What is the pH of 44.3 ml of a solution which is 0.26 M in NaF and 0.33 M in HF?
For HF use Ka=6.8x10^-4
Ka = 6.8*10^-4
pKa = - log (Ka)
= - log(6.8*10^-4)
= 3.167
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.167+ log {0.26/0.33}
= 3.064
Answer: 3.06
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