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Calculate the [H+] and pH of a 0.000185 M acetic acid solution. Keep in mind that the Ka of acetic acid is 1.76×10−5. This is a problem that will require you to use the quadratic. Start by setting up the quadratic equation before using the quadratic formula and find the coefficients. Then, solve the quadratic. Quadratic formula: ax2+bx+c=0; enter the values of a, b, and c

Answer :

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Answer:

[H⁺] = 0.000048936M

pH = 4.31

Explanation:

Acetic acid, CH₃COOH, dissociates in water as follows:

CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)

And Ka is defined as:

Ka = 1.76x10⁻⁵ =  [H⁺] [CH₃COO⁻] / [CH₃COOH]

Where [] are equilibrium concentrations of the species.

The 0.000185M of acetic acid will decreases X, and X of [H⁺] and [CH₃COO⁻] will be produced. That means Ka is:

1.76x10⁻⁵ =  [X] [X] / [0.000185 - X]

3.256x10⁻⁹ - 1.76x10⁻⁵X = X²

3.256x10⁻⁹ - 1.76x10⁻⁵X - X² = 0

Solving for X:

X = -0.000066M → False solution. There is no negative concentrations.

X = 0.000048936

As [H⁺] = X,

[H⁺] = 0.000048936M

And pH = -log [H⁺]

pH = 4.31

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