What is the pH of a mixture of 5ml of 0.1mol/litre sodium acetate and 4ml of 0.1mol/litre acetic acid? pKa= 4.76.... What is the pH on adding 1ml of 0?1mol/litre HCl to the mixture abouc
To determine the pH of the mixture of sodium acetate and acetic acid, we need to calculate the pH using the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the pH value
pKa is the acid dissociation constant
[A-] is the concentration of the conjugate base (sodium acetate in this case)
[HA] is the concentration of the acid (acetic acid in this case)
Step 1: Calculate the moles of sodium acetate and acetic acid:
Given: 5ml of 0.1mol/L sodium acetate; 4ml of 0.1mol/L acetic acid
Moles of sodium acetate = Volume (liters) × Concentration (mol/L) = 0.005 L × 0.1 mol/L = 0.0005 mol
Moles of acetic acid = Volume (liters) × Concentration (mol/L) = 0.004 L × 0.1 mol/L = 0.0004 mol
Step 2: Calculate the total moles of the conjugate base and the acid:
Total moles of the conjugate base = Moles of sodium acetate = 0.0005 mol
Total moles of the acid = Moles of acetic acid = 0.0004 mol
Step 3: Calculate the concentrations of the conjugate base and the acid:
Concentration of the conjugate base = Total moles / Total volume
Concentration of the acid = Total moles / Total volume
Since the total volume of the mixture is 5ml + 4ml = 9ml = 0.009 L:
Concentration of the conjugate base = 0.0005 mol / 0.009 L ≈ 0.0556 mol/L
Concentration of the acid = 0.0004 mol / 0.009 L ≈ 0.0444 mol/L
Step 4: Use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.0556/0.0444)
pH = 4.76 + log(1.25) ≈ 4.93
Therefore, the pH of the mixture of sodium acetate and acetic acid is approximately 4.93.
Now, to determine the pH after adding 1ml of 0.1mol/L HCl to the mixture:
Step 1: Calculate the moles of HCl added:
Given: 1ml of 0.1mol/L HCl
Moles of HCl added = Volume (liters) × Concentration (mol/L) = 0.001 L × 0.1 mol/L = 0.0001 mol
Step 2: Calculate the new moles of the conjugate base and the acid:
New moles of the conjugate base = Moles of sodium acetate = 0.0005 mol
New moles of the acid = Moles of acetic acid + Moles of HCl added = 0.0004 mol + 0.0001 mol = 0.0005 mol
Step 3: Calculate the new concentrations of the conjugate base and the acid:
Concentration of the conjugate base remains the same at 0.0556 mol/L
Concentration of the acid = New moles / Total volume = 0.0005 mol / (0.009 L + 0.001 L) ≈ 0.0500 mol/L
Step 4: Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.0556/0.0500)
pH = 4.76 + log(1.112) ≈ 4.91
Therefore, after adding 1ml of 0.1mol/L HCl to the mixture, the pH is approximately 4.91.
To calculate the pH of a mixture of sodium acetate and acetic acid, we need to consider the dissociation of acetic acid and the hydrolysis of sodium acetate.
First, let's calculate the concentrations of sodium acetate and acetic acid in the mixture:
- Sodium acetate: 0.1 mol/L * 5 mL/9 mL = 0.0556 mol/L
- Acetic acid: 0.1 mol/L * 4 mL/9 mL = 0.0444 mol/L
Next, we need to determine the reaction that occurs between sodium acetate and water. Sodium acetate hydrolyzes in water to form acetic acid and hydroxide ions:
CH3COONa + H2O ⇌ CH3COOH + OH-
Now, let's calculate the concentrations of acetic acid and hydroxide ions formed during the hydrolysis:
- Acetic acid: 0.0444 mol/L + 0.0556 mol/L = 0.1 mol/L
- Hydroxide ions: The concentrations of acetic acid and hydroxide ions will be equal since the reaction has a 1:1 stoichiometry.
Now, we can calculate the pOH of the solution using the hydroxide ion concentration:
pOH = -log [OH-] = -log [0.1 mol/L] = 1
Since pOH + pH = 14, we can calculate the pH of the solution:
pH = 14 - pOH = 14 - 1 = 13
Therefore, the pH of the mixture of sodium acetate and acetic acid is 13.
Now, let's consider the effect of adding 1 mL of 0.1 mol/L HCl to the mixture. HCl is a strong acid that will fully dissociate in water, leading to an increase in the concentration of hydrogen ions (H+).
Since the original volume of the mixture is 9 mL and we are adding an additional 1 mL, the final volume becomes 10 mL.
To calculate the change in concentration of hydrogen ions, we can use the equation:
[H+]final = ([H+]initial * Vinitial + [HCl] * Vadded) / Vfinal
Using the initial concentration of acetic acid ([H+]initial = 0.1 mol/L) and adding 1 mL of 0.1 mol/L HCl, we can calculate the final concentration of hydrogen ions:
[H+]final = (0.1 mol/L * 9 mL + 0.1 mol/L * 1 mL) / 10 mL = 0.09 mol/L
Now, let's find the pH of the solution after adding HCl. The pH is calculated using the formula:
pH = -log [H+]
pH = -log [0.09 mol/L] ≈ -log [9 x 10^-2] ≈ -(-1.05) ≈ 1.05
Therefore, on adding 1 mL of 0.1 mol/L HCl to the mixture, the pH of the solution will be approximately 1.05.