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Ray Of Light [21]
3 years ago
11

Write the Ka expression for an aqueous solution of hydrofluoric acid: (Note that either the numerator or denominator may contain

more than one chemical species. Enter the complete numerator in the top box and the complete denominator in the bottom box. Remember to write the hydronium ion out as , and not as ) Ka
Chemistry
1 answer:
damaskus [11]3 years ago
4 0

Answer:

Ka = ( [H₃O⁺] . [F⁻] ) / [HF]

Explanation:

HF is a weak acid which in water, keeps this equilibrium

HF (aq)  +  H₂O (l)  ⇄  H₃O⁺ (aq)  +  F⁻ (aq)      Ka

2H₂O (l)  ⇄  H₃O⁺ (l)  +  OH⁻ (aq)   Kw

HF is the weak acid

F⁻ is the conjugate stron base

Let's make the expression for K

K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]

K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]

K . [H₂O] = Ka

Ka, the acid dissociation constant, includes Kwater.

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3 years ago
If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula
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3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
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Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

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3 years ago
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IRISSAK [1]

Answer:

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3 years ago
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