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.
strong base, NaOH, and week acid, HF.
>7. approx 8.5
c. Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
To determine whether the solution contains 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, we need to compare the strength of the acid (HF) and the base (NaOH).
a. HF is a weak acid since it has a dissociation constant (Ka) less than 1. NaOH, on the other hand, is a strong base. Therefore, the solution contains a weak acid with a strong base.
b. At the equivalence point, the moles of acid (HF) will be equal to the moles of base (NaOH). This means that all the HF will be neutralized by the NaOH, resulting in the formation of sodium fluoride (NaF) and water (H2O). NaF is the conjugate base of the weak acid HF. The solution will be a salt solution of NaF, which undergoes hydrolysis.
Since NaF is the salt of a weak acid and a strong base, it will undergo hydrolysis and cause the solution to be basic. Therefore, the pH at the equivalence point will be greater than 7. The exact pH value can be calculated using the Henderson-Hasselbalch equation.
c. To calculate the pH after adding 20.0 mL of the 0.10 M NaOH, we need to determine the number of moles of NaOH added and the volume of the resulting solution.
Number of moles of NaOH added = concentration × volume
= 0.10 M × 0.020 L
= 0.002 mol
The initial volume was 100.0 mL, and an additional 20.0 mL of NaOH was added. Therefore, the total volume of the solution after the addition is 120.0 mL or 0.120 L.
To determine the concentration of the weak acid (HF) after the addition of NaOH, we can use the fact that the number of moles of HF that reacted with NaOH is equal to the moles of NaOH added:
Number of moles of HF reacted = Number of moles of NaOH added
= 0.002 mol
The number of moles of HF remaining can be calculated by subtracting the number of moles reacted from the initial moles of HF:
Number of moles of HF remaining = Initial moles of HF - moles of HF reacted
= (0.10 M) × (0.100 L) - (0.002 mol)
= 0.010 mol - 0.002 mol
= 0.008 mol
Now, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([conjugate base]/[weak acid])
The pKa value for HF is determined by its Ka value, which is 7.2 × 10^-4. Taking the negative logarithm of Ka, we get:
pKa = -log(7.2 × 10^-4) = 3.14
Plugging in the values into the Henderson-Hasselbalch equation:
pH = 3.14 + log([0.008 mol]/[0.008 mol])
= 3.14 + log(1)
= 3.14 + 0
= 3.14
Therefore, the pH after adding 20.0 mL of the 0.10 M NaOH is approximately 3.14.
To answer these questions, we need to understand the concept of titration and the reaction that occurs between HF (a weak acid) and NaOH (a strong base).
a. To determine if the solution contains 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, we consider the strength of the acid and base involved.
In this case, HF is a weak acid, and NaOH is a strong base. Therefore, the solution contains a weak acid (HF) with a strong base (NaOH).
b. At the equivalence point, the moles of the acid (HF) and the moles of the base (NaOH) are equal. This means that the HF and NaOH react in a 1:1 stoichiometric ratio. The reaction is as follows:
HF + NaOH -> NaF + H2O
Since NaF is a strong electrolyte and completely dissociates in water, it does not affect the pH significantly. Therefore, at the equivalence point, the pH will be equal to 7 because water is neutral (pH = 7).
c. To calculate the pH after adding 20.0 mL of 0.10 M NaOH, we need to find the moles of NaOH and HF that have reacted.
1. Calculate the moles of NaOH:
Moles of NaOH = (volume of NaOH in liters) x (molarity of NaOH)
= (20.0 mL) x (0.10 mol/L) / (1000 mL/L)
= 0.002 moles of NaOH
2. Use the balanced equation from part b to find the moles of HF that reacted. Since the ratio of HF to NaOH is 1:1, the moles of HF are also 0.002 moles.
3. Determine the initial moles of HF before the reaction.
Moles of HF = (volume of HF in liters) x (molarity of HF)
= (100.0 mL) x (0.10 mol/L) / (1000 mL/L)
= 0.010 moles of HF
4. Calculate the remaining moles of HF after the reaction.
Remaining moles of HF = Initial moles of HF - Moles of HF that reacted
= 0.010 moles - 0.002 moles
= 0.008 moles of HF
5. Calculate the concentration of HF after the reaction.
Concentration of HF = Remaining moles of HF / Volume of the solution in liters
= 0.008 moles / (100.0 mL + 20.0 mL) / (1000 mL/L)
≈ 0.072 M (approximately)
6. Finally, calculate the pH using the Henderson-Hasselbalch equation for weak acid solutions:
pH = pKa + log([A-] / [HA])
In this case, since HF is a weak acid and NaF (the conjugate base) is a strong electrolyte, we can assume that [A-] ≈ 0 and simplify the equation to:
pH ≈ pKa + log(0.072 / 0.10)
Using the given pKa value of HF (Ka = 7.2 × 10^-4), we can calculate the pH.
pH ≈ -log[(7.2 × 10^-4)] + log(0.072/0.10)
≈ -log[(7.2 × 10^-4) × (10^6)] + log(720/1000)
≈ -(-6 + 6) + log(0.72)
≈ 6 + log(0.72)
≈ 6 - 0.14
≈ 5.86
Therefore, the pH after adding 20.0 mL of 0.10 M NaOH is approximately 5.86.