Answer:
205mL of 0.140M KF solution
Step-by-step explanation:
pH = 2.70 solution.
It is possible to obtain the pH of the buffer of HF-KF using the H-H equation:
pH = pKa + log [KF] / [HF]
Where pH is desire pH = 2.70
pKa is pKa of HF = 3.17
[KF] could be taken as moles of KF
And [HF] moles of HF: 400.0mL = 0.4L * (0.212mol/L) = 0.0848 moles of HF
Replacing:
2.70 = 3.17 + log [KF] / [0.0848 moles HF]
-0.47 = log [KF] / [0.0848 moles HF]
0.3388 = [KF] / [0.0848 moles HF]
[KF] = 0.02873 moles of KF must be added.
In mL using concentration of KF (0.140M):
0.02873 moles KF * (1L / 0.140 mol) = 0.205L =
205mL of 0.140M KF solution