1)A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate to 100.0 mL of water. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
2) A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) to 100.0 mL of water and then titrating with strong base until a buffer with a 1:1 buffer ratio is produced. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
3) 100.0 mL of pure water.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
To predict the final pH in each scenario, we need to consider the principles of acid-base reactions and the concept of buffers.
1) In the first scenario, a buffer solution is prepared by adding 0.400 mol of acetic acid (a weak acid) and 0.400 mol of sodium acetate (its conjugate base) to 100.0 mL of water. The initial pH of the buffer solution is the same as the pKa of acetic acid, which is 4.74.
When 55.0 mL of 1.10 M NaOH (a strong base) is added to the buffer solution, it will react with acetic acid to form water and acetate ions. The reaction is as follows:
CH3COOH + NaOH → H2O + CH3COONa
Since acetic acid is a weak acid, the reaction is not complete, and some acetic acid will remain in the solution. However, the addition of NaOH will increase the concentration of acetate ions (CH3COO-) in the solution.
To determine the final pH, we need to calculate the new concentrations of acetic acid and acetate ions. We can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, [A-] refers to the concentration of acetate ions and [HA] refers to the concentration of acetic acid.
First, we need to convert the volume of NaOH solution (55.0 mL) to moles:
moles NaOH = volume (L) x concentration (M) = 0.055 L x 1.10 M = 0.0605 mol NaOH
Since the NaOH reacts with acetic acid in a 1:1 ratio, the moles of acetic acid that will react is also 0.0605 mol.
The initial concentration of acetic acid is 0.400 mol / 0.100 L = 4.0 M.
The initial concentration of acetate ions is also 0.400 mol / 0.100 L = 4.0 M.
After the reaction with NaOH, the concentration of acetic acid will decrease by 0.0605 M, and the concentration of acetate ions will increase by 0.0605 M.
Now we can calculate the final pH using the Henderson-Hasselbalch equation:
pH = 4.74 + log ([(4.0 + 0.0605) / (4.0 - 0.0605)])
Simplifying the equation:
pH = 4.74 + log (4.0605 / 3.9395)
pH = 4.74 + log (1.0304)
Using a calculator, the final pH is approximately 4.79.
2) In the second scenario, a buffer solution is prepared by adding 0.400 mol of acetic acid to 100.0 mL of water and then titrating with a strong base until a buffer with a 1:1 buffer ratio is produced. The initial pH of the buffer solution is 4.74, the same as the pKa of acetic acid.
The concept of a 1:1 buffer ratio means that the initial concentrations of acetic acid and acetate ions are the same.
When 55.0 mL of 1.10 M NaOH is added, it will react with the acetic acid in the same way as in the previous scenario. However, since the initial concentrations of acetic acid and acetate ions are the same, the change in their concentrations will also be equal.
Using the same calculations as in scenario 1, the final concentration of acetic acid will be (4.0 M - 0.0605 M) = 3.9395 M, and the final concentration of acetate ions will be (4.0 M + 0.0605 M) = 4.0605 M.
Using the Henderson-Hasselbalch equation:
pH = 4.74 + log ([(3.9395) / (4.0605)])
Simplifying the equation:
pH = 4.74 + log (0.9701)
Using a calculator, the final pH is approximately 4.67.
3) In the third scenario, there is no buffer present, only pure water.
When 55.0 mL of 1.10 M NaOH is added, it will react with the water according to the following reaction:
H2O + NaOH → Na+ + OH- + H2O
The NaOH reacts with water to produce sodium ions (Na+), hydroxide ions (OH-) and more water molecules. The concentration of OH- ions will increase, resulting in a higher pH.
To calculate the final pH, we need to determine the concentration of OH- ions.
moles NaOH = volume (L) x concentration (M) = 0.055 L x 1.10 M = 0.0605 mol NaOH
Since the reaction occurs in a 1:1 ratio, the concentration of OH- ions will be equal to the concentration of NaOH:
[OH-] = 0.0605 M
To convert this concentration to a pOH value:
pOH = -log [OH-] = -log (0.0605) = 1.22
Since pH + pOH = 14, the final pH can be calculated as:
pH = 14 - pOH = 14 - 1.22 = 12.78
Therefore, the final pH is approximately 12.78.