Consider a buffer solution comprised of HF and F-. The Ka of HF is 6.3 x...

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Consider a buffer solution comprised of HF and F-. The Ka of HF is 6.3 x...

Consider a buffer solution comprised of HF and F^-. The K_a of HF is 6.3 x 10^-4. If the pH of the buffer solution is 3.64 and [HF] = 0.110 M, how many moles of F^- are present? The volume of the buffer solution is 297 mL.

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Answer 1

Answer #1

Answer

0.08996mol

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where,

pH = pH of buffer solution , 3.64

pKa = pKa of weak acid , 3.20 (pKa = -logKa = -log(6.3×10^-4) = 3.20 )

[A^-] = concentration of conjucate base , concentration of F^-

HA = weak acid , concentration of HF , 0.110M

substituting the values

3.64 = 3.20 + log ( [F^-]/0.110M)

log([F^-]/0.110M) = 0.44

[F^-] /0.110M = 2.754

[F^-] = 0.3029M

Molarity = number of moles of solute per liter of solution

Number of moles of F^- = (0.3029mol/1000ml ) × 297ml = 0.08996mol

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Answer 2

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