When a 16.3 mL sample of a 0.382 M aqueous hydrofluoric acid solution is titrated with a 0.478 M aqueous sodium hydroxide solution, what is the pH at the midpoint in the titration

3.25  

Explanation:

We don't have to do calculations with volumes and concentrations.

We just have to remember that, at the midpoint in the titration of a weak acid by a strong base,

pH = pKₐ

For formic acid,

pKₐ = 3.25, so, at the midpoint,

pH = 3.25

The first graph below shows the titration curve for your acid. The endpoint is at about 13 mL, so the midpoint is at 6.5 mL, and the pH is about 3.2.

In the second graph, the concentration of formic acid has increased by 50% to 0.573 mol·L⁻¹, so the endpoint has moved to about 19.5 mL and the midpoint is around 9.8 mL. However, the pH at the midpoint is still about 3.2.

This shows that the pH at the midpoint depends only on the pKₐ of the acid.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

diaria maybe go see a doctor about that