a buffer composed of 0.50mol acetic acid and 0.50mol sodium acetate is diluted to a volume of 1.0L. The pH of the buffer is 4.74. How many moles of NaOH must be added to the buffer solution to increase its pH to 5.74?
I assume you are using pKa = 4.74 so that
pH = 4.74 + log 0.5/0.5 =
= 4.74 + log 1 = 4.74 + 0 = 4.74.
If we call acetic acid HAc and the acetate ion (the base) Ac, then
To make the pH 5.74 we calculate the B/A as follows:
5.74 = 4.74 + log B/A
You can go through the calculation but you should end up with (Ac)/(HAc) = 10
Then we write the equation
HAc + NaOH = NaAc + H2O
HAc = 0.5 to start
Ac = 0.5 to start.
We add x moles NaOH.
The HAc after the reaction will be 0.5 - x and the Ac will be 0.5+x
Just plug those values into the
(Ac)/(HAc) = 10 and solve for x.
I always like to put that value in and resolve the equation to see if I get 5.74.
Post your work if you get stuck.
thanks for your help, it worked out :)
To calculate the number of moles of NaOH that must be added to the buffer solution, we need to first determine the change in pH and then use the Henderson-Hasselbalch equation.
Step 1: Calculate the change in pH
The change in pH is given by the difference between the final pH (5.74) and the initial pH (4.74):
ΔpH = final pH - initial pH
ΔpH = 5.74 - 4.74
ΔpH = 1
Step 2: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH, pKa is the acid dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
Since the initial pH is 4.74, we can use this as the pH value in the Henderson-Hasselbalch equation. The pKa for acetic acid (CH3COOH) is 4.74. The concentration of acetic acid is 0.50 mol/1.0 L, and the concentration of sodium acetate (CH3COONa) is 0.50 mol/1.0 L.
Let's calculate the concentration of the conjugate base [A-]:
[HA] = 0.50 mol
[A-] = 0.50 mol
Now we can substitute these values into the Henderson-Hasselbalch equation:
4.74 = 4.74 + log([A-]/[HA])
0 = log([A-]/[HA])
Since the logarithm of 1 is zero, we can conclude that [A-]/[HA] = 1.
Step 3: Calculate the concentration of NaOH required to increase the pH by 1
To increase the pH by 1, we need to convert the logarithmic expression into a linear one. Since [A-]/[HA] = 1, this means that the concentrations of the conjugate base and acid are equal.
To maintain the 1:1 ratio of acetic acid to sodium acetate, we must add an equal amount of moles of NaOH to the buffer solution. Therefore, we need to determine the number of moles of NaOH required.
The initial number of moles of sodium acetate (CH3COONa) is 0.50 mol.
Total moles of NaOH required = 0.50 mol
So, 0.50 moles of NaOH must be added to the buffer solution to increase its pH to 5.74.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the acidic and basic components of the buffer:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH, pKa is the negative logarithm of the acid dissociation constant of the acidic component (acetic acid, in this case), [A-] is the concentration of the conjugate base (sodium acetate, in this case), and [HA] is the concentration of the acidic component (acetic acid, in this case).
Given:
- pH = 4.74
- [A-] = 0.50 mol
- [HA] = 0.50 mol
- pKa of acetic acid = 4.74 (since pH = pKa)
We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(4.74 - 4.74)
[A-]/[HA] = 1
This means that the ratio of [A-] to [HA] in the buffer is 1:1.
To increase the pH to 5.74, we need to add a base (NaOH) to the buffer solution. Sodium hydroxide (NaOH) will react with acetic acid (HA) to form sodium acetate (A-) and water.
The balanced chemical equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
Since the ratio of [A-] to [HA] is 1:1 and the volume of the buffer is 1.0 L, we need to add the same number of moles of NaOH as the moles of acetic acid present in the buffer.
Therefore, the answer is 0.50 moles of NaOH must be added to the buffer solution to increase its pH to 5.74.