Answer:
correct option is (a)
The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over
(C6H5COOH)
Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)
Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350
X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5
x = V2.275 x 10^-5
X = 1.5083 x 10^-5 moles per liter H+
pH = -log(H+) = -log 1.5083 x 10^-5 which
= 4.6215
