The ka of propanoic acid (c2h5cooh) is 1.34 × 10-5. calculate the ph of the solution and the concentrations of c2h5cooh and c2h5coo– in a 0.193 m propanoic acid solution at equilibrium.

Chemical reaction: C₂H₅COOH ⇄ C₂H₅COO⁻ + H⁺.
Ka(C₂H₅COOH) = 1,34·10⁻⁵.
c(C₂H₅COOH) = 0,193 M = 0,193 mol/L.
[C₂H₅COO⁻] = [H⁺] = x.
Ka = [C₂H₅COO⁻] · [H⁺] / [C₂H₅COOH].
1,34·10⁻⁵ = x² / (0,193 M - x).
Solve quadratic equation: [C₂H₅COO⁻] = [H⁺] = 0,00158 M.
[C₂H₅COOH] = 0,193 mol/L - 0,0158 M = 0,1772 M.
pH = -log[H⁺].
pH = -log(0,00158 M) = 2,8.

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priyankatutor priyankatutor · Answer 2 · 2021-10-19T12:27:32+02:00

The concentration of reactant and products at equilibrium is 0.00158 M and pH of the solution is 2.8

[tex]\rm \bold{ C_2H_5COOH\leftrightharpoons C_2H_5COO^- + H^+}[/tex]

[tex]\rm Ka= \bold{ \frac{[C_2H_5COO^-][ H^+]}{[ C_2H_5COOH]} }\\[/tex]

[tex]\rm Ka= \bold{ \frac{[x][x]}{[ 0.193} }\\[/tex]

Solve the equation,

The concetration of Hydrogen ion or [tex]\rm \bold{[ C_2H_5COO^- ]}[/tex] is 0.00158 M

[tex]\rm \bold {pH= -log [H^+]}\\\\\rm \bold{pH = -log 0.00158}\\\\\rm \bold{pH =2.8}[/tex]

Hence we can conclude that the concentration of reactant and products are 0.00158 M and pH of the solution is 2.8

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