Chem 14B Problem Set 9 soln

.pdf

School

University of California, Los Angeles *

*We aren’t endorsed by this school

Course

14B

Subject

Chemistry

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by DoctorStarGrasshopper4 on coursehero.com

Name__________________________________ UID_____________________________ Section________________________ Problem Set 9. Not turned in. 1. What is the p K a of the conjugate acid of hydrazine, given that the p K b of hydrazine is 5.77? Write the formula of the conjugate acid of hydrazine. Answer: this is a classic conjugate see saw situation. pKa + pKb = 14 pKa = 14 – pKb = 14-5.77 = 8.23 N 2 H 5 + 2. Choose the effective pH range of an aniline/anilinium chloride buffer. The value of the K b for aniline is 4.3 ´ 10 - 10 . Answer: We need the pKa of the conjugate acid/base pair. pKb = -log(4.3 x 10 -10 ) = 9.4 pKa = 14 - 9.4 = 4.6 The range is pKa +/- 1. So 3.6 – 5.6 3. For NH 3 , p K b = 4.74. What is the pH of an aqueous buffer solution that is 0.050 M NH 3 (aq) and 0.20 M NH 4 Cl(aq)? Answer: pH = pKa + log(A-/HA) pH = (14-4.74) + log(0.050/0.200) = 8.66 4. If a small amount of a strong base is added to buffer made up of a weak acid, HA, and the sodium salt of its conjugate base, NaA, the pH of the buffer solution does not change appreciably because A) the K a of HA is changed. B) No reaction occurs. C) the strong base reacts with A - to give HA, which is a weak acid. D) the strong base reacts with HA to give AOH and H + . E) the strong base reacts with HA to give A - , which is a weak base. Answer: E The fact that neutralizing a weak acid produces a weak base and vice versa, is the key to buffers.

5. A buffer solution contains 0.0200 M acetic acid and 0.0200 M sodium acetate. What is the pH after 2.0 mmol of HCl is added to 1.00 L of this buffer? p K a = 4.75 for acetic acid. Answer: The reaction being titrated is: Na + CHCOO - + HCL ⇌ CHCOOH + Na + + Cl - N(acetic acid) = n(sodium acetate) = 1 L x 0.0200M = 0.0200 mol If we add 2 mmol HCl = 0.002 mol to the reaction, it neutralized 2 mmol of the NaCHCOO (0.020-0.002=0.018), creating more CHCOOH (0.020+0.002=0.022). We can use an ICE table or plug this into Henderson Hasselbach: pH = 4.75 + log(0.018/0.22) = 4.66 6. What is the pH at the stoichiometric point for the titration of 0.100 M CH 3 COOH(aq) with 0.100 M KOH(aq)? The value of K a for acetic acid is 1.8 ´ 10 - 5 . Answer: pH = pKa + log(CH 3 COO - /CH 3 COOH) If we neutralize CH 3 COOH with equimolar KOH, we produce 0.1 M CH 3 COO - . Now this has to equilibrate: CH 3 COO - + H 2 O ⇌ CH 3 COOH + OH - The Kb of the conjugate base of acetic acid is Kb = 10 -14 /1.8 x 10 -5 = 5.56 x 10 -10 = x 2 /0.1-x x = 7.45 x 10 -6 = [OH - ] pOH = 5.12 pH = 14-5.12 = 8.87 7. One of the following graphs represents the titration of a strong acid with a strong base, while the other represents the titration of a weak acid with a strong base a) Which is which? Explain how you can tell. b) Identify the equivalence point in each titration. c) Identify the buffering region in the titration of the weak acid.

Answer: a) the left is the strong acid. The pH starts lower, there is no buffer region, and the pH is 7 at the equivalence point. b) They are both at 20 mL. The pH for the strong acid is 7.0 and for the weak acid is ~8.5 c) The buffering region is straddling 10 ml, the ½ stoichiometric point. Here pH = pKa = 5. This is a good buffer for pH range 4-6. 8. A 0.150 M NaOH (aq) solution is used to titrate a 25.00 mL 0.100 M aqueous formic acid (HCOOH) solution at 25 ˚C. At this temperature, the pKa of formic acid is 3.75. a) What is the original pH and percent ionization of the formic acid solution? b) What is the pH of the solution after 5.00 mL of 0.150 M NaOH is added to the analyte? c) What is the pH at the equivalence point? Answer: a) Ka = 10 -3.75 = 1.78 x 10 -4 = x 2 /(0.1-x) x = 0.004 = [H 3 O + ] pH = 2.37 % inonization = 0.004/0.100 = 4% b) First calculate the number of moles in the mix: c) n(HCOOH) = 0.025 L x 0.100 M = 0.0025 mol n(NaOH) = 0.005 L x 0.150 M = 0.00075 mol partial neutralization is n(HCOOH) – n(NaOH) = 0.0025 – 0.00075 = 0.00175 mol and n(HCOO-) = 0.00075 mol produced by neutralization. Find concentrations for new volume: 25 + 5 = 30 ml [HCOOH] = 0.00175/0.030 = 0.0583 M [HCOO-] = 0.00075/0.030 = 0.025 M Now we can use the Henderson Hasselbach equation: pH = pKa + log([HCOO-]/[HCOOH}) = 3.75 + log(0.025/0.0583) = 3.38 or we can do it the long way with an ICE table using these new concentrations as initial conditions: HCOOH ó H3O+ + HCOO- 0.0583 0 0.025 -x +x +x 0.0583 x 0.025+x Ka = 10 -3.75 = 1.78 x 10 -4 = [H 3 O + ][HCOO-]/[HCOOH] = x*(0.025+x)/(0.0583-x) x = 0.0004 = [H 3 O + ] pH = 3.38 (the difference between assuming x is small and solving the quadratic is really small and sensitive to rounding errors) The difference between pHs with HH vs ICE is negligible. (During 9am office hours, I was trying to do this in two steps, which was not the right way to do it because of the endless back and forth of each equilibrium. The “long” solution is probably what one of you was trying to convince me to do.) d) At the equivalence point n(HCOOH) = n(NaOH) producing n(HCOO - ) = 0.0025 mol We have to figure out the volume of NaOH added: V = 0.0025 mol/0.150 M = 0.0167 L [HCOO - ] = 0.0025 mol/(0.0167 + 0.025) L = 0.06 M Now we determine the pH of a 0.06 M solution of formate ion. pKb = 14-3.75 = 10.25 Kb = 10 -10.25 = 5.6 x 10 -11 = x 2 /(0.06-x) X= 1.84 x 10 -6 = [OH - ] pOH = 5.7

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version