In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270. Use the information she obtained to determine the Ka for this acid.

Ka(experiment) =

Ka = 6.87x10⁻⁵

Explanation:

The equilibrium of benzoic acid in water is:

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The equilibrium constant, Ka, is:

Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]

The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.

X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.

Concentrations in equilibrium are:

pH is defined as -log [H₃O⁺]. As pH = 2.270

2.270 = -log [H₃O⁺]

10^-2.270 = [H₃O⁺]

5.37x10⁻³M = [H₃O⁺] = X.

Replacing, concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M

[C₆H₅COO⁻] = 5.37x10⁻³M

[H₃O⁺] = 5.37x10⁻³M

Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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