(a) Rearrange the Arrhenius equation k = a(-#) = A so that T is the subject (function) of the formula. (b) Formic acid is a weak acid with a dissociation constant K, of 1.8 x 10-4. The K. relates the concentration of the H+ ions denoted [H+] and the amount of acid dissolved denoted N by the equation: [H*]? Ка N - [H+] %3D Given that there is 0.1 moles of formic acid dissolved, calculate the pH of the solution.

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(2.5) [CHEMISTRY]
(a) Rearrange the Arrhenius equation
k = A-R)
so that T is the subject (function) of the formula.
(b) Formic acid is a weak acid with a dissociation constant Ka of 1.8 x 10-4.
The Ka relates the concentration of the Ht ions denoted [H+] and the amount of
acid dissolved denoted N by the equation:
[H*]?
Ka =
N - [H+]
Given that there is 0.1 moles of formic acid dissolved, calculate the pH of the solution.
(Hint: an applied quadratic function)
Transcribed Image Text:(2.5) [CHEMISTRY] (a) Rearrange the Arrhenius equation k = A-R) so that T is the subject (function) of the formula. (b) Formic acid is a weak acid with a dissociation constant Ka of 1.8 x 10-4. The Ka relates the concentration of the Ht ions denoted [H+] and the amount of acid dissolved denoted N by the equation: [H*]? Ka = N - [H+] Given that there is 0.1 moles of formic acid dissolved, calculate the pH of the solution. (Hint: an applied quadratic function)
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