Find the ph of a 0.350 m aqueous benzoic acid (c6h5cooh) solution. for benzoic acid, ka = 6.5 × 10-5.

The solution would be using this:
C6H5COOH = H+ + C6H5COO- 
Ka = 6.5 x l0^-5 = (H+)(C6H5COO-) over (C6H5COOH) 
Let X = moles per liter (H+) and also moles per liter (C6H5COO-) 
Ka = 6.5 x l0^-5 = (X)(X) over .350 molar acid solution 6.5 x l0^-5 = X^2 over .350 
X^2 = 6.5 x l0^-5 times .350 which = 2.275 x l0^-5 
x = √2.275 x l0^-5
X = 1.5083 x l0^-5 moles per liter H+ 
pH = -log(H+) = -log 1.5083 x 10^-5 which = 4.8215

Because a model would make the investigation easier to understand