How many grams of CH3COOH are needed to make 2.0 liters of aqueous solution of pH=3.1?Be sure to back correct from equilibrium molecular acid molarity to starting molecular acid molarity.
The ion concentration is calculated:
[ion] = 10 ^ -pH = 10 ^ -3.1 = 7.9x10 ^ -4 M
You have the expression of Ka:
Ka = [CH3COO-] * [H3O +] / [CH3COOH]
1.8x10 ^ -5 = (1.9x19 ^ -4) ^ 2 / [CH3COOH]
It clears [CH3COOH] = 0.002 M
The mass of acetic acid required is calculated:
m AA = M * V * MM = 0.002 M * 2 L * 60 g / mol = 0.24 g
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