[Boxxxed]

Calculate the pH of vinegar knowing that it is a 5% solution of CH3COOH in water. Assume the density of vinegar is 1.00g/ml.

Ka = 1.74x10^-5

I'm not sure how to do this.

Would the reaction be CH3COOH + H2O -----> CH3COO- + H3O

Ka = [H3O][CH3COO] / [CH3COOH]

I'm not sure what to do next. I think I need to solve for x but I don't know what to do with the 5% value or density.

Do I assume there is 1 litre of solution?
      
Acid = 0.05 mol/L?

[Boxxxed]

Correction, is the concentration 1.04 M/L?

[contra]

I think that is the concentration you should be using.

[Borek]

I think that is the concentration you should be using.

Check your units.

[Borek]

5% solution of CH3COOH in water. Assume the density of vinegar is 1.00g/ml.

Contra was close. Imagine you have volume V of the solution - what is its mass? What part of this mass is acetic acid? How many moles of acetic acid? You know volume, you know moles, you can calculate concentration.

[contra]

I thought that's what I did Borek?

Assuming we have 1 liter of solution (1000mL), we have 50 grams of acetic acid. Convert that to moles of acetic acid and you have the molarity of the solution?
Am I missing something?

Edit: I see what you are saying now. For some reason the first time I did this I calculated it in terms of liters, then I thought I was wrong and switched to mLs.
Please forgive my stupidity

[Boxxxed]

Great, I used the right method but with a dumb error in adding up the molar mass. thanks again