0.1 M x 0.5L = 0.05 mols HCl.
Adding 25 mL 2M NaOH is
2M x 0.025 L = 0.05 mols NaOH.
What you want to do is to back off very slightly with the NaOH (you might try something like 24.95 mL which I calculate to give 0.00019 M or a pH of 3.7. Inching closer, 24.99 mL would leave H^+ of 4E-5 for pH 4.4. The problem here is two-fold.
hope it helps
Answer:
Kₐ = 6.7 x 10⁻⁴
Explanation:
First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:
HF + H₂O ⇄ H₃O⁺ + F⁻
Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]
Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.
pH = -log [ H₃O⁺ ]
1.68 = - log [ H₃O⁺ ]
Taking antilog to both sides of this equation:
10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]
[ F⁻ ] = 2.1 X 10⁻² M
Solving for Kₐ :
Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴
(Rounded to two significant figures, the powers of 10 have infinite precision )
Possibly decomposition but not sure
Answer:
WAIT
good luck on whatever this is for my dude.
