1)A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution.
The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places.
*A buffered solution resists a change in pH.*
2)Calculate the pH when 28.8 mL of 0.035 M HCl is added to 100.0 mL of the above buffer.
I know I have to create an ICE table, but how do i set it up for question number two?
You're adding HCl to the buffer solution of part 1.
question 1 gives NaAc = 0.10M
and HAc = 0.20M
You are taking 100 mL; theefore,
millimols NaAc = base = 0.1 x 100 = 10.
mmols HAc = acid = 0.20 x 100 = 20.
You are adding
28.8 mL x 0.035 = about 1 millimols HCl.
.........Ac^- + H^+ ==>HAc
I........10.....0.......20
add.............1..................
C........-1....-1.......+1
E.........9.....0.......21
To calculate the initial pH of a buffer solution, we need to consider the equilibrium of the weak acid and its conjugate base. In this case, we have a combination of sodium acetate (CH3COONa) as the conjugate base and acetic acid (CH3COOH) as the weak acid.
1) To calculate the initial pH of the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the desired value, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Given the concentrations of the sodium acetate and acetic acid in the buffer (0.10 M and 0.20 M, respectively), we can plug in the values into the Henderson-Hasselbalch equation.
pH = -log(1.8 x 10^-5) + log(0.10/0.20)
= -(-4.74) + log(0.5)
= 4.74 + (-0.30)
= 4.44
Therefore, the initial pH of the buffer solution is 4.44.
2) To calculate the pH after adding HCl to the buffer solution, we need to consider the reaction between the added HCl (a strong acid) and the weak acid (acetic acid) in the buffer.
Since the amount of HCl added is 28.8 mL of 0.035 M, we can calculate the moles of HCl:
moles of HCl = volume (L) x concentration (M)
= 0.0288 L x 0.035 M
= 0.001008 mol
Since HCl is a strong acid, it will completely dissociate in water, resulting in the same amount of H+ ions being released. Thus, the moles of H+ ions produced are also 0.001008 mol.
In the buffer solution, we have acetic acid and acetate ions present. The reaction between H+ ions and the acetate ions in the buffer can be represented as:
H+ + CH3COO- <--> CH3COOH
To set up the ICE (Initial, Change, Equilibrium) table, consider the initial concentrations of the reactants:
[H+] = 0.001008 mol
[CH3COO-] = 0.10 M
[CH3COOH] = 0.20 M
Since there is no change in the concentration of CH3COO- and CH3COOH initially, the change column remains 0. However, as H+ ions are consumed, they will result in a decrease in the concentration of CH3COO- and an increase in the concentration of CH3COOH. Let's assume the change in concentration of CH3COOH is x.
[H+] = 0.001008 - x
[CH3COO-] = 0.10 - x
[CH3COOH] = 0.20 + x
At equilibrium, the concentration of the weak acid (CH3COOH) plus the concentration of the conjugate base (CH3COO-) should remain constant:
[CH3COOH] + [CH3COO-] = 0.20 M + 0.10 M
0.20 + x + 0.10 - x = 0.30
Simplifying the equation, we find that:
0.30 = 0.30
This means that x = 0, and the concentrations of CH3COOH and CH3COO- remain unchanged.
Now, we can calculate the pH of the resulting solution by using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since [A-] = [CH3COO-] = 0.10 M and [HA] = [CH3COOH] = 0.20 M:
pH = -log(1.8 x 10^-5) + log(0.10/0.20)
= 4.74 + log(0.50)
= 4.74 - 0.30
= 4.44
Therefore, the pH of the resulting solution after adding HCl to the buffer remains at 4.44.