You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 93 mL of 1.00 M NaOH solution?
Let's call CH3COOH HA and the acetate ion will be A-
millimoles HA = 500 mL x 0.30M = 150.
mmoles A- = 500 mL x 0.20M = 100
We will add 93mL x 1.00M = 93 mmoles NaOH.
............HA + OH- ==> A- + H2O
initial....150...0.......100......
add..............93................
change.....-93...-93.....+93...+93
equil.......57....0......193
pH = pKa + log(base)/(acid)
Plug and chug.
My quickie answer is 5.3 but you should confirm that and don't estimate as I've done.
To determine the pH of the buffer solution after the addition of NaOH, we need to consider the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
First, let's calculate the moles of acetic acid and sodium acetate present in the initial 500.0 mL of the buffer solution.
Moles of acetic acid = Volume (L) * Concentration (mol/L)
Moles of acetic acid = 0.5 L * 0.30 mol/L = 0.15 mol
Moles of sodium acetate = Volume (L) * Concentration (mol/L)
Moles of sodium acetate = 0.5 L * 0.20 mol/L = 0.10 mol
Since acetic acid and sodium acetate are in a 1:1 molar ratio, the buffer solution contains the same number of moles of acetic acid and sodium acetate. Therefore, we have 0.15 mol of acetic acid and 0.10 mol of sodium acetate.
The reaction between acetic acid and sodium hydroxide can be represented as follows:
CH3COOH + NaOH → CH3COONa + H2O
Since it is a 1:1 stoichiometric reaction, the moles of acetic acid consumed will be equal to the moles of sodium hydroxide added to the solution.
Now let's calculate the moles of sodium hydroxide added to the solution.
Moles of sodium hydroxide = Volume (L) * Concentration (mol/L)
Moles of sodium hydroxide = 0.093 L * 1.00 mol/L = 0.093 mol
Since the moles of acetic acid consumed are equal to the moles of sodium hydroxide added, and the moles of acetic acid initially present in the solution are 0.15 mol, we can calculate the remaining moles of acetic acid after the reaction:
Remaining moles of acetic acid = Initial moles - Moles of acetic acid consumed
Remaining moles of acetic acid = 0.15 mol - 0.093 mol = 0.057 mol
Now we can calculate the concentration of acetic acid in the final solution, which is the remaining moles divided by the total volume in liters:
Concentration of acetic acid = Remaining moles / Total volume (L)
Concentration of acetic acid = 0.057 mol / (0.5 L + 0.093 L) = 0.087 M
Since acetic acid is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log([A-]/[HA])
The pKa of acetic acid is approximately 4.76. In this equation, [A-] represents the concentration of the acetate ion (CH3COO-) and [HA] represents the concentration of acetic acid (CH3COOH).
Using the equation, we can substitute the values:
pH = 4.76 + log(0.087/0.087)
pH = 4.76 + log(1)
pH = 4.76 + 0
pH = 4.76
Therefore, the pH of the buffer solution after the addition of NaOH will be approximately 4.76.
To determine the pH of the buffer solution after the addition of NaOH, we need to apply the Henderson-Hasselbalch equation. The equation is as follows:
pH = pKa + log ([A-]/[HA])
In this case, acetic acid (CH3COOH) is the weak acid (HA), and sodium acetate (CH3COONa) is its conjugate base (A-).
First, we need to calculate the initial concentrations of CH3COOH and CH3COONa in the buffer solution. We'll use the formula:
C = n/V
Where:
C = concentration
n = number of moles
V = volume
For CH3COOH (HA):
n = C x V
= 0.30 M x 0.500 L (500.0 mL converted to L)
= 0.150 moles
For CH3COONa (A-):
n = C x V
= 0.20 M x 0.500 L
= 0.100 moles
Next, we need to determine which component will be in excess after the NaOH solution is added. We compare the moles of NaOH and CH3COOH. The balanced equation for the reaction between NaOH and CH3COOH is:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that the molar ratio between CH3COOH and NaOH is 1:1. Therefore, the moles of NaOH added will be equal to the moles of CH3COOH consumed.
Moles of NaOH added = C x V
= 1.00 M x 0.093 L (93 mL converted to L)
= 0.093 moles
Since 1 mole of CH3COOH produces 1 mole of OH-, the moles of OH- will be the same as the moles of NaOH added. Meaning, the moles of CH3COOH consumed will also be 0.093 moles.
Using the equation:
Moles of CH3COOH remaining = initial moles - moles consumed
= 0.150 moles - 0.093 moles
= 0.057 moles
Moles of CH3COONa formed = moles of CH3COOH consumed = 0.093 moles
Now we can calculate the concentrations of CH3COOH and CH3COONa after the NaOH solution is added. We'll use the formula:
C = n/V
For CH3COOH (HA):
C = moles/V
= 0.057 moles / 0.500 L
= 0.114 M
For CH3COONa (A-):
C = moles/V
= 0.093 moles / 0.500 L
= 0.186 M
Lastly, we can substitute these values into the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-]/[HA])
The pKa value for acetic acid is 4.74.
pH = 4.74 + log (0.186 M / 0.114 M)
Calculating the ratio of ([A-]/[HA]):
[A-]/[HA] = 0.186 M / 0.114 M
= 1.632
Taking the log of 1.632:
log (1.632) = 0.212
Substituting the values into the equation:
pH = 4.74 + 0.212
= 4.952
Therefore, the pH of the solution after the addition of 93 mL of 1.00 M NaOH will be approximately 4.952.