What is the pH of the solution?
answer= 3.10
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In Moles
i: HF(0.05)+F⁻(0.045)→F⁻(0)+HF(0)
c: (-0.045)+(-0.045)→(+0.045)+(0.045)
e: HF(0.005) →F⁻(0.045)+HF(0.045)
Using
[H₃O⁺]= [(Initial Acid)/(Conj. Base)] * Ka
[H₃O⁺]= [(0.05)/( 0.045)] * 7.2×10^-4==0.0008
pH = - log [0.0008]==3.0969 rounded to 3.10
--answer
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