The ph of a 0.65m solution of hydrofluoric acid hf is measured to be 1.68. calculate the acid dissociation constant ka of hydrofluoric acid. round your answer to 2 significant digits.
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Ответ:
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 )
Ответ:
Molar mass of an unknown gas : 450.56 g/mol
Further explanationGraham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or
the effusion rates of two gases = the square root of the inverse of their molar masses:
or
M CO₂ = 44 g/mol
Carbon dioxide gas (CO2) effuses 3.2 times faster than an unknown gas, so r₁=3.2 r₂