A buffer solution contains 0.309 M hydrofluoric acid and 0.344 M potassium fluor
ID: 947765 • Letter: A
Question
A buffer solution contains 0.309 M hydrofluoric acid and 0.344 M potassium fluoride. If 0.0495 moles of hydrochloric acid are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding hydrochloric acid) (no Ka is given)
*Please show all work* I understand how buffers woork and how the addition of an acid or base shifts the equilibrium left or right based upon LeChatlier's principle, but my online homework is not explaining how it gets the solution very well. Thanks!
Explanation / Answer
No of mol of HF = 0.309*0.25 = 0.07725 mol
no of mol of KF = 0.344*0.25 = 0.086 mol
pka of HF = 3.17
pH = pka + log(KF-HCl/HF+HCl)
= 3.17+log((0.086-0.0495)/(0.07725+0.0495))
= 2.63
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