To prepare a pH 3.15 buffer from an acid/base pair with a pKa of 3.15, the A - /HA ratio must be 1. which of the following would produce this ratio
a. 2 mol HA and 2 mol A -
b. adding 0.5 mol HCL to 1 mol A -
c. Adding 1 mol NaOH to 2 mol HA
d. All of the above
The answer is D but can someone please explain!
Solution
pH = pKa
if [A - ] = [HA]
[A - ] / [HA] = 1.0
so pH = pKa = 3.15
a) moles of HA = moles of A -
so [A - ] / [HA] = 1.0
b) if we add 0.5 moles HCl
A - + HCl --------> HA + Cl -
50% A - becomes HA
so HA = A -
[A - ] /[HA] = 1.0
c) HA + NaOH --------> NaA + H2O
1 mole NaOH reacts with 1 mole HA
so 1 mole NaA - will form
1 mole HA will be left over
so HA =A -
[A - ] / [HA] = 1.0
so alll options are correct only.
.
To prepare a pH 3-15 buffer from an acid-base pair with a pKa of 3-15-.docx
1. To prepare a pH 3.15 buffer from an acid/base pair with a pKa of 3.15, the A -
/HA ratio must be
1. which of the following would produce this ratio
a. 2 mol HA and 2 mol A -
b. adding 0.5 mol HCL to 1 mol A -
c. Adding 1 mol NaOH to 2 mol HA
d. All of the above
The answer is D but can someone please explain!
Solution
pH = pKa
if [A -
] = [HA]
[A -
] / [HA] = 1.0
so pH = pKa = 3.15
a) moles of HA = moles of A -
so [A -
] / [HA] = 1.0
b) if we add 0.5 moles HCl
A -
+ HCl --------> HA + Cl -
50% A -
becomes HA
so HA = A -
[A -
] /[HA] = 1.0
2. c) HA + NaOH --------> NaA + H2O
1 mole NaOH reacts with 1 mole HA
so 1 mole NaA -
will form
1 mole HA will be left over
so HA =A -
[A -
] / [HA] = 1.0
so alll options are correct only.
![To prepare a pH 3.15 buffer from an acid/base pair with a pKa of 3.15, the A -
/HA ratio must be
1. which of the following would produce this ratio
a. 2 mol HA and 2 mol A -
b. adding 0.5 mol HCL to 1 mol A -
c. Adding 1 mol NaOH to 2 mol HA
d. All of the above
The answer is D but can someone please explain!
Solution
pH = pKa
if [A -
] = [HA]
[A -
] / [HA] = 1.0
so pH = pKa = 3.15
a) moles of HA = moles of A -
so [A -
] / [HA] = 1.0
b) if we add 0.5 moles HCl
A -
+ HCl --------> HA + Cl -
50% A -
becomes HA
so HA = A -
[A -
] /[HA] = 1.0](https://image.slidesharecdn.com/toprepareaph3-15bufferfromanacid-basepairwithapkaof3-15-230204002112-e7cbf9c4/85/to-prepare-a-ph-315-buffer-from-an-acidbase-pair-with-a-pka-of-315docx-1-320.jpg)
![c) HA + NaOH --------> NaA + H2O
1 mole NaOH reacts with 1 mole HA
so 1 mole NaA -
will form
1 mole HA will be left over
so HA =A -
[A -
] / [HA] = 1.0
so alll options are correct only.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)