A buffer is prepared by mixing 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF (Ka = 6.8x10^-4) Calculate the pH of this buffer after 5.0 mL of 0.05 M HCl is added.
45 mL x 0.15M NaF = 6.75 mmoles NaF.
35 mL x 0.10M HF = 3.5 mmoles HF.
5 mL x 0.05M HCl = 0.50 mmoles H^+.
...............F^- + H^+ ==> HF
initial......6.75....0......3.50
add HCl.............0.25........
change.......-0.25..-0.25....+0.25
equil........6.50....0.......3.75
Substitute these numbers into the HH equation and solve for pH. Check my work.
Why did the chemist become a comedian? Because he had all the right solutions! Okay, let's calculate the pH of this buffer after adding 5.0 mL of 0.05 M HCl.
To solve this problem, we'll use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, let's find the concentrations of NaF (A-) and HF (HA) in the buffer solution.
moles of NaF = Molarity x Volume
moles of NaF = 0.15 M x (45.0 mL/1000 mL) = 0.00675 moles
moles of HF = Molarity x Volume
moles of HF = 0.10 M x (35.0 mL/1000 mL) = 0.0035 moles
Next, calculate the concentrations:
[A-] = moles of NaF / Total Volume
[A-] = 0.00675 moles / (45.0 mL + 35.0 mL + 5.0 mL)
[A-] = 0.00675 moles / 0.085 L = 0.0794 M
[HA] = moles of HF / Total Volume
[HA] = 0.0035 moles / (45.0 mL + 35.0 mL + 5.0 mL)
[HA] = 0.0035 moles / 0.085 L = 0.041 M
Now, let's calculate pKa:
pKa = -log(Ka)
pKa = -log(6.8x10^-4)
pKa = 3.17
Finally, plug the values into the Henderson-Hasselbalch equation:
pH = 3.17 + log(0.0794 M / 0.041 M)
pH = 3.17 + log(1.935)
pH = 3.17 + 0.286
pH = 3.456
So, the pH of this buffer after adding 5.0 mL of 0.05 M HCl is approximately 3.456.
To solve this problem, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the weak acid and its conjugate base.
The Henderson-Hasselbalch equation is:
pH = pKa + log ([A-]/[HA])
First, let's calculate the initial concentrations of HF and NaF in the buffer solution.
Given:
Volume of NaF solution = 45.0 mL
Concentration of NaF = 0.15 M
Initial moles of NaF = volume × concentration
= 45.0 mL × 0.15 M
= 6.75 mmol
Similarly, for HF:
Volume of HF solution = 35.0 mL
Concentration of HF = 0.10 M
Initial moles of HF = volume × concentration
= 35.0 mL × 0.10 M
= 3.5 mmol
Since we have the moles of HF and NaF, we can calculate their initial concentrations.
Concentration of HF = moles / volume
= 3.5 mmol / 35.0 mL
= 0.100 M
Concentration of NaF = moles / volume
= 6.75 mmol / 45.0 mL
= 0.150 M
Now, let's calculate the moles of HCl added.
Given:
Volume of HCl = 5.0 mL
Concentration of HCl = 0.05 M
Moles of HCl added = volume × concentration
= 5.0 mL × 0.05 M
= 0.25 mmol
Now, we can calculate the new moles of HF and NaF in the buffer solution.
Moles of HF after addition of HCl = initial moles of HF - moles of HCl added
= 3.5 mmol - 0.25 mmol
= 3.25 mmol
Moles of NaF after addition of HCl = initial moles of NaF
= 6.75 mmol
With the new moles of HF and NaF, we can calculate their new concentrations.
Concentration of HF after addition of HCl = moles / volume
= 3.25 mmol / 35.0 mL
≈ 0.093 M
Concentration of NaF after addition of HCl = moles / volume
= 6.75 mmol / 45.0 mL
= 0.150 M
Now, let's calculate the ratio of [A-]/[HA].
[A-]/[HA] = concentration of NaF / concentration of HF
= (0.150 M) / (0.093 M)
≈ 1.61
Next, we need to calculate the pKa using the given Ka value.
Ka = [H+][A-] / [HA]
6.8 × 10^-4 = [H+](concentration of NaF) / (concentration of HF)
6.8 × 10^-4 = [H+](0.150 M) / (0.093 M)
[H+] = (6.8 × 10^-4)(0.093 M) / (0.150 M)
≈ 4.22 × 10^-4 M
Finally, let's calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log ([A-] / [HA])
= -log(4.22 × 10^-4) + log(1.61)
≈ -(-3.37) + 0.207
≈ 3.37 + 0.207
≈ 3.577
Therefore, the pH of the buffer after adding 5.0 mL of 0.05 M HCl is approximately 3.577.
To calculate the pH of the buffer solution after the addition of 5.0 mL of 0.05 M HCl, we need to consider the following steps:
Step 1: Calculate the amount of moles of HCl added.
- Volume of HCl added = 5.0 mL = 0.005 L
- Concentration of HCl added = 0.05 M
- Moles of HCl added = volume x concentration = 0.005 L x 0.05 mol/L
Step 2: Calculate the moles of HF and NaF initially present in the buffer solution.
- Moles of HF = initial volume x initial concentration = 35.0 mL x 0.10 mol/L
- Moles of NaF = initial volume x initial concentration = 45.0 mL x 0.15 mol/L
Step 3: Determine the change in moles of HF and NaF due to the reaction with HCl.
- The reaction between HCl and HF follows the equation: HF + HCl → H2O + FCl
- For every 1 mole of HF that reacts, 1 mole of NaF is formed.
- Since the stoichiometry of the reaction is 1:1, the moles of HCl consumed will be equal to the moles of F ions (NaF) formed.
Step 4: Calculate the final moles of HF and NaF in the buffer solution.
- Final moles of HF = initial moles of HF - moles of HCl consumed
- Final moles of NaF = initial moles of NaF + moles of HCl consumed
Step 5: Calculate the new concentrations of HF and NaF.
- Final concentration of HF = Final moles of HF / total volume of buffer solution
- Final concentration of NaF = Final moles of NaF / total volume of buffer solution
Step 6: Calculate the pH of the buffer solution using the Henderson-Hasselbalch equation.
- pH = pKa + log10([NaF] / [HF])
- pKa = -log10(Ka), where Ka is the acid dissociation constant of HF
Now, let's calculate each step in detail:
Step 1:
Moles of HCl added = 0.005 L x 0.05 mol/L = 0.00025 moles of HCl
Step 2:
Moles of HF = 35.0 mL x 0.10 mol/L = 0.0035 moles of HF
Moles of NaF = 45.0 mL x 0.15 mol/L = 0.00675 moles of NaF
Step 3:
Moles of HCl consumed = 0.00025 moles of HCl
Moles of F ions (NaF formed) = 0.00025 moles of HCl
Step 4:
Final moles of HF = 0.0035 moles of HF - 0.00025 moles of HCl = 0.00325 moles of HF
Final moles of NaF = 0.00675 moles of NaF + 0.00025 moles of HCl = 0.007 moles of NaF
Step 5:
Total volume of buffer solution = initial volume of HF + initial volume of NaF + volume of HCl added
= 35.0 mL + 45.0 mL + 5.0 mL
= 85.0 mL = 0.085 L
Final concentration of HF = 0.00325 moles of HF / 0.085 L = 0.038 mol/L
Final concentration of NaF = 0.007 moles of NaF / 0.085 L = 0.082 mol/L
Step 6:
pKa = -log10(6.8x10^-4) = 3.17
pH = 3.17 + log10(0.082 / 0.038) = 3.17 + 0.34 = 3.51
Therefore, the pH of the buffer after the addition of 5.0 mL of 0.05 M HCl is approximately 3.51.