Question

The ka of hydrofluoric acid (hf) at 25. 0 °c is 6. 8 × 10-4. What is the ph of a 0. 25 m aqueous solution of hf?.

Answer 1

HF <==> H+ + F-

Ka = 6.8x10-4 = [H+][F-] / [HF]

6.8x10-4 = (x)(x) / 0.25 - x (assume x is small relative to 0.25 and ignore it to avoid using quadratic)

6.8x10-4 = x2 / 0.25

x2 = 1.7x10-4

x = 1.3x10-2 M (this is ~ 5.2% of 0.25 so our assumption above was not valid. We must use quadratic)

6.8x10-4 = x2 /0.25 - x

1.7x10-4 - 6.8x10-4 x = x2

Use quadratic to find x = 0.0127 M = [H+]

pH = -log H+ = -log 0.0127

pH = 1.90 so answer is probably 1.88(B)

Step-by-step explanation:

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