- #1
Little Bear
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pH = pKa + log [base/acid] (H-H equation)
Using the H-H equation, calculate the theoretical volume of 0.2 M NaOH required to raise the pH of 0.1 M Tris HCl (pk = 8.1) to raise the pH to its pK value.
The way I see it is to get pH = pK so using H-H equation:
8.1 = 8.1 + log[0.2 M NaOH/0.1 M Tris HCl]
8.1-8.1 = log[0.2 M NaOH/0.1 M Tris HCl]
0 = log[0.2 M NaOH/0.1 M Tris HCl]
Then taking anti log,
1 = [0.2 M NaOH/0.1 M Tris HCl]
1 = 0.5
means
pH = 1, has base:acid ratio of 1/2
Thus if I have 0.1 mol/litre Tris HCL, I need half of 0.2 M NaOH.
But how do I convert these to volumes?
Please forgive me for asking this Q. I am really stuck. Am I on the right track? If so, how do I proceed to get the volumes. If not, what am I thinking that is wrong?
Please give up a detailed understanding of what I did wrong?
I would really appreciate it. Thanks.
Then I am suppose to draw a curve but how can I? with only one point at the theoretical volume of 0.2 M NaOH. I just don't understand. Please help me understand what they want from me.
Using the H-H equation, calculate the theoretical volume of 0.2 M NaOH required to raise the pH of 0.1 M Tris HCl (pk = 8.1) to raise the pH to its pK value.
The way I see it is to get pH = pK so using H-H equation:
8.1 = 8.1 + log[0.2 M NaOH/0.1 M Tris HCl]
8.1-8.1 = log[0.2 M NaOH/0.1 M Tris HCl]
0 = log[0.2 M NaOH/0.1 M Tris HCl]
Then taking anti log,
1 = [0.2 M NaOH/0.1 M Tris HCl]
1 = 0.5
means
pH = 1, has base:acid ratio of 1/2
Thus if I have 0.1 mol/litre Tris HCL, I need half of 0.2 M NaOH.
But how do I convert these to volumes?
Please forgive me for asking this Q. I am really stuck. Am I on the right track? If so, how do I proceed to get the volumes. If not, what am I thinking that is wrong?
Please give up a detailed understanding of what I did wrong?
I would really appreciate it. Thanks.
Then I am suppose to draw a curve but how can I? with only one point at the theoretical volume of 0.2 M NaOH. I just don't understand. Please help me understand what they want from me.