Answer: measure the mass (48.425g) of KCl
Explanation:
To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:
Molar Mass of KCl = 39 + 35.5 = 74.5g/mol
Number of mole (n) = 0.65
Mass conc of KCl = n x molar Mass
Mass conc of KCl = 0.65 x 74.5 = 48.425g
Therefore, to make 0.65M KCl, we must measure 48.425g
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 )
Just look it up on goog^le or a chart
