[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of [tex] FCH_{2}COOH [/tex]
[tex] FCH_{2}COOH(aq) + H_{2}O (l) <==> FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq) [/tex]
Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y = [tex] [H_{3}O^{+}] [/tex] =[tex] [FCH_{2}COO^{-}] [/tex] = 0.00770 M
[tex] pK_{a} = 2.59K_{a} = 10^{-2.59} = 0.00257 M [/tex]
[tex] K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770} [/tex]
[tex] 0.00257 = \frac{0.00005929}{x - 0.00770} [/tex]
0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is 0.031 M
The pH of the solution is 2.4.
Let us recall that the dissociation of the fluoroacetic acid follows the reaction;
FCH2COOH ⇄ FCH2COO- + H^+
Hence, Ka = [ FCH2COO-] [H^+]/[FCH2COOH]
Pka = 2.59
Ka = Antilog (-2.59) = 2.57 * 10^-3
So;
Let [ FCH2COO-] = [H^+] = x
2.57 * 10^-3 = x^2/ 0.00770
x = √2.57 * 10^-3 * 0.00770
x = 4.4 * 10^-3 M
pH = -log(4.4 * 10^-3 M) = 2.4
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