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pH Measurement and pH Titration Curve Prelaboratory Assignment Name and Drawer N

ID: 549941 • Letter: P

Question

pH Measurement and pH Titration Curve Prelaboratory Assignment Name and Drawer Number 1. A 10.0 mL portion of a 0.10 M HCI(aq) solution was diluted to 100 mL by the addition of 90.0 mL of water. A 10.0 mL portion of the diluted solution was further diluted by the addition of 90.0 mL of water. a. Calculate the pH of the original 0.10 M HCI(aq) solution by use of Equation (21-1). b. Calculate the concentration of H' and pH of the solution prepared after the first dilution. c. Calculate the concentration of H' and pH of the solution prepared after the second dilution. 2. A 10.0 mL portion of 0.10 M acetic acid HC3H,02 (Ka = 1.8 x 10-5) was diluted with 90.0 mL of deionized water. A 10.0 mL portion of the resulting diluted solution was diluted further by the addition of 90.0 mL of water. a. Write the chemical equation for the dissociation of acetic acid. b. Write the equation for the dissociation constant of acetic acid [Equation (21-7) can be used as a guide in writing the equation). c. Use the equation to calculate the concentration of Ht and CH,02" in the 0.10 M acetic acid solution (you might want to review your lecture notes and text for details of the calculation). d. Use the [H'] from 2c to calculate the pH of the solution by using Equation (21-1). e. Similarly calculate the pH of the solution prepared after the first dilution and the pH of the solution prepared after the second dilution. 3. A 10.00 mL portion of a solution of a weak monoprotic (one H') acid HA was titrated with 0.1894 M NaOH. The results of the titration are tabulated below. (a) write the balanced chemical equation for the titration reaction. (b) Plot the pH titration curve from the tabulated data. (c) From the titration curve, determine the end point of the titration. (d) Use the endpoint volume to calculate the number of moles of NaOH used in the titration. (e) Use the number of moles of NaOH and the balanced chemical equation to calculate the number of moles of HA that reacted during the titration. (f) Use the number of moles of HA and the volume of the original HA solution (10.00 mL) to calculate the original concentration of HA in the analyte. (Continued on next page) -143

Explanation / Answer

1)

a) as HCl is strong acid it dissociates compleely

[H+] = [HCl] = 0.10 M

pH = - log [H+]

pH = - log [0.10]

pH = 1.0

b) accirding to dilution law

M1V1 = M2V2

M1 = 0.10 M , V1 = 10 mL

M2 = ? , V2 = 100 mL

M2 = (M1V1 / V2)

M2 = (0.10 x 10 / 100)

M2 = 0.01 M

[H+] = [HCl] = 0.01 M

pH = - log [H+]

pH = - log [0.01]

pH = 2.0

c) M1V1 = M2V2

M1 = 0.01 M , V1 = 10 mL

M2 = ? , V2 = 100 mL

M2 = (0.01 x 10 / 100)

M2 = 0.001 M

[H+] = [HCl] = 0.001 M

pH = - log [H+]

pH = - log [0.001]

pH = 3.0