Calculate the change in pH if 10mL of 0.1M HCl is added to the buffer made by mixing 25 mL of 1.0 M CH3COOH and 25 mL of 0.5 M CH3COONa.
First, determine the number of moles for CH3COOH (HAc), CH3COONa (Ac^-), and HCl
Molarity=moles/Volume (L)
Moles=molarity*volume (L)
Moles of HCl=0.1M*0.010L=x
Moles of HAc=1.0M*0.025L=y
Moles of Ac^-=0.5M*0.025L=z
Calculate the pH of the buffer:
pH=pka+log[A^-/HA]
Where
pka=4.75
A^-=z
HA=y
and
pH=???
Solve for pH:
pH=4.75+log[z/y]
When HCl is added to the solution, HAc will increase and Ac^- will decrease.
Solve for the amount of change:
y+x=y1
z-x=z1
Use the Henderson-Hasselbalch equation and solve for the pH:
pH=4.75+log[y1/z1]
Subtract the pH of the buffer after addition of HCl from the pH of the buffer after addition of HCl.
pH=4.66
To calculate the change in pH when HCl is added to the buffer solution, we need to determine the new concentrations of the acids and bases in the buffer solution.
First, let's find the initial concentrations of CH3COOH and CH3COONa in the buffer solution:
Initial concentration of CH3COOH = 1.0 M × (25 mL / (25 mL + 25 mL)) = 0.5 M
Initial concentration of CH3COONa = 0.5 M × (25 mL / (25 mL + 25 mL)) = 0.25 M
Next, we will calculate the concentration of CH3COO- ion that results from the dissociation of CH3COONa:
CH3COO- concentration = 0.5 M × (25 mL / (25 mL + 25 mL)) = 0.25 M
Now, let's calculate the initial concentration of H+ ion from the dissociation of CH3COOH:
CH3COOH is a weak acid and only partially dissociates in water. The concentration of H+ ion can be determined using the acid dissociation constant, Ka, which is equal to 1.8 × 10^-5 for CH3COOH.
CH3COOH equilibrium equation: CH3COOH ⇌ CH3COO- + H+
Using an ICE (Initial, Change, Equilibrium) table, assume that x is the concentration of H+ ion that dissociates from CH3COOH:
Initial concentration CH3COOH: 0.5 M - x
Change: +x
Equilibrium concentration CH3COOH: 0.5 M - x
Equilibrium concentration CH3COO-: 0.25 M + x
Equilibrium concentration H+: x
Apply the equilibrium expression:
Ka = [CH3COO-][H+] / [CH3COOH]
Substitute the known values:
1.8 × 10^-5 = (0.25 M + x)(x) / (0.5 M - x)
Since CH3COOH is weak, we can assume that x is small compared to 0.5 M and neglect it in the denominator:
1.8 × 10^-5 ≈ (0.25 M)(x) / (0.5 M)
Simplifying:
1.8 × 10^-5 ≈ 0.5x / 2
x ≈ (1.8 × 10^-5) × (2 / 0.5)
x ≈ 7.2 × 10^-5 M
Therefore, the concentration of H+ ion in the buffer solution is approximately 7.2 × 10^-5 M.
Now, let's calculate the change in pH when 10 mL of 0.1 M HCl is added to the buffer solution. HCl is a strong acid that completely dissociates in water.
The moles of HCl added = concentration × volume = 0.1 M × 10 mL = 0.001 moles.
Using the Henderson-Hasselbalch equation for calculating pH in a buffer solution:
pH = pKa + log([CH3COO-] / [CH3COOH])
pKa = -log10(Ka) = -log10(1.8 × 10^-5) ≈ 4.74
Substituting the known values:
pH = 4.74 + log((0.25 M + 0.001 moles) / (0.5 M - 0.001 moles))
pH = 4.74 + log(0.251 / 0.499)
pH ≈ 4.74 + log(0.503)
Using logarithmic properties, we can rewrite this as:
pH ≈ 4.74 + log10(0.503)
pH ≈ 4.74 + (-0.298)
pH ≈ 4.442
Therefore, the initial pH of the buffer solution is approximately 4.442.