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.
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?
4 answers
can u solve it with like hydronium and water ionization?
No. The easiest way is to use the H-H equation.
will the solution have a pH less that 7 or more than 7
2 answers