
Answer:
the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Explanation:
Given the data in the question;
   H₂     +    F₂    ⇄   2HF
I Â Â 1.69 M Â Â Â Â 1.69 M Â Â Â Â Â 0
C   -x         -x        +2x
E   1.69-x     1.69-x      +2x
given that Kc = 115 Â Â Â Â
Kc = [ HF ]² / [H₂][F₂]
we substitute
115 = [ 2x ]² / [ 1.69-x  ][ 1.69-x ]
lets find the square root of both sides
10.7238 = 2x / [ 1.69-x  ]
10.7238[ 1.69-x  ] = 2x
18.123222 - 10.7238x = 2x
2x + 10.7238x = 18.123222
12.7238x = 18.123222
x = 18.123222 / 12.7238
x = 1.424356
Hence, equilibrium concentration of HF = 2x
that is;
HF = 2 × 1.424356
HF = 2.8487 ≈ 2.85 M
Therefore, the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.