Question

How many mL (to the nearest mL) of 0.140-M KF solution should be added to 660. mL of 0.169-M HF to prepare a pH = 3.30 solution?

Answer 1

Ka of HF =6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.1805

use formula for buffer

pH = pKa + log ([KF]/[HF])

3.3 = 3.1805 + log ([KF]/[HF])

log ([KF]/[HF]) = 0.1195

[KF]/[HF] = 1.3169

[KF] = 0.2226

volume , V = 660.0 mL

= 0.660 L

we have below equation to be used:

number of mol of KF,

n = Molarity * Volume

= 0.2226*0.660

= 0.1469 mol

Now use:

number of mol of KF = initial concentration / initial volume

0.1469 = 0.140/initial volume

initial volume = 0.953 L

initial volume = 953 mL

Answer: 953 mL