Question

SHORT ANSWER: Write your answer in the space provided. Show all your work. Attach the Excel files for all the graphs 1. (15 points) A 100.0 mL aliquot of 0.100 M weak base B (pKb 5.00) was titrated with 1.00 M HCl. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL. Show all your calculations.

Answer 1

Given that concentration of HCl = 1.00 M concentration of B = 0.100 M Volume of B = 100.0 mL pKb = 5 Ko of B = 1.0 x 10-5 a) Before Addition of any acid: B + H20 * BH + OH * N20 Initial 0.100 Change -X Equillibrium 0.100 - x K. =_X.X=1.0x10-5 0.100- x x +1.0x10--1.0x100 = 0' by solving we get x = 0.000995 :: [OH"]=0.000995M pOH =-log(0.000995)=3.00 pH = 14 – 3.00 = 11.00

b) Addition of 1 mL of HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCl = M*V = 1.00 M* 1 mL = 1 mmol Net moles of B = 10.0 – 1 = 9 mmol mmol + HCIE BHCH Initial Change Equillibrium According to Henderson's equation [acid] _ pOH = pK, + log -= 5-0.95424 [base] pOH = 4.05 pH =14-pOH =14-4.05 =9.95

c) Addition of 5 mL of HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCl = M*V = 1.00 M* 5.0 mL = 5 mmol Net moles of B = 10.0 - 5.0 = 5.0 mmol This is the mid point. At this point, pOH = pKb = 5.0 pH = 14 - 5.0 = 9.0

d) Addition of 9 mL of HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCI = M*V = 1.00 M* 9.0 mL = 9 mmol Net moles of B = 10.0 -9.0 = 1.0 mmol mmol B HCIE BHCH- Initial Change o qo Equillibrium According to Henderson's equation POH = pK; + loghacia = 5+log = 5+0.954243 [base] pOH = 5.954243 = 5.95 pH =14 - POH = 14-5.95 = 8.05

e) Addition of 9 mL of HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCl = M*V = 1.00 M* 9.9 mL = 9.9 mmol Net moles of B = 10.0 -9.9 = 0.1 mmol mmol + HCIS BHCl- Initial 10 Change -9.9 -9.9 +9.9 Equillibrium 0.1 9.9 = 5+1.995635 According to Henderson's equation 9.9 POH = pK, +log 5+log — [base] pOH = 7.00 pH = 14- pOH =14-7.00 = 7.00 0.1

(f) Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCI = M*V = 1.00 M* 10 mL = 10 mmol This is equivalence point At this point, there is no B is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 10 mmol [BH*CI+]= =0.0909M Total volume (100+10) mL K, of BH* -- 1x10-14 K 1.0x10-5 =1x10-9 Dissociation of BH+ is given by mmol mmol BH BHT ܥ B + Initial 0.0909 Change -X Equillibrium 0.0909 - X -=1x10-9 0.0909-X x is small, so 0.0909- x= 0.0909 .x.x 0.0900 =1x10-9 x² = 9.091x10-11 x=9.53x106M :: [H*]=9.53x100M pH = -log(9.53x10^)= 5.02

g) Addition of 10.1 ml HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCl = M*V = 1.00 M* 10.1 mL = 10.1 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HCl in the solution. Net moles of HCl = 10.1 – 10.0 = 0.1 mmol Total volume of the solution = 100 + 10.1 = 110.1 mL Conc. of HCI = [H+] = moles/ volume = 0.1 mmol / 110.1 mL = 0.000908 M pH =- log (H+] = -log[0.000908] = 3.04 b) Addition of 12 mL HCI: Number of moles of B = M*V = 0.100 M* 100.0 mL = 10.0 mmol Number of moles of HCl = M*V = 1.00 M* 12 mL = 12 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HCl in the solution. Net moles of HCl = 12 - 10.0 = 2 mmol Total volume of the solution = 100 + 12 = 112 mL Conc. of HCl = [H+] = moles/ volume = 2 mmol / 112 mL = 0.017857 M pH =- log [H+] = -log[0.017857] = 1.75

volume of HCl	pH
0	11
1	9.95
5	9
9	8.05
9.9	7
10	5.02
10.1	3.04
12	1.75

ру 8.05 5.02 3.04 - 1.75 1 2 4 6 8 10 12 14