOH as shown on slide 13. (ans: pH= 8.72) Hint!!! How do you calculate Kb?

Strong Base/Polyprotic Acid Titration Curves (e. g. 1M NaOH added to 0 Strong Base/Polyprotic Acid Titration Curves (e.g. 0.1M NaOH added to 0.1M H3PO4)

From Last Image Initial point: pH of 0.1M H3PO4 First midpoint: [H3PO4] = [H2PO4-], pH at 1st midpoint = pKa # 1 First equivalence point: No H3PO4 remains. Second midpoint: [H2PO4-] = [HPO42-], pH at 1st midpoint = pKa # 2 Second equivalence point: No H2PO4- remains Third midpoint: [HPO42-] = [PO43-] , pH at 1st midpoint = pKa # 3 Third equivalence point: Only phosphate remains