A 50mL of 0.50M hydrofluoric acid (Ka = 7.2 x 10^-4) is titrated with a 0.20M
sodium hydroxide solution. What is the pH of the solution when 35mL of base
has been added?
To work this problem you must determine where you are on the titration curve. Are you before the equivalence point, at the eq pt or after the eq pt? If before you will use the Henderson-Hasselbalch equation. If at the eq pt you use the hydrolysis of the NaF salt produced in the reaction. If after the eq pt it is just a diluted solution of the excess NaOH. So let's see where we are.
millimoles HF at the beginning = mL x M = 50 mL x 0.50 M = 25
millimoles NaOH added = mL x M = 35 mL x 0.20 M = 7
....................HF + NaOH ==> NaF + H2O
initial...........25..........0...............0..........0
add............................7....................................
change......-..-7.........-7...............+7.................
equilibrium....18..........0................7...................
So you can see that not all of the HF you had at the beginning has reacted and you have some salt produced; therefore, you have a buffer of a week acid (HF) and its salt (NaF) so you use the Henderson-Hasselbalch equation. pH = pKa + log [(base)/(acid)]
You will need to convert Ka for HF to pKa Remember pKa = -log Ka.
For (base) use M NaF at that point so M = millimoles/mL and for (acid) use M HF at that point as M = millimoles/mL.
Post your work if you get stuck.
No. The easiest way is to use the H-H equation.
can u solve it with like hydronium and water ionization?
will the solution have a pH less that 7 or more than 7
To find the pH of the solution after adding 35 mL of a 0.20M sodium hydroxide (NaOH) solution to a 50 mL sample of 0.50M hydrofluoric acid (HF), we need to carry out the following steps:
Step 1: Determine the moles of hydrofluoric acid (HF) present in the 50 mL solution.
To calculate the moles of HF, we use the formula:
moles = concentration (M) × volume (L)
Given:
- Concentration of HF: 0.50 M
- Volume of HF solution: 50 mL = 0.050 L
moles of HF = 0.50 M × 0.050 L = 0.025 moles of HF
Step 2: Determine the moles of sodium hydroxide (NaOH) added to the solution.
We can calculate the moles of NaOH by using the same formula:
moles = concentration (M) × volume (L)
Given:
- Concentration of NaOH: 0.20 M
- Volume of NaOH added: 35 mL = 0.035 L
moles of NaOH = 0.20 M × 0.035 L = 0.007 moles of NaOH
Step 3: Determine the moles of hydrofluoric acid (HF) remaining after the reaction.
Since the reaction is a 1:1 ratio between HF and NaOH (from the balanced equation), the moles of HF remaining after the reaction will be the initial moles minus the moles of NaOH used in the reaction.
moles of HF remaining = initial moles of HF - moles of NaOH used
= 0.025 moles - 0.007 moles
= 0.018 moles of HF
Step 4: Use the Henderson-Hasselbalch equation to calculate the pH.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this case, HF acts as a weak acid (HA) and its conjugate base, the fluoride ion (F-), acts as A-. The pKa of hydrofluoric acid is given as 7.2 × 10^-4.
Using the equation, we can substitute the values:
pH = -log(Ka) + log([A-]/[HA])
= -log(7.2 × 10^-4) + log([0.018]/[0.018])
Simplifying further:
pH = -log(7.2 × 10^-4) + log(1)
= -(-3.86) (since log(1) = 0)
= 3.86
Therefore, the pH of the solution after adding 35 mL of the 0.20M NaOH solution is approximately 3.86.