You have to design a buffer based on one of the systems below
System 1: HA1 and A1-1 K= 4 E-3
System 2: HA2 and A2-1 K= 5 E-4
System 3: HA3 and A3-1 K= 6 E-5
For the following parts assume that the buffer above in system2 has been constructed with [HA] =1M and that there is no change in volume when acids or bases are added
a) Pick the best system to make a buffer with a pH of 3.5.
b) For which ever system you have chosen find the [A^-1]/[HA] required to make the pH equal to 3.5
c) Find the new pH if .03 moles of HCl is added to 100 ml of buffer in system 2
d) Find the maximum mass of NaOH which could be added to 100 ml of the original buffer without changing the pH by more than .7 pH units
This problem is a little confusing because of the wording; i.e., it isn't easy to tell when we are talking about system 2 and when we aren't.
The best buffer to use is one that has pKa = pH you want for the solution. So convert Ka for each system to pKa. I get something like
for 4E-3 = 2.4
for 5E-4 = 3.3
for 6E-5 = 4.2
Take your pick from my initial remarks.
b. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
Set pH = 3.50 and solve for base/acid ratio.
c. System 2 is
pH = 3.30 + log (base/acid)
Adding 0.03 moles HCl to the buffer. The buffer, to resist a change in pH, will react H^+ with the A^- base, as
A^- + H^+ ==> HA
Therefore, addition of 0.03 mole HCl will decrease A^- by 0.03 and increase HA by 0.03. Make those change in log(base/acid) and calculate new pH.
d. Set pH in the H-H equation to 3.5 + 0.7 = ??(I suppose this is what is meant--the problem doesn't say 0.7 from what initial value) and calculate b/a ratio. That will tell you what the new HA and A must be and you add NaOH to accomplish that.
To design a buffer, we need to choose the system that has a pKa closest to the desired pH. The pKa can be calculated using the equation: pKa = -log(Ka).
a) First, let's calculate the pKas of each system:
System 1: pKa1 = -log(4E-3)
System 2: pKa2 = -log(5E-4)
System 3: pKa3 = -log(6E-5)
The desired pH is 3.5. To compare the pKa values, we need to convert them to pH values using the equation: pH = -log[H+].
pH1 = -log(4E-3) ≈ 2.4
pH2 = -log(5E-4) ≈ 3.3
pH3 = -log(6E-5) ≈ 4.2
Comparing the pH values to the desired pH of 3.5, we can see that System 2 has the closest pH. Therefore, system 2 is the best choice to make a buffer with a pH of 3.5.
b) The ratio of [A^-1]/[HA] required to achieve a particular pH can be determined using the Henderson-Hasselbalch equation: pH = pKa + log([A^-1]/[HA]).
Given that the buffer in System 2 has been constructed with [HA] = 1M, we need to find the [A^-1]/[HA] ratio that gives a pH of 3.5.
pH = 3.5
pKa2 = -log(5E-4)
Rearranging the Henderson-Hasselbalch equation, we get:
log([A^-1]/[HA]) = pH - pKa2
log([A^-1]/[HA]) = 3.5 - (-log(5E-4))
Using the logarithmic identity log(a/b) = log(a) - log(b), the equation simplifies to:
log([A^-1]/[HA]) = 3.5 + log(5E-4)
Now, we can calculate the ratio [A^-1]/[HA] by taking the antilog (10^x) on both sides:
[A^-1]/[HA] = 10^(3.5 + log(5E-4))
c) To find the new pH if 0.03 moles of HCl is added to 100 ml of the buffer in System 2, we need to consider the stoichiometry of the reaction. In this case, since the buffer is constructed with [HA], adding HCl will convert some of the [HA] into [A^-1]. Thus, the [HA] concentration will decrease, and the [A^-1] concentration will increase.
First, convert the volume of buffer from ml to liters:
Volume = 100 ml = 0.1 L
Given that 0.03 moles of HCl is added, we can calculate the change in moles of [HA] and [A^-1]. Since the buffer was initially constructed with [HA] = 1M, the moles of [HA] will decrease by 0.03 moles, and the moles of [A^-1] will increase by 0.03 moles.
The new concentrations of [HA] and [A^-1] can be calculated by dividing the moles by the new volume of the buffer, which is still 0.1 L.
[HA] = (1 - 0.03) moles / 0.1 L
[A^-1] = 0.03 moles / 0.1 L
Finally, we can use the Henderson-Hasselbalch equation to find the new pH:
pH = pKa2 + log([A^-1]/[HA])
Substitute the values of pKa2, [A^-1], and [HA] into the equation to find the new pH.
d) To find the maximum mass of NaOH that can be added to 100 ml of the original buffer without changing the pH by more than 0.7 pH units, we can consider the neutralization reaction between NaOH and HA. The moles of NaOH added must be equal to the moles of HA in the buffer to maintain the pH within the given range.
First, calculate the moles of HA in the initial buffer using its concentration ([HA] = 1M) and volume (100 ml = 0.1 L).
Moles of HA = 1M x 0.1 L
Since NaOH reacts in a 1:1 ratio with HA, the moles of NaOH required to neutralize the HA is equivalent to the moles of HA:
Moles of NaOH = Moles of HA
Finally, use the molecular weight of NaOH to calculate the maximum mass that can be added without changing the pH by more than 0.7 pH units.
Maximum mass = Moles of NaOH x Molecular weight of NaOH
Remember to convert the mass into a suitable unit according to the given values.