A buffer solution contains 0.392
M hydrofluoric acid
and 0.357
M potassium fluoride
.
If 0.0229 moles
of sodium hydroxide are added
to 125 mL of this buffer, what is the
pH of the resulting solution?
(Assume that the volume does not change upon
adding sodium hydroxide.)
pH =
pKa of HF = 3.10.
125 mL 0.392 M HF = (125x0.392) mLxM HF = 49 mLx(mol/L) = 49 (mL/L)xmol = 49 x 10-3 mol = 0.049 mol.
125 mL 0.357 M KF = (125x0.357) x 10-3 mol = 0.044625 mol.
0.0229 moles of NaOH will consume that much moles of acid, so, moles of acid = (0.049 - 0.0229) mol = 0.0261 mol
and produce 0.0229 moles of NaF, so, moles of F- = (0.044625 + 0.0229) mol = 0.067525 mol.
According to Henderson-Hasselbalch equation,
since [ ] (concentration) is proportional to no. of moles, so, we can substitute [ ] by no. of moles in HH equation,
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