How many moles of NaOH must be added to 1.0 L of 2.6 M HC3H5O3 to produce a solution buffered at each pH?
1)pH= 4.28
2)pH=pKa
To answer this question, we need to understand the concept of a buffer solution and how it works.
A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. The buffer capacity depends on the concentrations of the weak acid and its conjugate base.
To calculate the amount of moles of NaOH needed to produce a buffer solution at a certain pH, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the desired pH of the buffer solution, pKa is the acid dissociation constant of the weak acid component of the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Let's calculate the amount of moles of NaOH needed for each scenario:
1) pH = 4.28:
In this case, we are given the desired pH. We need to find the pKa value of the weak acid component of the buffer. Once we have the pKa, we can use the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] and calculate the amount of moles of NaOH needed.
2) pH = pKa:
In this case, we are given that the desired pH is equal to the pKa value. This means that the concentration of the weak acid and its conjugate base will be equal in the buffer solution. We can use the Henderson-Hasselbalch equation to determine the ratio of [A-] to [HA] and calculate the amount of moles of NaOH needed.
Please provide the pKa value for the weak acid component of the buffer to proceed with the calculations.
To determine the number of moles of NaOH needed to produce a buffered solution at a specific pH, you need to consider the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]),
where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
1) pH = 4.28:
In this case, we need to calculate the concentration of the conjugate base ([A-]) and the acid ([HA]).
Using the Henderson-Hasselbalch equation, we can rewrite it as follows:
4.28 = pKa + log([A-]/[HA]).
Since the pH is close to the pKa, we can make the assumption that [A-] and [HA] are approximately equal.
Therefore, [A-]/[HA] ≈ 1.
Now, let's calculate the concentration of [A-] and [HA]:
2.6 M = [HA] + [A-].
Since [A-] ≈ [HA], let's assume they are both equal to x.
Therefore, 2.6 M = x + x,
or 2.6 M = 2x.
Solving for x:
x = 2.6 M / 2 = 1.3 M.
Hence, you would need to add 1.3 moles of NaOH.
2) pH = pKa:
For a solution buffered at pH = pKa, we know that the concentration of the conjugate base ([A-]) and the acid ([HA]) will be equal.
So, in this case, [A-] = [HA].
Using the Henderson-Hasselbalch equation:
pKa = pKa + log([A-]/[HA]).
Since [A-] = [HA], we can substitute them in:
pKa = pKa + log([A-]/[A-]),
which simplifies to:
0 = log(1).
Since log(1) = 0, this equation is true for any value of pKa.
Therefore, the number of moles of NaOH required to produce a solution buffered at pH = pKa is zero.
HC2H3O2 = HAc
moles HAc = M x L = 2.6 x 1 = 2.6
Let x = mols OH to be added.
..........HAc + OH^- --> Ac^- + H2O
I.........2.6....0.......0.......0
add..............x...............
C.........-x....-x.......x.......x
E.........2.6-x..0.......x.......x
I don't know what value you are using for pKa.
pH = pKa + log (base)/(acid)
4.28 = pKa + log (x)/(2.6-x)
Solve for x
pH = pKa is done the same way.
Post your work if you get stuck.