Question

If a buffer system containing hydrofluoric acid and fluoride (pKa = 3.17) initially has a pH of 3.22 and base is added until a new pH of 4.69 is obtained, how has the ratio of fluoride to hydrofluoric acid changed?

Answer 1

A. from 1.12 : 1 to 33.1 : 1

Step-by-step explanation:

#Plato

Use the Henderson-Hasselbalch Equation:

`pH = pKa + log(([A^-])/([HA]))`

Plug in values given (let's start with the initial pH of 3.22)

3.22 = 3.17 + log(x) | Since we don't know the concentrations, make it x.

Subtract 3.17 from 3.22 (we're finding x)

.05 = log(x)

Take the inverse of log to find x

x = 1.12

Now, "if x is less than 1.0, then there is more acid than base; if greater than 1.0, there is more base than acid." So we have more base than acid so the initial ratio is 1.12:1.

To find the new pH, plug in values into the equation again:

4.69 = 3.17 + log(x)

You'll find that x = 33.1. Once again, x is greater than 1 so there is more base than acid. Plus, you can see that the solution's pH went up from 3.22 to 4.69 so it got more basic.

Thus, the ratio went from 1.12 : 1 to 33.1 : 1

Answer 2

First, let's write the equation:
HF + H2O ==> F- + H3O+ k = (F- * H3O+ )/ HF

since wa have a buffer: we can use this law
pH = pKa - log (base / acid) = pKa - log (F- / HF)
pH - pKa = - log (F- / HF)
pka - pH = log (F- / HF)
so :
`x^(pka - pH)` = F- / HF

when pH was 3.22:
F- / HF =
`x^(3.17 - 3.22)` = 0.891

and when the pH became 4.69:
F- / HF =
`x^(3.17 - 4.69)` = 0.030