If you add 0.35grams NaF to 150mL of a 0.30M HF, what is the pH of the resulting solution?
To start off would I use the value of 1.4e-11 for F^-^ and 7.2e-4 for HF?
Chemistry(Please help) - DrBob222, Tuesday, April 10, 2012 at 12:04am
This is a buffer problem.
pH = pKa + log(base)/(acid)
Use Ka for HF
Convert g NaF in 150 mL to M, then lug in the numbers.
Chemistry(Please help) - Hannah, Tuesday, April 10, 2012 at 10:39am
So it would be pH=4.74 + log base/acid and I use 7.2e-4 for HF but after I convert g NaF in 150mL to M is this what I use for the base?
yes
To determine the pH of the resulting solution, you need to use the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([base]/[acid])
In this case, NaF is the base and HF is the acid. To start solving the problem, you will need the value of pKa for HF. If you have been given the pKa value of HF as 7.2e-4, then you can proceed with that.
The next step is to convert the mass of NaF (0.35 grams) to moles. You can do this by using the molar mass of NaF, which is 41.99 g/mol.
moles NaF = mass NaF / molar mass NaF
= 0.35 g / 41.99 g/mol
After that, you need to convert the volume of the solution (150 mL) to liters:
Volume (L) = Volume (mL) / 1000
= 150 mL / 1000
= 0.150 L
Now calculate the concentration of the NaF solution in moles per liter (Molarity):
Molarity NaF = moles NaF / Volume (L)
= moles NaF / 0.150 L
With this value, you can substitute it as [base] in the Henderson-Hasselbalch equation.
pH = 7.2e-4 + log([NaF]/[HF])
= 7.2e-4 + log([base]/[acid])
Replace [base] with the molarity of the NaF solution, and [acid] with the molarity of the HF solution (0.30 M), and solve for pH.
pH = 7.2e-4 + log([NaF]/0.30)
= 7.2e-4 + log(Molarity NaF / 0.30)
By substituting the corresponding values, you should be able to find the pH of the resulting solution.