The pH will become a little more acid, by changing 0.225 units.
Explanation:
Step 1: Data given
Volume of acetic acid buffer = 1.80 * 10² mL
pH of the buffer = 5.00
Molarity of acid and conjugate base = 0.1.00 M
A student adds 5.10 mL ( = 0.0051 L) of a 0.440 M HCl solution
The pKa of acetic acid is 4.74
Step 2: The Henderson-Hasselbalch equation
pH = pKa + log ([conjugate base]/[weak acid])
with acetic acid (CH3COOH) as the weak acid and CH3COO- as it's conjugate base. The pH =5 and the pKa of acetic acid is 4.74
5 = 4.74 + log ([conjugate base]/[weak acid])
log ([CH3COO-]/[(CH3COOH]) = 0.26
[CH3COO-]/[(CH3COOH] = 10^0.26 = 1.82
This means the buffer contains 1.82 times more conjugate base (CH3COO-) than weak acid (CH3COOH)
Since both the conjugate base and the weak acid, share the same volume we can say:
(Number of moles of CH3COO-) = 1.82* (Number of moles of CH3COOH)
Step 3: Calculate number of moles of acid and conjugate base
As said, the total molarity of acid and conjugate base is 0.100 M
[CH3COOH] + [CH3COO-] = 0.100 M
Since molarity = Number of moles/ volume (mol/L), We can write this as:
(n(CH3COOH)/180 * 10^-3 L ) + (n(CH3COO-)/180 * 10^-3 L ) = 0.100 moles / L
OR
n(CH3COOH) + n(CH3COO-) = 0.018
1.82*n(CH3COOH) + n(CH3COOH) = 0.018
n(CH3COOH) = 0.00638 moles
n(CH3COO-) = 1.82 * 0.00638 moles = 0.0116 moles
Step 4: Adding 5.10 mL of a 0.440 M HCl solution:
When adding HCl, the HCl will react with the acetate anions to form acetic acid and chloride anions ( Cl-):
HCl(aq) + CH3COO- (aq) → CH3COOH(aq) + Cl-(aq)
⇒For 1 mole of HCl consumed, we need 1 mole of CH3COO-, to produce 1 mole of CH3COOH and 1 mole of Cl-
Step 5: Calculate moles of HCl
Number of moles = Molarity * Volume
Number of moles HCl = 0.440 M * 5.10 *10^-3 L = 0.002244 moles
Step 6: The limiting reactant
HCl is the limiting reactant and will be completely consumed by the reaction. 0.002244 moles is consumed. There will remain 0 moles.
Step 7: Calculate remain moles
For CH3COO- , there is also 0.002244 moles consumed.
There will remain 0.0116 moles - 0.002244 moles = 0.009356 moles
For CH3COOH, there will be produced 0.002244 moles. There will be in total: 0.00638 moles + 0.002244 moles = 0.008624 moles
The total volume of the solution is: 180 mL + 5.10 mL = 185.1 mL = 0.1851 L
Step 8: Calculate the concentrations
Concentration = Number of moles / volume
The concentrations of acetic acid and acetate ions will be:
[CH3COOH] = 0.008624 moles / 0.1851 L
[CH3COOH] = 0.0466 M
[CH3COO-] = 0.009356 moles / 0.1851 L
[CH3COO-] = 0.0505 M
Step 9: Calculate the new pH
pH = 4.74 + log(0.0505 M/ 0.0466 M)
pH = 4.775
ΔpH = 5 - 4.775 = 0.225
The pH will become a little more acid, by changing 0.225 units.