Asked by David
Q1: You wish to prepare a buffer solution with pH = 11.10. What volume of 6.0 M HCl would you add to 500 mL of 0.10 M (C2H5)2NH to prepare the buffer? You may assume that the solution’s volume remains constant.
Q2: What is the resulting pH when 20 mL of 0.20 M NaOH is added to the buffer prepared in Q1?
Q3: Imagine diluting 250 mL of the buffer prepared in Q1 with distilled water to a final volume of 500 mL. What is the resulting pH when 20 mL of 0.20 M NaOH is added to the diluted buffer?
Q2: What is the resulting pH when 20 mL of 0.20 M NaOH is added to the buffer prepared in Q1?
Q3: Imagine diluting 250 mL of the buffer prepared in Q1 with distilled water to a final volume of 500 mL. What is the resulting pH when 20 mL of 0.20 M NaOH is added to the diluted buffer?
Answers
Answered by
DrBob222
What pKa are you using for diethylamine. I found 11.02.
millimoles R2NH = mL x M = 500 x 0.1 = 50.
.........R2NH + H^+ ===> R2NH2^+
I.........50....0.........0
Add.............x...........
C.........-x...-x.........x
E.........50-x..0.........x
pH = pKa + log(base/acid)
11.10 = 11.02 + log(50-x/x)
Solve for x = millimols acid added.
M = mmols/mL or
mL = mmols/M = ?mmols/6 = mL 6M HCl.
I will leave the other parts for you. Post your work if get stuck.
)
millimoles R2NH = mL x M = 500 x 0.1 = 50.
.........R2NH + H^+ ===> R2NH2^+
I.........50....0.........0
Add.............x...........
C.........-x...-x.........x
E.........50-x..0.........x
pH = pKa + log(base/acid)
11.10 = 11.02 + log(50-x/x)
Solve for x = millimols acid added.
M = mmols/mL or
mL = mmols/M = ?mmols/6 = mL 6M HCl.
I will leave the other parts for you. Post your work if get stuck.
)
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