B. (20 points) A 25.00 ml of 0.500M solution of a weak base trim ethyTa ine (CH3)3N)is titrated with a 0.125 M HCL. Kb of (CH3)3N at 25 °C is 6.3 x 10 a. Calculate the pH of the solution after adding 75.00 mL of the acid 0. 23L KG HCI PH-2.0 2 b. Calculate the the solution after adding 100.00 mL of the acid. 0.125M-?? O12S mo l 1.9 pH
Solution
millimoles of amine = 25.00 x 0.500 = 12.50
a) millimoles of HCl added = 75.00 x 0.125 = 9.375
12.50 - 9.375 = 3.125 millimoles base left
9.375 millimoles salt formed
[salt] = 9.375 / 100 = 0.09375 M
[base] = 3.125 / 100 = 0.03125 M
pKb = - log Kb = - log [6.3 x 10 -5 ] = 4.20
pOH = pKb + log [salt] / [base]
pOH = 4.20 + log [0.09375] / [0.03125]
pOH = 4.68
pH = 14.0 - 4.68
pH = 9.32
b) millimoles of HCl added = 100 x 0.125 = 12.5
all base convert to salt
[salt] = 12.50 / 125 = 0.1 M
pOH = 1/2 [pKw + pKb + log C]
pOH = 1/2 [14 + 4.20 + log 0.10]
pOH = 8.60
pH = 14 - 8.60
pH = 5.40
.
B- (20 points) A 25-00 ml of 0-500M solution of a weak base trim ethyT.docx
1. B. (20 points) A 25.00 ml of 0.500M solution of a weak base trim ethyTa ine (CH3)3N)is titrated
with a 0.125 M HCL. Kb of (CH3)3N at 25 °C is 6.3 x 10 a. Calculate the pH of the solution
after adding 75.00 mL of the acid 0. 23L KG HCI PH-2.0 2 b. Calculate the the solution after
adding 100.00 mL of the acid. 0.125M-?? O12S mo l 1.9 pH
Solution
millimoles of amine = 25.00 x 0.500 = 12.50
a) millimoles of HCl added = 75.00 x 0.125 = 9.375
12.50 - 9.375 = 3.125 millimoles base left
9.375 millimoles salt formed
[salt] = 9.375 / 100 = 0.09375 M
[base] = 3.125 / 100 = 0.03125 M
pKb = - log Kb = - log [6.3 x 10 -5
] = 4.20
pOH = pKb + log [salt] / [base]
pOH = 4.20 + log [0.09375] / [0.03125]
pOH = 4.68
pH = 14.0 - 4.68
pH = 9.32
b) millimoles of HCl added = 100 x 0.125 = 12.5
all base convert to salt
[salt] = 12.50 / 125 = 0.1 M
![B. (20 points) A 25.00 ml of 0.500M solution of a weak base trim ethyTa ine (CH3)3N)is titrated
with a 0.125 M HCL. Kb of (CH3)3N at 25 °C is 6.3 x 10 a. Calculate the pH of the solution
after adding 75.00 mL of the acid 0. 23L KG HCI PH-2.0 2 b. Calculate the the solution after
adding 100.00 mL of the acid. 0.125M-?? O12S mo l 1.9 pH
Solution
millimoles of amine = 25.00 x 0.500 = 12.50
a) millimoles of HCl added = 75.00 x 0.125 = 9.375
12.50 - 9.375 = 3.125 millimoles base left
9.375 millimoles salt formed
[salt] = 9.375 / 100 = 0.09375 M
[base] = 3.125 / 100 = 0.03125 M
pKb = - log Kb = - log [6.3 x 10 -5
] = 4.20
pOH = pKb + log [salt] / [base]
pOH = 4.20 + log [0.09375] / [0.03125]
pOH = 4.68
pH = 14.0 - 4.68
pH = 9.32
b) millimoles of HCl added = 100 x 0.125 = 12.5
all base convert to salt
[salt] = 12.50 / 125 = 0.1 M](https://image.slidesharecdn.com/b-20pointsa25-00mlof0-500msolutionofaweakbasetrimethyt-230204011505-7d129be0/85/b-20-points-a-2500-ml-of-0500m-solution-of-a-weak-base-trim-ethytdocx-1-320.jpg)
![pOH = 1/2 [pKw + pKb + log C]
pOH = 1/2 [14 + 4.20 + log 0.10]
pOH = 8.60
pH = 14 - 8.60
pH = 5.40](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)