Experiment 4 - Determining the K. of a Weak Acid with sample results) Introduction Acids are typically molecular compoun

Post by **answerhappygod** » Mon May 02, 2022 7:40 pm

: Experiment 4 Determining The K Of A Weak Acid With Sample Results Introduction Acids Are Typically Molecular Compoun 1 (43.57 KiB) Viewed 33 times

: Experiment 4 Determining The K Of A Weak Acid With Sample Results Introduction Acids Are Typically Molecular Compoun 2 (45.37 KiB) Viewed 33 times

: Experiment 4 Determining The K Of A Weak Acid With Sample Results Introduction Acids Are Typically Molecular Compoun 3 (44.54 KiB) Viewed 33 times

: Experiment 4 Determining The K Of A Weak Acid With Sample Results Introduction Acids Are Typically Molecular Compoun 4 (34.14 KiB) Viewed 33 times

Experiment 4 - Determining the K. of a Weak Acid with sample results) Introduction Acids are typically molecular compounds that contain at least one ionizable proton, and they donate it to substances known as bases that are capable of accepting that proton. In the absence of an actual base, however, water can play that role, and vice versa, in the absence of an actual acid water can play that role in the presence of a base. This is because water is amphoteric, it can accept as well as donate a proton. When water accepts a proton from an acid it forma a hydronium ion, or hydrated proton H0 What remains of the acid is called its conjugate base. Consider the behavior of a monoprotic strong acid HA dissolved in water WA(aq) + W,000 -- 1,09(aq) + AC) feq. 1) Acid Water Hydronium Conj. Base Notice the arrow in the chemical equation, it goes in one direction, this is because a strong acid is a strong electrolyte, therefore it completely dissociates when dissolved in water. A weak acid, however, is a weak electrolyte and only partially dissociates in water HA(49) + H2000 - 120*60p) +A"(44) Inq-2 The dissociation of a weak acid can also be represented in the following way NA(49) H(aq) + 1" (aq) feq. 39 The two chemical equations are functionally equivalent, therefore, in chemical terms 4,0 (2) and H lag) represent the same species. Weak acids establish an equilibrium whose numerical value is called the acid dissociation constant K. Monoprotio weak acids can only establish one equilibrium and therefore only have one acid dissociation constant; polyprotic acids, however, can establish multiple equilibria, each with its own corresponding dissociation constant. For a monoprotic weak acid the dissociation equilibrium expression is: leg4 MAI The following are the acid dissociation constant expressions for H.co.a diprotic acid and notice how each expression is based on the dissociation reaction that takes place 1.601 feq. 5) Ma- INCA, leq, 6) 1
Water also dissociates partially, in a process called autoionization, and establishes its own equilibrium called the lon-product constant of water. The dissociation can also be represented in two ways: 24.00) H30*(aq) + OH(aq) feq. H.00) - *() + OH() K =[H0*-ON") = 1 x 10-4 (at 25 °C) (eq. 9 feq. 8) In pure water (H.01 - OH and the pH - 7.00, however, when in mixed with an acid, the amount of hydronium ion increases, and the amount of hydroxide decreases in an inversely proportional manner that maintains equilibrium. The opposite is true if water is mixed with a base; the amount of hydroxide increases and the amount of hydronium ion decreases pH = -log". [+] = 10 leg. 10) When H:0") - [OHN --log(1 x 10") - 7 feq. 11) When H.0" >[OH-], pH <7 and when [H.0*1 [OH1, PH > 7 (eq. 12) The pH of a weak acid depends on its concentration and on its k. [*]VK:(HA). (approximate relationship when [**] [A]) fcq. 13) These relationships make it theoretically possible to determine the K, of an unknown weak acid through pH measurements and mathematical conversion. In practical terms, however, there is a very specific pH that can be quickly used to reliably determine a weak acid's K. If a salt of the conjugate base A is added to a solution of HA, the equilibrium begins to shift to the reactant side, and the amount of H' decreases. Eventually, when enough A is added to the mixture its concentration becomes equal to that of HA When this happens, the numerical value of Hi equals K. of the weak acid and the pH equals the weak acid's pk. This is the approach that will be taken to determine the K. of a monoprotic weak acid. In this experiment the K. of a weak acid will be determined by creating the necessary conditions in the solution so that its pH equals the pk of the weak acid.
Experimental Materials Sodium acetate Acetic acid (0.10 M and 0.02 M) Hot plate (x2) Graduated cylinder 25 ml. (x2) Magnetic stirring bar x 21 Beader 50 mL (x2) pH meter Dropper (x2) Experimental Procedure 1. Label two 50 ml beakers "A" and "B" 2. Determine the volume in mL of 0.10 Macetic acid needed to make 25.0 mL or 0.020 M concentration, transfer that volume to a 25 ml graduated cylinder, then fill the graduated cylinder up to the 25 mL mark with distilled water. 3. Determine the molar mass of sodium acetate (CH.CO.Na). 4. Using a scale, measure up to but no more than 0.040 g of sodium acetate, record the mass down to the milligram level, and transfer to the 50 ml beaker labelled "A". 5. Determine the number of moles of sodium acetate contained in the sample and record that number 6. Determine the volume of 0.020 M acetic acid that contains the same number of moles of acetic acid as the sample of sodium acetate, 7. Use a graduated cylinder to transfer the volume determined in step 6 to the 50 ml beaker labelled "A", add a magnetic stirring bar to the mixture, then place the 50 ml. beaker labelled "A" on a hot plate to stir 8. Once all the sodium acetate has been dissolved measure the pH of the solution with the pH meter. Note: be sure to rinse the pH meter probe with distilled water and wipe it dry with a "Kim Wipe" before and after each use. 9. Repeat steps 4 - 8 with the following changes: a. Use up to but no more than 0.200 g of sodium acetate (change in step 4 b. Use 0.10 Macetic acid (cha tep c. Use the beaker labeled "B" (change in step 7) 10. Convert the pH value to the experimental of the weak acid.
Data Sheet 1. Volume of 0.10 M acetic acid needed for 25.0 mL of 0.020 M. ml. 山 2. Molar mass of sodium acetate 3. Mass of sodium acetate transferred to beaker "A". 0.03628 4. Moles of sodium acetate measured. mol 5. Volume of 0.020 M acetic acid needed. ml 6. pH of beaker "A" mixture. pH - 4.76 7. Mass of sodium acetate transferred to beaker "B". 0.1991 8. Moles of sodium acetate measured. mol 9. Volume of 0.10 M acetic acid needed. ml. 10. pH of beaker "B" mixture. pH 4.74 11. Accepted pk of acetic acid. pk - 4.74 12. Calculated K. from pH of beaker "A" K- 13. Calculated K, from pH of beaker "B" 14. Accepted K of acetic acid K. - 1.8x10 Questions to consider when preparing the laboratory report: 1. How close were the pH measurements in both beakers to each other and based on your understanding should the two measurements have been close or very different from each other? 2. Which measurement produced a pH closer to the accepted pk of acetic acid of 4.74 and which would you have expected to have a pH closer to the pk.?