Here\'s the part 1 info Using the calculations from your prelab, prepare 100 mL
ID: 884282 • Letter: H
Question
Here's the part 1 info
Using the calculations from your prelab, prepare 100 mL of 0.25M Tris and 100 mL 0.25M Tris-HCl. Be sure to use the 10 mM KCl provided as the solvent.
Using the solutions you just made, create 10 mL samples of the following buffer ratios:
100% Tris-HCl
75% Tris-HCl, 25% Tris
50% Tris-HCl, 50% Tris
25% Tris-HCl, 75% Tris
100% Tris
Complete the following table using the molarities you calculated for each buffer in Part 1. Using a 10 mL aliquot of the buffer, show the change in mmoles when 0.01 mL of 2.0M HCl is added to the buffer. Predict the change in pH after the addition.
Complete the following table using the molarities you calculated for each buffer in Part 1. Using a 10 mL aliquot of the buffer, show the change in mmoles when 0.01 mL of 2.0M NaOH is added to the buffer. Predict the change in pH after the addition.
Buffer # Initial moles of Tris-HCL Initial moles of Tris Post Addition of HCLmmol of Tris-HCl Post Addition of HCL
mmol of Tris Predicted pH Buffer 1 0.0001875 6.25e-05
1
2
3 Buffer 2 0.000125 0.000125
4
5
6 Buffer 3 8.75e-05 0.0001625
7
8
9
Explanation / Answer
0.01 mL of 2.0M HCl =0.00002 moles=0.02mmol
the blanks=
1) 0.1875mmoles+0.02mmol=0.2075 mmol
2)6.25e-05-0.00002 moles=4.25e^-05moles=0.0425mmol
3) pH=Pka + log [base]/[acid]=8.214 + log 0.0425/0.2075=7.5254
4) 0.125+0.02mmol=0.145mmol
5)0.125-0.02mmol=0.105mmol
6)pH=8.314+log [0.105]/0.145=8.214-0.1402=8.0738
7)8.75e-02mmol+0.02mmol=0.1075mmol
8)0.1625mmol-0.02mmol=0.1425mmol
9)pH=8.214+ log 0.1425/0.1075=8.3363
table 2
0.01 mL of 2.0M NaOH=0.00002 moles=0.02mmol
10)0.1875mmoles-0.02mmol=0.1675mmol
11)6.25e-05+0.00002 moles=4.25e^-05moles=0.0825mmol
12) pH=8.314+log0.0825/0.1675=7.9064
13)0.125-0.02mmol=0.105mmol
14)0.125-0.02mmol=0.145mmol
15)pH=8.314+log [0.145]/0.105=8.214-0.1402=8.3309
16)8.75e-02mmol-0.02mmol=0.0675mmol
17)0.1625mmol+0.02mmol=0.1825mmol
18)pH=8.214+ log 0.1825/0.0675=8.6457