Question

I need help with the problem in the last photo.. I thought I’d post my lab explanation and data if that helps you get a better understanding, but it’s just the question at the end. I know I need to use the Henderson Hasselbach equation.. so... 4.70 = pKa + log( [acetate-ion] / [acetic-ion] ) and solve for pKa, then Ka.. but how do I find the concentrations to put in the log fraction?

Thanks, in advance!

Learning Objectives: 1. To test the acidic and basic properties of ionic compounds. 2 To create a buffer solution and calculate its pH. 3. To determine the acid dissociation constant of acetic acid using the Henderson-Hasselbach equation. 4. To test a buffer's capacity by reacting it with a strong acids and a strong base. Discussion: Buffers are solutions that resist changes in pH when small quantities of an acid or base are added to them. These solutions contain a weak acid and its conjugate base or a weak base and its conjugate acid. Because these mixtures contain both the acidic and basic species, they can easily neutralize small quantities of either H' or OH. Buffers are also resistant to dilution effects, maintaining the same pH even when water is added to them. Part A: Acidic and Basic Properties of lonic Compouds The first part of the experiment involves using a pH meter to determine the pH of several salt solutions A pH meter consists of a meter and two electrodes. The measurement of pH requires two electrodes, a sensing electrode that is sensitive to HaO concentration and a reference electrode. This is because the pH meter is really a voltmeter that measures the electrical potential of a solution. The reference electrode is an electrode that develops a known potential that is essentially independent of the contents of the solution into which it is placed.. The glass electrode is sensitive to the HaO concentration of the solution into which it is placed. Part B: Buffers A buffer system will be prepared from solutions of conjugate acid- base pairs. The pH of the buffer system will be determined using the pH meter. The Ka of the weak acid used in the buffer will be calculated using the pH data. The effect of adding an acid and a base to the system will be monitored. For a weak acid H,O(aa HA (a)H,O +A (aal Reaction: And K [HaO'ITA] HA Knowing the experimentally determined pH of the solution the [HsO'] can be calculated from pH= -log|HaO"] The Ka of the buffer system can be calculated from the Henderson-Hasselbalch equation log [A1 HA pH pk +

Answer 1

In som beak en sodium acetate was added 3.5ng & 8.8m of 3.0M acetic acid + SS2 ml di stilled ouat er. in total volume of the solution u (8.8 + 55.2) m = 64 mi . Ker 8 8 we of 3.om acetic and (CH3COOH) moles of acetic acid present E 3.0 mol x & 8.8 X 700 MM X 5 = 0.0264 mol en In 64 m solution Eettacoor] = 0.0284 mol & 100m = 0.4125 mol/l buna = 0:4115 M now, number of moler of sodium acetale (CH3COONa). In 3.511 g molar man of 2 CHCOONa. "x 82.0348 +82.034 91mor] = 0.04218 mol In 6 mi solution [CHCOONa. = 0.0428 mal & 10000 = 0.669 melse = 0.6694 3.511 px 1 mol

wow, from Henderson -Hanselbalch. equation are ger [conjugale sait] pH= pka & dog I weak and - pH = pkat log [eth Coona] at 10 TcHtz cont] PH - pka & Logo 11004125 There, pt=4.707 & 4.7 = pkat 021 = pka = 4.7 0.21 = 4.49 pka = 4.99 =) – log ka = 44 pka = alog, kaf & logka un 449 in & - 10-449 = 3.2x10-5 in Experimental ka value of CHg coolt ils 3.2x10-5 > Jiu wilg 3:2 x10-5