[4Players1]

Hey everyone

Im stuck on following question:
0.036 mol of acetic acid (CH3COOH) are contained in 750 ml solution. Calculate the pH of the solution under the assumption that 1.8% of acid particles have reacted with water (In technical terms this is called dissociation 0.018)

To help us  the equation pH = 0.5 (pKa - log c) was set, but this doesnt really help me all that much :/
Also we got this equation : CH3COOH + H2O -> CH3COO-+ H3O +
Can someone give me an understandable approach for this task and the result?


Thanks in advance!

[Borek]

Ignore the equation, simply use the information given. If 1.8% of 0.036 mole of acid dissociated, how many moles of H⁺ were produced?