1.15 mol HF is added to enough 0.202-M NaF solution to give a final volume of 2.7 L. What is the pH of the resulting sol...
1.15 mol HF is added to enough 0.202-M NaF solution to give a final volume of 2.7 L. What is the pH of the resulting solution given that the Ka of HF is 3.5x10-4 under these conditions?
2.76
1.75
0.37
3.50
3.13
Homework Answers
[HF] = mol of HF / volume in L
= 1.15 mol / 2.7 L
= 0.426 M
Ka = 3.5*10^-4
pKa = - log (Ka)
= - log(3.5*10^-4)
= 3.456
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.456+ log {0.202/0.426}
= 3.132
Answer: 3.13
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