J.R. S. answered 03/24/18
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This is a buffer system, so you can use the Henderson Hasselbalch equation, but first you must know or look up the Ka or pKa of HF. The value I found is Ka = 6.6x10-4 and pKa = 3.18.
Henderson Hasselbalch:
pH = pKa + log [salt]//[acid]
[salt]: (200 ml)(0.1 M) = (300 ml)(x M) and x = 0.0667 M
[acid]: (100 ml)(0.1 M) = (300 ml)(x M) and x = 0.0333 M
pH = 3.18 + log (0.0667/0.0333)
pH = 3.18 + 0.301
pH = 3.48
Can't tell if the correct answer is D or E because it will depend on the value of pKa that was used.