A buffer that contains 0.36 M base, B, and 0.27 M of its conjugate acid, BH + ,

Question

A buffer that contains 0.36 M base, B, and 0.27 M of its conjugate acid, BH+, has a pH of 8.18. What is the pH after 0.0027 mol HCl is added to 0.25 L of this solution?

So my work:

pH = pKa + log [B] / [BH+]

                                8.18 = pKa + log (0.36) / (0.25)

                                pKa =-4.08660

Moles BH+ = (0.25 L) (0.25 mol BH+ / L) = 0.0625 mol BH+ (unrounded)

                                Moles B = (0.36 L) (0.25 mol B / L) = 0.10 mol B

PH=pKa + log [B]/{BH+]

yeah, I got lost .... from here.I think on the right track. Please show all work so I can understand it.

Explanation / Answer

pOH = pKb + log (Salt/Base)

pOH = 14 - pH = 5.82

=> 5.82 = pKb + log (0.27 / 0.36)

=> pKb = 5.945

pH after adding acid

pOH = pKb + log (salt + X / base - X)

X = acid added

pOH = 5.945 + log (0.27*0.25 + 0.0027 / 0.36*0.25 - 0.0027)

=> pOH = 5.85

=> pH = 14 - pOH = 8.15

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