How many grams of CH3COOH are needed to make 4.2 liters of aqueous solution of pH=3.4?Be sure to back correct from equilibrium molecular acid molarity to starting molecular acid molarity.

I can tell you how to solve the problem BUT I have no idea what your second sentence says.

CH3COOH is a weak acid, therefore, one must determine CH3COOH from the Ka and the known H3O^+.
CH3COOH + H2O ==> H3O^+ + CH3COO^-
Ka = (H3O^+)(CH3COO^-)/(CH3COOH)
If pH = 3.4, use pH = -log(H3O^+) to solve for (H3O^+). Plug that into the Ka expression for (H3O^+) and for (CH3COO^-)
For (CH3COOH), plug in X-(H3O^+) and solve for X. That will give you the molarity of CH3COOH you need for that pH.
You want 4.2L so moles CH3COOH needed will be M x L.
moles = grams/molar mass. Solve for grams.

To determine the number of grams of CH3COOH needed to make a certain volume of aqueous solution with a given pH, we need to use the concept of acidity and pH.

Acidity is a measure of the concentration of hydrogen ions (H+) in a solution. In the case of CH3COOH (acetic acid), it is a weak acid that partially ionizes in water, releasing hydrogen ions. The pH of a solution is a logarithmic measure of the concentration of H+ ions.

To solve this problem, we need to follow these steps:

Step 1: Convert pH to H+ concentration
- The pH value of 3.4 indicates an H+ concentration of 10^(-3.4) M. This is derived from the formula pH = -log[H+].

Step 2: Convert H+ concentration to CH3COOH molarity
- Since CH3COOH is a weak acid, it partially ionizes in water according to the equation CH3COOH ⇌ H+ + CH3COO-.
- The initial concentration of CH3COOH is equal to the concentration of H+ ions.
- Therefore, the molarity of CH3COOH is 10^(-3.4) M.

Step 3: Calculate moles of CH3COOH
- To convert molarity to moles, multiply the molarity (in mol/L) by the volume in liters. In this case, the volume is 4.2 L.
- Moles of CH3COOH = Molarity × Volume = 10^(-3.4) mol/L × 4.2 L

Step 4: Calculate the mass of CH3COOH
- The molar mass of CH3COOH is 60.052 g/mol.
- Mass of CH3COOH = Moles × Molar mass = (10^(-3.4) mol/L × 4.2 L) × 60.052 g/mol

Using these steps, you can calculate the number of grams of CH3COOH needed to make 4.2 liters of the aqueous solution with a pH of 3.4.