A buffer contains 0.50 mol CH3CO2H and 0.50 mol CH3CO2- diluted with water to 1.0 L. How many moles of NaOH are required to increase the pH of the buffer to 5.00? (pKa for CH3CO2H = 4.74)
To calculate the moles of NaOH required to increase the pH of the buffer, we need to understand the buffer system and its response to the addition of a strong base.
A buffer solution consists of a weak acid (CH3CO2H in this case) and its conjugate base (CH3CO2-) in similar amounts. It helps maintain the pH of a solution by resisting changes to its acidity or alkalinity.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the weak acid and the concentration ratio of the weak acid and its conjugate base. The equation is:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution (5.00 in this case)
- pKa is the acid dissociation constant, given as 4.74 in the question
- [A-] is the concentration of the conjugate base (CH3CO2-)
- [HA] is the concentration of the weak acid (CH3CO2H)
We have the values for [HA] and [A-], both of which are 0.50 mol. To use the Henderson-Hasselbalch equation, we can solve for log([A-]/[HA]):
5.00 = 4.74 + log(0.50/0.50)
This simplifies to:
0.26 = log(1)
Since log(1) = 0, we have:
0.26 = 0
This equation is not possible, which means it is not possible to achieve a pH of 5.00 with this buffer solution. No amount of NaOH can be added to the buffer to achieve that pH.
Therefore, the answer is that no moles of NaOH are required to increase the pH of the buffer to 5.00.
To increase the pH of a buffer, we need to add a base, in this case, NaOH. The reaction will be as follows:
CH3CO2H + OH- → CH3CO2- + H2O
To determine the amount of NaOH required, we need to consider the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Given that the desired pH is 5.00, we can rearrange the equation to solve for log ([A-]/[HA]):
log ([A-]/[HA]) = pH - pKa
log ([A-]/[HA]) = 5.00 - 4.74
log ([A-]/[HA]) = 0.26
We have equal moles of CH3CO2H and CH3CO2-, so let's use the variable x for both of them. Hence, at equilibrium, [A-] = x and [HA] = x.
Plugging these values into the equation, we have:
log (x/x) = 0.26
log (1) = 0.26
This equation simplifies to:
0 = 0.26
This implies that the concentration of CH3CO2H and CH3CO2- will not change significantly when the pH is increased to 5.00.
Therefore, we don't need to add any NaOH to the buffer to achieve the desired pH of 5.00.