A solution is prepared by titrating a 100.0 mL sample of 0.10 M HF (Ka = 7.2 × 10-4) with 0.10 M NaOH
a. Does the solution contain a strong acid with a strong base, a strong acid with a weak base, a weak acid with a strong base, or a weak acid with a weak base?
b. Estimate the pH at the equivalence point. Will it be equal to 7, greater than 7, or less than 7?.
c. What is the pH after 20.0 mL of the 0.10 M NaOH is added? Show all calculations.
a. The solution contains a weak acid (HF) with a strong base (NaOH).
b. The pH at the equivalence point can be estimated using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]).
In this case, the weak acid is HF and its conjugate base is F-. The pKa for HF is calculated using the Ka value: pKa = -log(Ka) = -log(7.2 × 10^-4).
Since HF is a weak acid and NaOH is a strong base, at the equivalence point [A-] = [HA]. Therefore, the equation simplifies to: pH = pKa + log(1) = pKa.
So, the pH at the equivalence point will be equal to the pKa value, which can be calculated based on the given Ka value.
c. To find the pH after adding 20.0 mL of the 0.10 M NaOH, we need to determine the moles of NaOH added and then calculate the concentrations of HF and F-.
1. Moles of NaOH added:
Molarity (M) = moles (mol) / volume (L)
Moles of NaOH = Molarity (0.10 M) * Volume (20.0 mL) / 1000 (conversion from mL to L)
2. Moles of HF remaining:
Moles of HF = Initial moles of HF - Moles of NaOH added
Initial moles of HF = Molarity (0.10 M) * Volume (100.0 mL) / 1000 (conversion from mL to L)
3. Concentration of HF:
Volume of solution = Initial volume of HF - Volume of NaOH added
Concentration of HF = Moles of HF remaining / Volume of solution
4. Concentration of F- (conjugate base):
Concentration of F- = Moles of NaOH added / Volume of solution
5. Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([F-] / [HF])
Please provide the value of Ka for HF to proceed further calculations.
To determine the answer to these questions, we need to understand the concepts of a strong acid, weak acid, strong base, and weak base, as well as the process of titration.
a. Understanding the nature of the given acid (HF) and base (NaOH) will help us determine the type of solution.
A strong acid is an acid that completely ionizes in water, releasing all of its hydrogen ions. Examples include hydrochloric acid (HCl) and nitric acid (HNO3).
A weak acid, on the other hand, does not completely ionize in water and only releases a fraction of its hydrogen ions. Acetic acid (CH3COOH) and carbonic acid (H2CO3) are examples of weak acids.
Similarly, a strong base fully ionizes in water, releasing all of its hydroxide ions. Examples include sodium hydroxide (NaOH) and potassium hydroxide (KOH).
A weak base does not fully ionize and only releases a portion of its hydroxide ions. Ammonia (NH3) and sodium carbonate (Na2CO3) are examples of weak bases.
In this case, we are given HF (hydrofluoric acid) and NaOH (sodium hydroxide). HF is a weak acid, and NaOH is a strong base. Therefore, the solution contains a weak acid with a strong base.
b. To estimate the pH at the equivalence point of a weak acid-strong base titration, we need to understand the concept of a dissociation constant (Ka) and the equilibrium equation of the weak acid.
The Ka value of HF (hydrofluoric acid) is given as 7.2 × 10^-4. The equilibrium equation for HF in water is:
HF + H2O ⇌ H3O+ + F-
At the equivalence point, the number of moles of the weak acid (HF) will be equal to the moles of the strong base (NaOH) added. In this case, 0.10 moles of NaOH will react with 0.10 moles of HF.
At the equivalence point, all of the HF will be neutralized, leaving only the conjugate base F- in solution. The concentration of the F- ions will be equal to the initial concentration of NaOH (0.10 M).
Since F- is the conjugate base of a weak acid, it will react with water to some extent, producing hydroxide ions (OH-) and regenerating a small amount of HF. However, the concentration of F- will be significantly greater than the concentration of the hydrofluoric acid.
Considering the presence of the strong base (NaOH), the pH at the equivalence point will be greater than 7.
c. To determine the pH after adding 20.0 mL of 0.10 M NaOH, we need to calculate the excess amount of NaOH remaining after neutralizing the initial 100.0 mL of 0.10 M HF.
The number of moles of NaOH initially present in the solution is:
Moles of NaOH = volume (L) x concentration (M)
Moles of NaOH = (0.02 L) x (0.10 M)
Moles of NaOH = 0.002 moles
The number of moles of HF initially present in the solution is:
Moles of HF = volume (L) x concentration (M)
Moles of HF = (0.100 L) x (0.10 M)
Moles of HF = 0.010 moles
Since the stoichiometric ratio between NaOH and HF is 1:1, the excess moles of NaOH remaining after neutralization are:
Excess moles of NaOH = Moles of NaOH - Moles of HF
Excess moles of NaOH = 0.002 - 0.010
Excess moles of NaOH = -0.008 moles
Since we have a negative value for excess moles of NaOH, it means that all the HF has been neutralized, and there is still 0.008 moles of excess NaOH in solution.
To calculate the new concentration of NaOH after adding 20.0 mL, we need to take into account the new total volume of the solution, which is the initial 100.0 mL of HF plus the 20.0 mL of NaOH:
Total volume = Initial volume + Volume added
Total volume = 100.0 mL + 20.0 mL
Total volume = 120.0 mL = 0.120 L
The new concentration of NaOH after adding 20.0 mL can be calculated as:
New concentration = Excess moles of NaOH / Total volume (L)
New concentration = 0.008 moles / 0.120 L
New concentration = 0.067 M
Now, we can calculate the pOH of the solution using this new concentration of NaOH. Since pOH + pH = 14 (at 25°C), we can find the pH by subtracting the pOH from 14.
pOH = -log10[OH-]
pOH = -log10[0.067]
pOH = 1.18
pH = 14 - pOH
pH = 14 - 1.18
pH = 12.82
Therefore, the pH after adding 20.0 mL of 0.10 M NaOH to the solution is approximately 12.82.