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A buffer solution contains  0.392 M  hydrofluoric acid and  0.357 M  potassium fluoride .   If  0.0229 moles of  sodium hydroxide are added...

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 =

0 0
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Answer #1

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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