Respuesta :
Answer:
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.
The change in the concentration of the the conjugate acid–base pair
gives the new pH, from which the change in pH can be calculated.
Response:
- The change in pH, ΔpH is approximately 0.224
Which method can be used to calculate the change in the pH of the solution?
Given:
The volume of acetic acid = 1.80 × 10² mL
The Henderson-Hasselbalch equation is presented as follows;
[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base] }{[Weak \ base]}}[/tex]
Which gives;
[tex]5 = \mathbf{ 4.740 + log\dfrac{[Conjugate \ base] }{[Weak \ base]}}[/tex]
[tex]log\dfrac{[Conjugate \ base] }{[Weak \ base]} = 5 - 4.740 = \mathbf{ 0.26}[/tex]
Therefore;
[tex]\dfrac{[Conjugate \ base] }{[Weak \ base]} = 10^{0.26} \approx \mathbf{ 1.8197}[/tex]
[Conjugate base] + [Conjugate acid] = 0.100 M
Which gives;
[tex]\mathbf{ \dfrac{ n\cdot CH_3COO^-}{0.18} + \dfrac{1.82 \cdot n \cdot CH_3COOH}{0.18}} = 0.100 \, moles /L[/tex]
n·CH₃COO⁻ + 1.8197·n·CH₃COOH = 0.18 × 0.100 moles/L
2.8197·n = 0.18 × 0.100 moles/L
[tex]n = \dfrac{0.18 \times 0.100}{2.8197} \approx 0.006384[/tex]
Therefore;
The number of moles of CH₃COOH ≈ 0.006384 moles
The number of moles of CH₃COO⁻ = 1.8197 × n ≈ 1.8197 × 0.006384 moles ≈ 0.01162 moles
Number of moles of HCl in 5.10 mL 0.440 M HCl is found as follows;
Number of moles of HCl = 5.10 × 10⁻³ L × 0.440 M = 0.002244 moles
Number of moles of CH₃COO⁻ remaining = 0.01162 - 0.002244 = 0.009376
Number of moles of CH₃COOH ≈ 0.002244 + 0.006384 = 0.008628
New volume of the solution = 180 mL + 5.1 mL = 185.1 mL
[tex][Conjugate \, base] = [CH_3COO^-] = \dfrac{0.009376}{0.1851} \approx \mathbf{0.05065}[/tex]
[tex][Conjugate \, acid] = [CH_3COOH] = \dfrac{0.008628}{0.1851} \approx 0.0466[/tex]
Which gives;
[tex]pH = 4.74 + log \left(\dfrac{0.05065}{0.0466} \right) \approx 4.776[/tex]
- The change in pH of the solution, ΔpH = 5.00 - 4.776 = 0.224
Learn more about the Henderson-Hasselbalch equation here:
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