A solution that is 0.220 M HF and 0.240 M NaF reacts to produce a mixture...

Question

Question

A solution that is 0.220 M HF and 0.240 M NaF reacts to produce a mixture with what pH? The pKa of HF is 3.20.
Please show all steps.

Answer 1

Answer #1

Given that,

pKa of HF = 3.20

Concentration of HF (acid) = [HF] = 0.220 M

Concentration of NaF = [NaF] = 0.240 M

Here, in the mixture, salt NaF dissociates to Na+ and F- and hence the concentration of the conjugate base F- is same as that of NaF

Thus, the concentration of conjugate base [F-] = 0.240

We have Henderson-Hasselbalch equation,

pH = pKa + log [conjugate base]/[acid]

Therefore,

pH = 3.20 + log (0.220)/(0.240)

pH = 3.20 + log (0.9167)

pH = 3.20 - 0.0378

pH = 3.1622

Thus, the required pH of the mixture of 0.220 M HF and 0.240 M NaF is 3.1622

Answer 2

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